What is the "trick" to writing a Quine? - quine

I read Ken Thompson's classic paper Reflections on Trusting Trust in which he prompts users to write a Quine as an introduction to his argument (highly recommended read).
A quine is a computer program which takes no input and produces a copy of its own source code as its only output.
The naive approach is simply to want to say:
print "[insert this program's source here]"
But one quickly sees that this is impossible. I ended up writing one myself using Python but still have trouble explaining "the trick". I'm looking for an excellent explanation of why Quines are possible.

The normal trick is to use printf such that the format-string represents the structure of the program, with a place-holder for the string itself to get the recursion you need:
The standard C example from http://www.nyx.net/~gthompso/quine.htm illustrates this quite well:
char*f="char*f=%c%s%c;main(){printf(f,34,f,34,10);}%c";main(){printf(f,34,f,34,10);}
edit: After writing this, I did a bit of searching: http://www.madore.org/~david/computers/quine.html gives a very good, more theoretical, description of what exactly quines are and why they work.

Here's one I wrote that uses putchar instead of printf; thus it has to process all its own escape codes. But it is %100 portable across all C execution character sets.
You should be able to see that there is a seam in the text representation that mirrors a seam in the program text itself, where it changes from working on the beginning to working on the end. The trick to writing a Quine is getting over this "hump", where you switch to digging your way out of the hole! Your options are constrained by the textual representation and the language's output facilities.
#include <stdio.h>
void with(char *s) {
for (; *s; s++) switch (*s) {
case '\n': putchar('\\'); putchar('n'); break;
case '\\': putchar('\\'); putchar('\\'); break;
case '\"': putchar('\\'); putchar('\"'); break;
default: putchar(*s);
}
}
void out(char *s) { for (; *s; s++) putchar(*s); }
int main() {
char *a[] = {
"#include <stdio.h>\n\n",
"void with(char *s) {\n",
" for (; *s; s++) switch (*s) {\n",
" case '\\",
"n': putchar('\\\\'); putchar('n'); break;\n",
" case '\\\\': putchar('\\\\'); putchar('\\\\'); break;\n",
" case '\\\"': putchar('\\\\'); putchar('\\\"'); break;\n",
" default: putchar(*s);\n",
" }\n}\n",
"void out(char *s) { for (; *s; s++) putchar(*s); }\n",
"int main() {\n",
" char *a[] = {\n",
NULL }, *b[] = {
"NULL }, **p;\n",
" for (p = a; *p; p++) out(*p);\n",
" for (p = a; *p; p++) {\n",
" putchar('\\\"');\n",
" with(*p);\n",
" putchar('\\\"'); putchar(','); putchar('\\",
"n');\n",
" }\n",
" out(\"NULL }, *b[] = {\\",
"n\");\n",
" for (p = b; *p; p++) {\n",
" putchar('\\\"');\n",
" with(*p);\n",
" putchar('\\\"'); putchar(','); putchar('\\",
"n');\n",
" }\n",
" for (p = b; *p; p++) out(*p);\n",
" return 0;\n",
"}\n",
NULL }, **p;
for (p = a; *p; p++) out(*p);
for (p = a; *p; p++) {
putchar('\"');
with(*p);
putchar('\"'); putchar(','); putchar('\n');
}
out("NULL }, *b[] = {\n");
for (p = b; *p; p++) {
putchar('\"');
with(*p);
putchar('\"'); putchar(','); putchar('\n');
}
for (p = b; *p; p++) out(*p);
return 0;
}
A common trick is to jump start the quine by writing a program to read a textfile and output an array of numbers. Then you modify it to use a static array, and run the first program against the new (static array) program, producing an array of number that represents the program. Insert that into the static array, run it again until it settles down, and that gets you a quine. But, it's tied to a specific character set (== not 100% portable). A program like the above (and not the classic printf hack) will work the same on ASCII or EBCDIC (the classic printf hack fails in EBCDIC because it contains hard-coded ASCII).
edit:
Reading the question again, carefully (finally), it appears you're actually looking for more philosophy less technique. The trick that lets you out of the infinite regress is the two-fer. You've got to get both the encoded program and the expanded program out of the same data: using the same data 2 ways. This data thus only describes the part of the program surrounding its future manifestation, the frame. The image within this frame is a straight copy of the original.
This is how you would naturally go about producing a recursive drawing by hand: the tv of a tv of tv. At some point you get tired and just sketch some glare over the screen, because the recursion has been sufficiently established.
edit:
I'm looking for an excellent explanation of why Quines are possible.
The "possibility" of a Quine goes into the depths of the mathematical revolutions of the 19th and 20th centuries. The "classic" quine by W. V. O. Quine, is the sequence of words (IIRC)
yields false when appended to itself
which is a paradox, akin to David's request for something that "makes me happy when sad, and makes me sad when happy" answered by the medallion inscribed on both sides: "this too shall pass".
The same sort of knot was investigated by the Pioneers of modern mathematical logic such as Frege, Russell and Whitehead, Łukasiewicz, and of course, our boys Turing, Church and Thue. The trick that makes it possible to transpose the Quine from the realm of wordplay to a programmatic demonstration (untwisting the paradox part along the way), was Gödel's method of encoding the arithmetic operations themselves as numbers, so an entire mathematical expression can be encoded into a single (enormous) integer. In particular, a mathematical function that performs a decoding of this representation can be expressed in the same (numerical) form. This number (a Gödel-encoded function) is both code and data.
This power-trio (Code, Representation, Data), can be transposed to different repesentations. By choosing a different Representation (or a chain like: bytes-> ASCII-> hexadecimal-> integer), alters the behavior of the Code, which alters the appearance of the Data.

Related

Parallel Dynamic Programming with CUDA

It is my first attempt to implement recursion with CUDA. The goal is to extract all the combinations from a set of chars "12345" using the power of CUDA to parallelize dynamically the task. Here is my kernel:
__device__ char route[31] = { "_________________________"};
__device__ char init[6] = { "12345" };
__global__ void Recursive(int depth) {
// up to depth 6
if (depth == 5) return;
// newroute = route - idx
int x = depth * 6;
printf("%s\n", route);
int o = 0;
int newlen = 0;
for (int i = 0; i<6; ++i)
{
if (i != threadIdx.x)
{
route[i+x-o] = init[i];
newlen++;
}
else
{
o = 1;
}
}
Recursive<<<1,newlen>>>(depth + 1);
}
__global__ void RecursiveCount() {
Recursive <<<1,5>>>(0);
}
The idea is to exclude 1 item (the item corresponding to the threadIdx) in each different thread. In each recursive call, using the variable depth, it works over a different base (variable x) on the route device variable.
I expect the kernel prompts something like:
2345_____________________
1345_____________________
1245_____________________
1234_____________________
2345_345_________________
2345_245_________________
2345_234_________________
2345_345__45_____________
2345_345__35_____________
2345_345__34_____________
..
2345_245__45_____________
..
But it prompts ...
·_____________
·_____________
·_____________
·_____________
·_____________
·2345
·2345
·2345
·2345
...
What I´m doing wrong?
What I´m doing wrong?
I may not articulate every problem with your code, but these items should get you a lot closer.
I recommend providing a complete example. In my view it is basically required by Stack Overflow, see item 1 here, note use of the word "must". Your example is missing any host code, including the original kernel call. It's only a few extra lines of code, why not include it? Sure, in this case, I can deduce what the call must have been, but why not just include it? Anyway, based on the output you indicated, it seems fairly evident the launch configuration of the host launch would have to be <<<1,1>>>.
This doesn't seem to be logical to me:
I expect the kernel prompts something like:
2345_____________________
The very first thing your kernel does is print out the route variable, before making any changes to it, so I would expect _____________________. However we can "fix" this by moving the printout to the end of the kernel.
You may be confused about what a __device__ variable is. It is a global variable, and there is only one copy of it. Therefore, when you modify it in your kernel code, every thread, in every kernel, is attempting to modify the same global variable, at the same time. That cannot possibly have orderly results, in any thread-parallel environment. I chose to "fix" this by making a local copy for each thread to work on.
You have an off-by-1 error, as well as an extent error in this loop:
for (int i = 0; i<6; ++i)
The off-by-1 error is due to the fact that you are iterating over 6 possible items (that is, i can reach a value of 5) but there are only 5 items in your init variable (the 6th item being a null terminator. The correct indexing starts out over 0-4 (with one of those being skipped). On subsequent iteration depths, its necessary to reduce this indexing extent by 1. Note that I've chosen to fix the first error here by increasing the length of init. There are other ways to fix, of course. My method inserts an extra _ between depths in the result.
You assume that at each iteration depth, the correct choice of items is the same, and in the same order, i.e. init. However this is not the case. At each depth, the choices of items must be selected not from the unchanging init variable, but from the choices passed from previous depth. Therefore we need a local, per-thread copy of init also.
A few other comments about CUDA Dynamic Parallelism (CDP). When passing pointers to data from one kernel scope to a child scope, local space pointers cannot be used. Therefore I allocate for the local copy of route from the heap, so it can be passed to child kernels. init can be deduced from route, so we can use an ordinary local variable for myinit.
You're going to quickly hit some dynamic parallelism (and perhaps memory) limits here if you continue this. I believe the total number of kernel launches for this is 5^5, which is 3125 (I'm doing this quickly, I may be mistaken). CDP has a pending launch limit of 2000 kernels by default. We're not hitting this here according to what I see, but you'll run into that sooner or later if you increase the depth or width of this operation. Furthermore, in-kernel allocations from the device heap are by default limited to 8KB. I don't seem to be hitting that limit, but probably I am, so my design should probably be modified to fix that.
Finally, in-kernel printf output is limited to the size of a particular buffer. If this technique is not already hitting that limit, it will soon if you increase the width or depth.
Here is a worked example, attempting to address the various items above. I'm not claiming it is defect free, but I think the output is closer to your expectations. Note that due to character limits on SO answers, I've truncated/excerpted some of the output.
$ cat t1639.cu
#include <stdio.h>
__device__ char route[31] = { "_________________________"};
__device__ char init[7] = { "12345_" };
__global__ void Recursive(int depth, const char *oroute) {
char *nroute = (char *)malloc(31);
char myinit[7];
if (depth == 0) memcpy(myinit, init, 6);
else memcpy(myinit, oroute+(depth-1)*6, 6);
myinit[6] = 0;
if (nroute == NULL) {printf("oops\n"); return;}
memcpy(nroute, oroute, 30);
nroute[30] = 0;
// up to depth 6
if (depth == 5) return;
// newroute = route - idx
int x = depth * 6;
//printf("%s\n", nroute);
int o = 0;
int newlen = 0;
for (int i = 0; i<(6-depth); ++i)
{
if (i != threadIdx.x)
{
nroute[i+x-o] = myinit[i];
newlen++;
}
else
{
o = 1;
}
}
printf("%s\n", nroute);
Recursive<<<1,newlen>>>(depth + 1, nroute);
}
__global__ void RecursiveCount() {
Recursive <<<1,5>>>(0, route);
}
int main(){
RecursiveCount<<<1,1>>>();
cudaDeviceSynchronize();
}
$ nvcc -o t1639 t1639.cu -rdc=true -lcudadevrt -arch=sm_70
$ cuda-memcheck ./t1639
========= CUDA-MEMCHECK
2345_____________________
1345_____________________
1245_____________________
1235_____________________
1234_____________________
2345__345________________
2345__245________________
2345__235________________
2345__234________________
2345__2345_______________
2345__345___45___________
2345__345___35___________
2345__345___34___________
2345__345___345__________
2345__345___45____5______
2345__345___45____4______
2345__345___45____45_____
2345__345___45____5______
2345__345___45____5_____5
2345__345___45____4______
2345__345___45____4_____4
2345__345___45____45____5
2345__345___45____45____4
2345__345___35____5______
2345__345___35____3______
2345__345___35____35_____
2345__345___35____5______
2345__345___35____5_____5
2345__345___35____3______
2345__345___35____3_____3
2345__345___35____35____5
2345__345___35____35____3
2345__345___34____4______
2345__345___34____3______
2345__345___34____34_____
2345__345___34____4______
2345__345___34____4_____4
2345__345___34____3______
2345__345___34____3_____3
2345__345___34____34____4
2345__345___34____34____3
2345__345___345___45_____
2345__345___345___35_____
2345__345___345___34_____
2345__345___345___45____5
2345__345___345___45____4
2345__345___345___35____5
2345__345___345___35____3
2345__345___345___34____4
2345__345___345___34____3
2345__245___45___________
2345__245___25___________
2345__245___24___________
2345__245___245__________
2345__245___45____5______
2345__245___45____4______
2345__245___45____45_____
2345__245___45____5______
2345__245___45____5_____5
2345__245___45____4______
2345__245___45____4_____4
2345__245___45____45____5
2345__245___45____45____4
2345__245___25____5______
2345__245___25____2______
2345__245___25____25_____
2345__245___25____5______
2345__245___25____5_____5
2345__245___25____2______
2345__245___25____2_____2
2345__245___25____25____5
2345__245___25____25____2
2345__245___24____4______
2345__245___24____2______
2345__245___24____24_____
2345__245___24____4______
2345__245___24____4_____4
2345__245___24____2______
2345__245___24____2_____2
2345__245___24____24____4
2345__245___24____24____2
2345__245___245___45_____
2345__245___245___25_____
2345__245___245___24_____
2345__245___245___45____5
2345__245___245___45____4
2345__245___245___25____5
2345__245___245___25____2
2345__245___245___24____4
2345__245___245___24____2
2345__235___35___________
2345__235___25___________
2345__235___23___________
2345__235___235__________
2345__235___35____5______
2345__235___35____3______
2345__235___35____35_____
2345__235___35____5______
2345__235___35____5_____5
2345__235___35____3______
2345__235___35____3_____3
2345__235___35____35____5
2345__235___35____35____3
2345__235___25____5______
2345__235___25____2______
2345__235___25____25_____
2345__235___25____5______
2345__235___25____5_____5
2345__235___25____2______
2345__235___25____2_____2
2345__235___25____25____5
2345__235___25____25____2
2345__235___23____3______
2345__235___23____2______
2345__235___23____23_____
2345__235___23____3______
2345__235___23____3_____3
2345__235___23____2______
2345__235___23____2_____2
2345__235___23____23____3
2345__235___23____23____2
2345__235___235___35_____
2345__235___235___25_____
2345__235___235___23_____
2345__235___235___35____5
2345__235___235___35____3
2345__235___235___25____5
2345__235___235___25____2
2345__235___235___23____3
2345__235___235___23____2
2345__234___34___________
2345__234___24___________
2345__234___23___________
2345__234___234__________
2345__234___34____4______
2345__234___34____3______
2345__234___34____34_____
2345__234___34____4______
2345__234___34____4_____4
2345__234___34____3______
2345__234___34____3_____3
2345__234___34____34____4
2345__234___34____34____3
2345__234___24____4______
2345__234___24____2______
2345__234___24____24_____
2345__234___24____4______
2345__234___24____4_____4
2345__234___24____2______
2345__234___24____2_____2
2345__234___24____24____4
2345__234___24____24____2
2345__234___23____3______
2345__234___23____2______
2345__234___23____23_____
2345__234___23____3______
2345__234___23____3_____3
2345__234___23____2______
2345__234___23____2_____2
2345__234___23____23____3
2345__234___23____23____2
2345__234___234___34_____
2345__234___234___24_____
2345__234___234___23_____
2345__234___234___34____4
2345__234___234___34____3
2345__234___234___24____4
2345__234___234___24____2
2345__234___234___23____3
2345__234___234___23____2
2345__2345__345__________
2345__2345__245__________
2345__2345__235__________
2345__2345__234__________
2345__2345__345___45_____
2345__2345__345___35_____
2345__2345__345___34_____
2345__2345__345___45____5
2345__2345__345___45____4
2345__2345__345___35____5
2345__2345__345___35____3
2345__2345__345___34____4
2345__2345__345___34____3
2345__2345__245___45_____
2345__2345__245___25_____
2345__2345__245___24_____
2345__2345__245___45____5
2345__2345__245___45____4
2345__2345__245___25____5
2345__2345__245___25____2
2345__2345__245___24____4
2345__2345__245___24____2
2345__2345__235___35_____
2345__2345__235___25_____
2345__2345__235___23_____
2345__2345__235___35____5
2345__2345__235___35____3
2345__2345__235___25____5
2345__2345__235___25____2
2345__2345__235___23____3
2345__2345__235___23____2
2345__2345__234___34_____
2345__2345__234___24_____
2345__2345__234___23_____
2345__2345__234___34____4
2345__2345__234___34____3
2345__2345__234___24____4
2345__2345__234___24____2
2345__2345__234___23____3
2345__2345__234___23____2
1345__345________________
1345__145________________
1345__135________________
1345__134________________
1345__1345_______________
1345__345___45___________
1345__345___35___________
1345__345___34___________
1345__345___345__________
1345__345___45____5______
1345__345___45____4______
1345__345___45____45_____
1345__345___45____5______
1345__345___45____5_____5
1345__345___45____4______
1345__345___45____4_____4
1345__345___45____45____5
1345__345___45____45____4
1345__345___35____5______
1345__345___35____3______
1345__345___35____35_____
1345__345___35____5______
1345__345___35____5_____5
1345__345___35____3______
1345__345___35____3_____3
1345__345___35____35____5
1345__345___35____35____3
1345__345___34____4______
1345__345___34____3______
1345__345___34____34_____
1345__345___34____4______
1345__345___34____4_____4
1345__345___34____3______
1345__345___34____3_____3
1345__345___34____34____4
1345__345___34____34____3
1345__345___345___45_____
1345__345___345___35_____
1345__345___345___34_____
1345__345___345___45____5
1345__345___345___45____4
1345__345___345___35____5
1345__345___345___35____3
1345__345___345___34____4
1345__345___345___34____3
1345__145___45___________
1345__145___15___________
1345__145___14___________
1345__145___145__________
1345__145___45____5______
1345__145___45____4______
1345__145___45____45_____
1345__145___45____5______
1345__145___45____5_____5
1345__145___45____4______
1345__145___45____4_____4
1345__145___45____45____5
1345__145___45____45____4
1345__145___15____5______
1345__145___15____1______
1345__145___15____15_____
1345__145___15____5______
1345__145___15____5_____5
1345__145___15____1______
1345__145___15____1_____1
1345__145___15____15____5
1345__145___15____15____1
1345__145___14____4______
1345__145___14____1______
1345__145___14____14_____
1345__145___14____4______
1345__145___14____4_____4
1345__145___14____1______
1345__145___14____1_____1
1345__145___14____14____4
1345__145___14____14____1
1345__145___145___45_____
1345__145___145___15_____
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1345__145___145___45____5
1345__145___145___45____4
1345__145___145___15____5
1345__145___145___15____1
1345__145___145___14____4
1345__145___145___14____1
1345__135___35___________
1345__135___15___________
1345__135___13___________
1345__135___135__________
1345__135___35____5______
1345__135___35____3______
1345__135___35____35_____
1345__135___35____5______
1345__135___35____5_____5
1345__135___35____3______
1345__135___35____3_____3
1345__135___35____35____5
1345__135___35____35____3
1345__135___15____5______
1345__135___15____1______
1345__135___15____15_____
1345__135___15____5______
1345__135___15____5_____5
1345__135___15____1______
1345__135___15____1_____1
1345__135___15____15____5
1345__135___15____15____1
1345__135___13____3______
1345__135___13____1______
1345__135___13____13_____
1345__135___13____3______
1345__135___13____3_____3
1345__135___13____1______
1345__135___13____1_____1
1345__135___13____13____3
1345__135___13____13____1
1345__135___135___35_____
1345__135___135___15_____
1345__135___135___13_____
1345__135___135___35____5
1345__135___135___35____3
1345__135___135___15____5
1345__135___135___15____1
1345__135___135___13____3
1345__135___135___13____1
1345__134___34___________
1345__134___14___________
1345__134___13___________
1345__134___134__________
1345__134___34____4______
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1345__134___34____4_____4
1345__134___34____3______
1345__134___34____3_____3
1345__134___34____34____4
1345__134___34____34____3
1345__134___14____4______
1345__134___14____1______
1345__134___14____14_____
1345__134___14____4______
1345__134___14____4_____4
1345__134___14____1______
1345__134___14____1_____1
1345__134___14____14____4
1345__134___14____14____1
1345__134___13____3______
1345__134___13____1______
1345__134___13____13_____
1345__134___13____3______
1345__134___13____3_____3
1345__134___13____1______
1345__134___13____1_____1
1345__134___13____13____3
1345__134___13____13____1
1345__134___134___34_____
1345__134___134___14_____
1345__134___134___13_____
1345__134___134___34____4
1345__134___134___34____3
1345__134___134___14____4
1345__134___134___14____1
1345__134___134___13____3
1345__134___134___13____1
1345__1345__345__________
1345__1345__145__________
1345__1345__135__________
1345__1345__134__________
1345__1345__345___45_____
1345__1345__345___35_____
1345__1345__345___34_____
1345__1345__345___45____5
1345__1345__345___45____4
1345__1345__345___35____5
1345__1345__345___35____3
1345__1345__345___34____4
1345__1345__345___34____3
1345__1345__145___45_____
1345__1345__145___15_____
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1345__1345__145___45____4
1345__1345__145___15____5
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1345__1345__135___35_____
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1345__1345__135___13_____
1345__1345__135___35____5
1345__1345__135___35____3
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1345__1345__135___15____1
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1345__1345__134___34____3
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1245__245________________
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1245__245___25___________
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1245__245___45____5______
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1245__245___45____45_____
1245__245___45____5______
1245__245___45____5_____5
1245__245___45____4______
1245__245___45____4_____4
1245__245___45____45____5
1245__245___45____45____4
1245__245___25____5______
1245__245___25____2______
1245__245___25____25_____
1245__245___25____5______
1245__245___25____5_____5
1245__245___25____2______
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1245__245___24____4______
1245__245___24____4_____4
1245__245___24____2______
1245__245___24____2_____2
1245__245___24____24____4
1245__245___24____24____2
1245__245___245___45_____
1245__245___245___25_____
1245__245___245___24_____
1245__245___245___45____5
1245__245___245___45____4
1245__245___245___25____5
1245__245___245___25____2
1245__245___245___24____4
1245__245___245___24____2
1245__145___45___________
1245__145___15___________
1245__145___14___________
1245__145___145__________
1245__145___45____5______
1245__145___45____4______
1245__145___45____45_____
1245__145___45____5______
1245__145___45____5_____5
1245__145___45____4______
...
1235__1235__235___25_____
1235__1235__235___23_____
1235__1235__235___35____5
1235__1235__235___35____3
1235__1235__235___25____5
1235__1235__235___25____2
1235__1235__235___23____3
1235__1235__235___23____2
1235__1235__135___35_____
1235__1235__135___15_____
1235__1235__135___13_____
1235__1235__135___35____5
1235__1235__135___35____3
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1235__1235__135___15____1
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1235__1235__125___25_____
1235__1235__125___15_____
1235__1235__125___12_____
1235__1235__125___25____5
1235__1235__125___25____2
1235__1235__125___15____5
1235__1235__125___15____1
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1235__1235__125___12____1
1235__1235__123___23_____
1235__1235__123___13_____
1235__1235__123___12_____
1235__1235__123___23____3
1235__1235__123___23____2
1235__1235__123___13____3
1235__1235__123___13____1
1235__1235__123___12____2
1235__1235__123___12____1
1234__234________________
1234__134________________
1234__124________________
1234__123________________
1234__1234_______________
1234__234___34___________
1234__234___24___________
1234__234___23___________
1234__234___234__________
1234__234___34____4______
1234__234___34____3______
1234__234___34____34_____
1234__234___34____4______
1234__234___34____4_____4
1234__234___34____3______
1234__234___34____3_____3
1234__234___34____34____4
1234__234___34____34____3
1234__234___24____4______
1234__234___24____2______
1234__234___24____24_____
1234__234___24____4______
1234__234___24____4_____4
1234__234___24____2______
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1234__234___24____24____4
1234__234___24____24____2
1234__234___23____3______
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1234__234___23____3______
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1234__234___234___34_____
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1234__134___34___________
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1234__134___34____34_____
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1234__134___34____3______
1234__134___34____3_____3
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1234__1234__123___12____1
========= ERROR SUMMARY: 0 errors
$
The answer given by Robert Crovella is correct at the 5th point, the mistake was in the using of init in every recursive call, but I want to clarify something that can be useful for other beginners with CUDA.
I used this variable because when I tried to launch a child kernel passing a local variable I always got the exception: Error: a pointer to local memory cannot be passed to a launch as an argument.
As I´m C# expert developer I´m not used to using pointers (Ref does the low-level-work for that) so I thought there was no way to do it in CUDA/c programming.
As Robert shows in its code it is possible copying the pointer with memalloc for using it as a referable argument.
Here is a kernel simplified as an example of deep recursion.
__device__ char init[6] = { "12345" };
__global__ void Recursive(int depth, const char* route) {
// up to depth 6
if (depth == 5) return;
//declaration for a referable argument (point 6)
char* newroute = (char*)malloc(6);
memcpy(newroute, route, 5);
int o = 0;
int newlen = 0;
for (int i = 0; i < (6 - depth); ++i)
{
if (i != threadIdx.x)
{
newroute[i - o] = route[i];
newlen++;
}
else
{
o = 1;
}
}
printf("%s\n", newroute);
Recursive <<<1, newlen>>>(depth + 1, newroute);
}
__global__ void RecursiveCount() {
Recursive <<<1, 5>>>(0, init);
}
I don't add the main call because I´m using ManagedCUDA for C# but as Robert says it can be figured-out how the call RecursiveCount is.
About ending arrays of char with /0 ... sorry but I don't know exactly what is the benefit; this code works fine without them.

Ambiguous process calcChecksum

CONTEXT
I'm using a code written to work with a GPS module that connects to the Arduino through serial communication. The module starts each packet with a header (0xb5, 0x62), continues with the information you requested and ends with to bytes of checksum, CK_A, and CK_B. I don't understand the code that calculates that checksum. More info about the algorithm of checksum (8-Bit Fletcher Algorithm) in the module protocol (https://www.u-blox.com/sites/default/files/products/documents/u-blox7-V14_ReceiverDescriptionProtocolSpec_%28GPS.G7-SW-12001%29_Public.pdf), page 74 (87 with index).
MORE INFO
Just wanted to understand the code, it works fine. In the UBX protocol, I mentioned there is also a piece of code that explains how it works (isn't write in c++)
struct NAV_POSLLH {
//Here goes the struct
};
NAV_POSLLH posllh;
void calcChecksum(unsigned char* CK) {
memset(CK, 0, 2);
for (int i = 0; i < (int)sizeof(NAV_POSLLH); i++) {
CK[0] += ((unsigned char*)(&posllh))[i];
CK[1] += CK[0];
}
}
In the link you provide, you can find a link to RFC 1145, containing that Fletcher 8 bit algorithm as well and explaining
It can be shown that at the end of the loop A will contain the 8-bit
1's complement sum of all octets in the datagram, and that B will
contain (n)*D[0] + (n-1)*D[1] + ... + D[n-1].
n = sizeof byte D[];
Quote adjusted to C syntax
Try it with a couple of bytes, pen and paper, and you'll see :)

String comparison in the core language

Taking this simple comparison loopValue == "Firstname", is the following statement true?
If the internal operand inspecting the first char does not match the compared string, it will early abort
So taking the rawer form loopValue and "Firstname" are both []byte. And it would walk the array kind of like so as callback loop for truth:
someInspectionFunc(loopValue, "Firstname", func(charA, charB) {
return charA == charB
})
... making it keep on going until it bumps false and checks if the number of iterations was equal to both their lengths. Also does it check length first?
if len(loopValue) != len("Firstname") {
return false
}
I can't really find an explanation in the go source-code on GitHub as it's a bit above me.
The reason I'm asking this is because I'm doing big data processing and am benchmarking and doing cpu, memory and allocation pprof to squeeze some more juice out of the process. From that process it kind of made me think how Go (but also just C in general) would do this under the hood. Is this fully on an assembly level or does the comparison already happen in native Go code (kind of like sketched in the snippets above)?
Please let me know if I'm being too vague or if I missed something. Thank you
Update
When I did a firstCharater match in big strings of json, before really comparing I got about 3.7% benchmarking gain on 100k heavy entries:
<some irrelevant inspection code>.. v[0] == firstChar && v == lookFor {
// Match found when it reaches here
}
the code above (especially on long strings) is faster than just going for v == lookFor.
The function is handled in assembly. The amd64 version is:
TEXT runtime·eqstring(SB),NOSPLIT,$0-33
MOVQ s1str+0(FP), SI
MOVQ s2str+16(FP), DI
CMPQ SI, DI
JEQ eq
MOVQ s1len+8(FP), BX
LEAQ v+32(FP), AX
JMP runtime·memeqbody(SB)
eq:
MOVB $1, v+32(FP)
RET
And it's the compiler's job to ensure that the strings are of equal length before that is called. (The runtime·memeqbody function is actually where the optimized memory comparisons happen, but there's probably no need to post the full text here)
The equivalent Go code would be:
func eqstring_generic(s1, s2 string) bool {
if len(s1) != len(s2) {
return false
}
for i := 0; i < len(s1); i++ {
if s1[i] != s2[i] {
return false
}
}
return true
}

loop on prompt with a yes or no?

Good afternoon,
I'm trying to accomplish a task that i know should be doable. however my attempts seem to fail every time. My endeavor is to learn to code in Objective -c and have been making good progress. what i would like to do is add a loop to my current application that asks at the end if i would like to run again or some thing to that regard, and reply with a yes or no. if no the program ends and if yes it jumps back to the top of the project to start all over. kinda like what i have below? forgive me please if its not quite perfect, im still getting used to programing and am finding it incredibly fun.
#include <stdio.h>
int main(void)
{
char loop = yes;
while (loop = yes)
{
.
.
.
}
printf ("would you like to continue (yes/no)/n");
scanf ("%s", loop);
}
The printf and scanf need to be moved up inside the curly braces of the while loop. Also, you want \n instead of /n in the printf. Finally, you're going to get a string back with that scanf() call, so you'll want to declare loop as a char array, and then in the while loop, check the first element of that array for a 'y' or 'n' or something like that. You might also want to look at getchar() instead of scanf() for that sort of thing.
Not compiled here, but should work:
#include <stdio.h>
int main(void)
{
char buffer[256];
do {
.
.
.
printf ("would you like to continue (yes/no)/n");
scanf ("%s", buffer);
} while (strcmp(buffer,"yes") != 0);
}
One wouldn't do anything like that in a real world application, but for demonstration purpose it should be ok.
I made your variable an array, because strings are arrays of characters in C. Length is set to 256 bytes (255 characters + 0-byte as delimiter). I changed the loop to do-while to make it run at least once. For string comparison you need to call a function. strcmp returns 0 for identical strings. Finally, the question belongs in the loop.
It is plain C though, using nothing of Objective-C.
int main() {
char A = 'n';
char B = 'y';
char Answer;
printf("Does the subject have a glazed over look? (y/n): \n");
scanf("%c",&Answer);
if (Answer=='N'||Answer=='y'|| Answer=='N'||Answer=='Y')
printf("Good\n");
else
printf("Please enter 'y' or 'n' \n ");
return 0;
}
#include <stdio.h>
int main(void)
{
avi;
char loop[10];
while (loop = yes)
{
.
.
.
}
printf ("would you like to continue (yes/no)/n");
scanf ("%s", loop);
if(strcpm(loop,"YES")==0) goto avi:
}

How do I get user input without using scanf in a console app?

I'm trying to allow users to simply hit Enter without typing anything, and use this to mean accepting a default value. scanf isn't doing what I want and the app still 'blocks': the next line of code doesn't run.
The only way is to actually type something THEN press Enter.
I tried using NSFileHandle and fileHandleWithStandardInput; however, it seems that the user is now forced to hit Ctrl-D to indicate EOF.
Someone suggested using fgets, but I cannot work out what to pass as 3rd parameter (of FILE* type). Tried stdin but it doesn't 'block'.
How do I accept input from a user, using Objective-C, and at the same time allow the user to simply hit Enter without being forced to type anything? How do I read a single line, even if that line is blank?
Assuming the code doesn't block and the next line runs immediately (as you seemed to indicate early in the question and in a comment), you have a common problem when mixing non-line-based and line-based input.
What happens is you have a newline left in the buffer, and fgets sees that, reads it, and returns, instead of doing what you really want: ignoring it, and then reading a line.
The solution is to simply do the ignoring part yourself, and then call fgets:
#include <stdio.h>
#include <string.h>
FILE* ignoreline(FILE* stream) {
for (int c; (c = fgetc(stream)) != EOF;) {
if (c == '\n') break;
}
return stream;
}
void example_use() {
char buf[1000];
ignoreline(stdin);
fgets(buf, sizeof buf, stdin);
// or, since it returns the stream, can be more compact:
fgets(buf, sizeof buf, ignoreline(stdin));
}
int main() { // error handling omitted
int n;
printf("Enter a number: ");
scanf("%d", &n);
char buf[1000];
printf("Enter a line: ");
ignoreline(stdin); // comment this line and compare the difference
fgets(buf, sizeof buf, stdin);
*strchr(buf, '\n') = '\0';
printf("You entered '%s'.\n", buf);
return 0;
}
Note that it is also common and encouraged to "pair" the ignoreline with the scanf (or other non-line-based input) to turn that into line-based input. You may want to modify it, in that case, so you can tell the difference between input of "42 abc" and "42" (in the "Enter a number" case). Some people just use fgets everywhere, then parse that line with sscanf, and while that works, it's not necessary.
I use getch(); in library conio.h
simply the program waits for any key to be pressed
If you're using Windows, you can use the ReadConsoleInput function (see MSDN for more on this) :
INPUT_RECORD keyin;
DWORD r;
while (ReadConsoleInput(GetStdHandle(STD_INPUT_HANDLE),&keyin,1,&r)) {
if (keyin.EventType!=KEY_EVENT) continue;
if (keyin.Event.KeyEvent.wVirtualKeyCode==VK_SPACE) break; ///use these VK codes to get any key's input
if (keyin.Event.KeyEvent.wVirtualKeyCode==VK_F1)
{
printf("You pressed F1\n");
}
if (keyin.Event.KeyEvent.wVirtualKeyCode==VK_F2)
{
printf("You pressed F2\n",);
}
}//end while loop
You don't need to hit enter after each key then.This works like a dream for me...
use getchar() to take input without using scanf function...