Does anyone know how to convert between little and big endians in CataputC?
Part of the design, I am working on, needs to use litle endian and rest uses big endian.
I am searching for something equivalent to "downto" "to" keywords in vhdl.
After a while I realized that you simply have to reverse it in loop and Catapult synthesis tool accepts it as signal reverse. So I made myself a useful template.
template<int W> ac_int<W> reversed(ac_int<W> in) {
ac_int<W> out;
for (int i = 0; i < W; i++)
out[i] = in[W - i - 1];
return out;
}
Related
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_______________
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2345__345___35___________
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2345__345___45____5______
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2345__345___45____4______
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2345__345___45____45____5
2345__345___45____45____4
2345__345___35____5______
2345__345___35____3______
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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______
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2345__345___34____4______
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2345__345___34____3______
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2345__345___345___45_____
2345__345___345___35_____
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2345__345___345___45____5
2345__345___345___45____4
2345__345___345___35____5
2345__345___345___35____3
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2345__345___345___34____3
2345__245___45___________
2345__245___25___________
2345__245___24___________
2345__245___245__________
2345__245___45____5______
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2345__245___45____4______
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2345__245___25____5______
2345__245___25____2______
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2345__245___25____2______
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2345__245___25____25____5
2345__245___25____25____2
2345__245___24____4______
2345__245___24____2______
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2345__245___24____4______
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2345__245___24____2______
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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______
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2345__235___35____5______
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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______
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2345__234___24____2______
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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______
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1345__345___45____45____5
1345__345___45____45____4
1345__345___35____5______
1345__345___35____3______
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1345__345___35____3______
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1345__345___35____35____5
1345__345___35____35____3
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1345__345___34____3______
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1345__345___345___45____5
1345__345___345___45____4
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1345__145___45___________
1345__145___15___________
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1345__145___45____4______
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1345__145___45____45____5
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1345__145___15____5______
1345__145___15____1______
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1345__145___15____5______
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1345__145___15____1______
1345__145___15____1_____1
1345__145___15____15____5
1345__145___15____15____1
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1345__145___14____1______
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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
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1345__145___145___15_____
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1345__145___145___45____5
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1345__145___145___15____5
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1345__145___145___14____1
1345__135___35___________
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1345__135___35____5______
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1345__135___35____3______
1345__135___35____3_____3
1345__135___35____35____5
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1345__135___15____5______
1345__135___15____1______
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1345__135___15____5______
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1345__135___15____1______
1345__135___15____1_____1
1345__135___15____15____5
1345__135___15____15____1
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1345__134___14____1______
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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______
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1345__134___13____3______
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1345__134___13____1______
1345__134___13____1_____1
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1345__134___13____13____1
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1345__134___134___13____1
1345__1345__345__________
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1345__1345__135__________
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1345__1345__345___45_____
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1345__1345__135___35_____
1345__1345__135___15_____
1345__1345__135___13_____
1345__1345__135___35____5
1345__1345__135___35____3
1345__1345__135___15____5
1345__1345__135___15____1
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1345__1345__135___13____1
1345__1345__134___34_____
1345__1345__134___14_____
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1245__245________________
1245__145________________
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1245__1245_______________
1245__245___45___________
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1245__245___245___45_____
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...
1235__1235__235___25_____
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1235__1235__123___13____3
1235__1235__123___13____1
1235__1235__123___12____2
1235__1235__123___12____1
1234__234________________
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1234__1234_______________
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1234__234___34____34_____
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========= 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.
void KeyExpansion(unsigned char key[N_KEYS], unsigned int* w)
{
unsigned int temp;
for(int i=0; i< N_KEYS; i++)
{
w[i] = (key[N_KEYS*i]<<24) + (key[N_KEYS*i+1]<<16) + (key[N_KEYS*i+2]<<8) + key[N_KEYS*i+3];
}
for(int i = 4; i< EXPANDED_KEY_COUNT; i++)
{
temp = w[i-1];
if(i % 4 == 0)
temp = SubWord(RotWord(temp)) ^ Rcon[i/4];
w[i] = temp ^ w[i-4] ;
}
}
Big-O helps us do analysis based on the input. The issue with your question is that there seems to be several inputs, which may or may not relate with each other.
Input variables look like N_KEYS, and EXPANDED_KEY_COUNT. We also don't know what SubWord() or RotWord() do based on what is provided.
Since SubWord() and RotWord() aren't provided, lets assume they are constant for easy calculations.
You have basic loops and iterate over each value, so its pretty straight forward. This means you have O(N_KEYS) + O(EXPANDED_KEY_COUNT). So the overall time complexity depends on two inputs, and would be bound by the larger.
If SubWord() or RotWord() do anything special that aren't constant time, then that would affect the time complexity of O(EXPANDED_KEY_COUNT) portion of code. You could adjust the time complexity by multiplied against it. But by the names of the methods, it sounds like their time complexity would be based on the length of the string, would would be yet another different input variable.
So this isn't a clear answer, because the question isn't fully clear, but I tried to break things down for you as best as I could.
-(void)InitWithPwd:(char *)pPwd
{
char szResult[17];
//generate md5 checksum
CC_MD5(pPwd, strlen(pPwd),&szResult[0]);
szResult[16] = 0;
m_csPasswordHash[0]=0;
for(int i = 0;i < 16;i++)
{
char sz[3] = {'\0'};
//crash in blow row. The first pass is ok. The third pass crash.
//I can't understand.
sprintf(&sz[0],"%2.2x",szResult[i]);
strcat(m_csPasswordHash,sz);
}
m_csPasswordHash[32] = 0;
printf("pass:%s\n",m_csPasswordHash);
m_ucPacketType = 1;
}
I want to get the md5 of the password. But above code crash again and again. I can't understand why.
Your buffer (sz) is too small, causing sprintf() to generate a buffer overflow which leads to undefined behavior, in your case a crash.
Note that szResult[1] might be a negative value when viewed as an int (which happens when passing a char-type value to sprintf()), which can cause sprintf() to disregard your field width and precision directives in order to format the full value.
Here is an example showing this problem. The example code is written in C, but that shouldn't matter for this case.
This solves the problem by making sure the incoming data is considered unsigned:
sprintf(sz, "%02x", (unsigned char) szResult[i]);
I have the following code which progressively goes through a string of bits and rearrange them into blocks of 20bytes. I'm using 32*8 blocks with 40 threads per block. However the process takes something like 36ms on my GT630M. Are there any further optimization I can do? Especially with regard to removing the if-else in the inner most loop.
__global__ void test(unsigned char *data)
{
__shared__ unsigned char dataBlock[20];
__shared__ int count;
count = 0;
unsigned char temp = 0x00;
for(count=0; count<(streamSize/8); count++)
{
for(int i=0; i<8; i++)
{
if(blockIdx.y >= i)
temp |= (*(data + threadIdx.x*(blockIdx.x + gridDim.x*(i+count)))&(0x01<<blockIdx.y))>>(blockIdx.y - i);
else
temp |= (*(data + threadIdx.x*(blockIdx.x + gridDim.x*(i+count)))&(0x01<<blockIdx.y))<<(i - blockIdx.y);
}
dataBlock[threadIdx.x] = temp;
//do something
}
}
It's not clear what your code is trying to accomplish, but a couple obvious opportunities are:
1) if possible, use 32-bit words instead of unsigned char.
2) use block sizes that are multiples of 32.
3) The conditional code may not be costing you as much as you expect. You can check by compiling with --cubin --gpu-architecture sm_xx (where xx is the SM version of your target hardware), and using cuobjdump --dump-sass on the resulting cubin file to look at the generated assembly. You may have to modify the source code to loft the common subexpression into a separate variable, and/or use the ternary operator ? : to hint to the compiler to use predication.
I'm having trouble implementing realloc in a very basic way.
I'm trying to expand the region of memory at **ret, which is pointing to an array of structs
with ret = realloc(ret, newsize); and based on my debug strings I know newsize is correctly increasing over the course of the loop (going from the original size of 4 to 8 to 12 etc.), but when I do sizeof(ptr) it's still returning the original size of 4, and the things I'm trying to place into the newly allocated space can't be found (I think I've narrowed it down to realloc() which is why I'm formatting the question like this)
I can post the function in it's entirety if the problem isn't immediately evident to you, I'm just trying to not "cheat" with my homework too much (the code is kind of messy right now anyway, with heavy use of printf() for debug).
[EDIT] Alright, so based on your answers I'm failing at debugging my code, so I guess I'll post the whole function so you can tell me more about what I'm doing wrong.
(You can ignore the printf()'s since most of that is debug that isn't even working)
Booking **bookingSelectPaid(Booking **booking) {
Booking **ret = malloc(sizeof(Booking*));
printf("Initial address of ret = %p\n", ret);
size_t i = 0;
int numOfPaid = 0;
while (booking[i] != NULL)
{
if (booking[i]->paid == 1)
{
printf("Paying customer! sizeof(Booking*) = %d\n", (int)sizeof(Booking*));
++numOfPaid;
size_t newsize = sizeof(Booking*) * (numOfPaid + 1);
printf("Newsize = %d\n", (int)newsize);
Booking **temp = realloc(NULL, (size_t)newsize);
if (temp != NULL)
printf("Expansion success! => %p sizeof(new pointer) = %d ret = %p\n", temp, (int)sizeof(temp), ret);
ret = realloc(ret, newsize);
ret[i] = booking[i];
ret[i+1] = NULL;
}
++i;
printf("Sizeof(ret) = %d numOfPaid = %d\n", (int)sizeof(ret), numOfPaid);
}
return ret; }
[EDIT2] --> http://pastebin.com/xjzUBmPg
[EDIT3] Just to be clear, the printf's, the temp pointer and things of that nature are debug, and not part of the intended functionality. The line that is puzzling me is either the one with realloc(ret, newsize); or ret[i] = booking[i]
Basically I know for sure that booking contains a table of structs that ends in NULL, and I'm trying to bring the ones that have a specific value set to 1 (paid) onto the new table, which is what my main() is trying to get from this function... So where am I going wrong?
I think the problem here is that your sizeof(ptr) only returns the size of the pointer, which will depend on your architecture (you say 4, so that would mean you're running a 32-bit system).
If you allocate memory dynamically, you have to keep track of its size yourself.
Because sizeof(ptr) returns the size of the pointer, not the allocated size
Yep, sizeof(ptr) is a constant. As the other answer says, depends on the architecture. On a 32 bit architecture it will be 4 and on a 64 bit architecture it will be 8. If you need more help with questions like that this homework help web site can be great for you.
Good luck.