what will happen when the integer crosses its limit? The output is 3595 , and how it will come? And it is 2 byte type ?
#include<stdio.h>
#include<conio.h>
void main()
{
int n=12,res=1;
clrscr();
while(n>3)
{
n+=3;
res*=3;
}
printf("%d",n*res);
getch();
}
The program will have undefined behavior.
The condition you gave is non terminating. It's a loop where the condition will never be terminated in a well defined manner.
You will go on multiplying and then once it will overflow. And then if you get a negative result in n or <=3 then it will stop. And in the mean time res has also overflown. As a result you will not be sure how this program behaves. We can't be sure of what the result will be.
The behaviour is undefined - you should not rely on anything specific. Common manifestations on int overflow are:
Wraparound such that 1 + INT_MAX becomes INT_MIN. This is what every Windows PC I have encountered does. The bit pattern produced by the operation matches the unsigned cousin exactly.
Clamping such that 1 + INT_MAX becomes INT_MAX. I last observed this on a machine (with signed magnitude int) running a variant of UNIX in the 1990s.
Related
(123L).toInt() produces 123 but Long.MAX_VALUE.toInt() produces -1. Clearly this is not correct. Without having to write a lot of boilerplate code, is there a way to get Kotlin to throw an exception when a value is out of range/bounds for the target type?
TL;DR you can make a custom extension function that checks if the value is between Int.MIN_VALUE and Int.MAX_VALUE
fun Long.toIntThrowing() : Int {
if (this < Int.MIN_VALUE || this > Int.MAX_VALUE) {
throw RuntimeException()
}
return this.toInt()
}
The "weird" behavior you are observing is happening, because in Kotlin, Long is represented as a 64 bit signed integer, while Int is represented as a 32 bit signed integer.
While 123L is easily representable by a 32 bit integer, Long.MAX_VALUE will overflow the Integer (almost) twice, resulting in the behavior you are observing.
I believe the example below will illustrate it better:
println((2147483647L).toInt()) // the max 32 bit signed int
println((2147483648L).toInt()) // 1 more overflows it to the min (negative) 32 bit signed int
println((2147483649L).toInt()) // 2 more...
println((Long.MAX_VALUE - 1).toInt())
println((Long.MAX_VALUE).toInt())
results in :
2147483647
-2147483648
-2147483647
-2
-1
From: https://discuss.kotlinlang.org/t/checked-and-unsigned-integer-operations/529/2
Exceptions on arithmetic overflow: this will likely make arithmetics significantly slower, and we don’t see how to avoid it without changes to the JVM, nor are we ready to accept the slowdown
If you are running on the JVM you may use Math.toIntExact:
Returns the value of the long argument; throwing an exception if the value overflows an int.
There doesn't seem to be a pure Kotlin way, but at least you can nicely wrap it:
fun Long.toIntExact() = Math.toIntExact(this)
I have confusion in this particular line-->
result = (double) hi * (1 << 30) * 4 + lo;
of the following code:
void access_counter(unsigned *hi, unsigned *lo)
// Set *hi and *lo to the high and low order bits of the cycle
// counter.
{
asm("rdtscp; movl %%edx,%0; movl %%eax,%1" // Read cycle counter
: "=r" (*hi), "=r" (*lo) // and move results to
: /* No input */ // the two outputs
: "%edx", "%eax");
}
double get_counter()
// Return the number of cycles since the last call to start_counter.
{
unsigned ncyc_hi, ncyc_lo;
unsigned hi, lo, borrow;
double result;
/* Get cycle counter */
access_counter(&ncyc_hi, &ncyc_lo);
lo = ncyc_lo - cyc_lo;
borrow = lo > ncyc_lo;
hi = ncyc_hi - cyc_hi - borrow;
result = (double) hi * (1 << 30) * 4 + lo;
if (result < 0) {
fprintf(stderr, "Error: counter returns neg value: %.0f\n", result);
}
return result;
}
The thing I cannot understand is that why is hi being multiplied with 2^30 and then 4? and then low added to it? Someone please explain what is happening in this line of code. I do know that what hi and low contain.
The short answer:
That line turns a 64bit integer that is stored as 2 32bit values into a floating point number.
Why doesn't the code just use a 64bit integer? Well, gcc has supported 64bit numbers for a long time, but presumably this code predates that. In that case, the only way to support numbers that big is to put them into a floating point number.
The long answer:
First, you need to understand how rdtscp works. When this assembler instruction is invoked, it does 2 things:
1) Sets ecx to IA32_TSC_AUX MSR. In my experience, this generally just means ecx gets set to zero.
2) Sets edx:eax to the current value of the processor’s time-stamp counter. This means that the lower 64bits of the counter go into eax, and the upper 32bits are in edx.
With that in mind, let's look at the code. When called from get_counter, access_counter is going to put edx in 'ncyc_hi' and eax in 'ncyc_lo.' Then get_counter is going to do:
lo = ncyc_lo - cyc_lo;
borrow = lo > ncyc_lo;
hi = ncyc_hi - cyc_hi - borrow;
What does this do?
Since the time is stored in 2 different 32bit numbers, if we want to find out how much time has elapsed, we need to do a bit of work to find the difference between the old time and the new. When it is done, the result is stored (again, using 2 32bit numbers) in hi / lo.
Which finally brings us to your question.
result = (double) hi * (1 << 30) * 4 + lo;
If we could use 64bit integers, converting 2 32bit values to a single 64bit value would look like this:
unsigned long long result = hi; // put hi into the 64bit number.
result <<= 32; // shift the 32 bits to the upper part of the number
results |= low; // add in the lower 32bits.
If you aren't used to bit shifting, maybe looking at it like this will help. If lo = 1 and high = 2, then expressed as hex numbers:
result = hi; 0x0000000000000002
result <<= 32; 0x0000000200000000
result |= low; 0x0000000200000001
But if we assume the compiler doesn't support 64bit integers, that won't work. While floating point numbers can hold values that big, they don't support shifting. So we need to figure out a way to shift 'hi' left by 32bits, without using left shift.
Ok then, shifting left by 1 is really the same as multiplying by 2. Shifting left by 2 is the same as multiplying by 4. Shifting left by [omitted...] Shifting left by 32 is the same as multiplying by 4,294,967,296.
By an amazing coincidence, 4,294,967,296 == (1 << 30) * 4.
So why write it in that complicated fashion? Well, 4,294,967,296 is a pretty big number. In fact, it's too big to fit in an 32bit integer. Which means if we put it in our source code, a compiler that doesn't support 64bit integers may have trouble figuring out how to process it. Written like this, the compiler can generate whatever floating point instructions it might need to work on that really big number.
Why the current code is wrong:
It looks like variations of this code have been wandering around the internet for a long time. Originally (I assume) access_counter was written using rdtsc instead of rdtscp. I'm not going to try to describe the difference between the two (google them), other than to point out that rdtsc does not set ecx, and rdtscp does. Whoever changed rdtsc to rdtscp apparently didn't know that, and failed to adjust the inline assembler stuff to reflect it. While your code might work fine despite this, it might do something weird instead. To fix it, you could do:
asm("rdtscp; movl %%edx,%0; movl %%eax,%1" // Read cycle counter
: "=r" (*hi), "=r" (*lo) // and move results to
: /* No input */ // the two outputs
: "%edx", "%eax", "%ecx");
While this will work, it isn't optimal. Registers are a valuable and scarce resource on i386. This tiny fragment uses 5 of them. With a slight modification:
asm("rdtscp" // Read cycle counter
: "=d" (*hi), "=a" (*lo)
: /* No input */
: "%ecx");
Now we have 2 fewer assembly statements, and we only use 3 registers.
But even that isn't the best we can do. In the (presumably long) time since this code was written, gcc has added both support for 64bit integers and a function to read the tsc, so you don't need to use asm at all:
unsigned int a;
unsigned long long result;
result = __builtin_ia32_rdtscp(&a);
'a' is the (useless?) value that was being returned in ecx. The function call requires it, but we can just ignore the returned value.
So, instead of doing something like this (which I assume your existing code does):
unsigned cyc_hi, cyc_lo;
access_counter(&cyc_hi, &cyc_lo);
// do something
double elapsed_time = get_counter(); // Find the difference between cyc_hi, cyc_lo and the current time
We can do:
unsigned int a;
unsigned long long before, after;
before = __builtin_ia32_rdtscp(&a);
// do something
after = __builtin_ia32_rdtscp(&a);
unsigned long long elapsed_time = after - before;
This is shorter, doesn't use hard-to-understand assembler, is easier to read, maintain and produces the best possible code.
But it does require a relatively recent version of gcc.
I'm doing an exercise from a textbook and the book is outdated, so I'm sort of figuring out how it fits into the new system as I go along. I've got the exact text, and it's returning
'Implicit conversion loses integer precision: 'time_t' (aka 'long') to 'unsigned int''.
The book is "Cocoa Programming for Mac OS X" by Aaron Hillegass, third edition and the code is:
#import "Foo.h"
#implementation Foo
-(IBAction)generate:(id)sender
{
// Generate a number between 1 and 100 inclusive
int generated;
generated = (random() % 100) + 1;
NSLog(#"generated = %d", generated);
// Ask the text field to change what it is displaying
[textField setIntValue:generated];
}
- (IBAction)seed:(id)sender
{
// Seed the randm number generator with time
srandom(time(NULL));
[textField setStringValue:#"Generator Seeded"];
}
#end
It's on the srandom(time(NULL)); line.
If I replace time with time_t, it comes up with another error message:
Unexpected type name 'time_t': unexpected expression.
I don't have a clue what either of them mean. A question I read with the same error was apparently something to do with 64- and 32- bit integers but, heh, I don't know what that means either. Or how to fix it.
I don't have a clue what either of them mean. A question I read with the same error was apparently something to do with 64- and 32- bit integers but, heh, I don't know what that means either. Or how to fix it.
Well you really need to do some more reading so you understand what these things mean, but here are a few pointers.
When you (as in a human) count you normally use decimal numbers. In decimal you have 10 digits, 0 through 9. If you think of a counter, like on an electric meter or a car odometer, it has a fixed number of digits. So you might have a counter which can read from 000000 to 999999, this is a six-digit counter.
A computer represents numbers in binary, which has two digits 0 and 1. A Binary digIT is called a BIT. So thinking about the counter example above, a 32-bit number has 32 binary digits, a 64-bit one 64 binary digits.
Now if you have a 64-bit number and chop off the top 32-bits you may change its value - if the value was just 1 then it will still be 1, but if it takes more than 32 bits then the result will be a different number - just as truncating the decimal 9001 to 01 changes the value.
Your error:
Implicit conversion looses integer precision: 'time_t' (aka 'long') to 'unsigned int'
Is saying you are doing just this, truncating a large number - long is a 64-bit signed integer type on your computer (not on every computer) - to a smaller one - unsigned int is a 32-bit unsigned (no negative values) integer type on your computer.
In your case the loss of precision doesn't really matter as you are using the number in the statement:
srandom(time(NULL));
This line is setting the "seed" - a random number used to make sure each run of your program gets different random numbers. It is using the time as the seed, truncating it won't make any difference - it will still be a random value. You can silence the warning by making the conversion explicit with a cast:
srandom((unsigned int)time(NULL));
But remember, if the value of an expression is important such casts can produce mathematically incorrect results unless the value is known to be in range of the target type.
Now go read some more!
HTH
Its just a notification. You are assigning 'long' to 'unsigned int'
Solution is simple. Just click the yellow notification icon on left ribbon of that particular line where you are assigning that value. it will show a solution. Double click on solution and it will do everything automatically.
It will typecast to match the equation. But try next time to keep in mind the types you are assigning are same.. hope this helps..
I have this code:
unsigned int k=(len - sizeof(MSG_INFO));
NSLog(#"%d",k);
for( unsigned int ix = 0; ix < k; ix++)
{
m_pOutPacket->m_buffer[ix] = (char)(pbuf[ix + sizeof(MSG_INFO)]);
}
The problem is, when:
len = 0 and sizeof(MSG_INFO)=68;
k=-68;
This condition gets into the for loop and is continuing for infinite times.
Your code says: unsigned int k. So k isn't -68, it's unsigned. This makes k a very big number, based around a 4 byte int, it would be 4294967210. This is obviously quite a lot more than 0, so it's going to take your for loop a while to get that high, although it would terminate eventually.
The reason you think that it's -86, is that when you print it out with a function like NSLog, it has no direct knowledge about the arguments passed in, it determines how to treat the arguments, based around the format string, supplied as the first argument.
You're calling:
This:
NSLog(#"%d",k);
This tells NSLog to treat the argument as a signed int (%d). You should be doing this:
NSLog(#"%u",k);
So that NSLog treats the argument as the type that it is: unsigned (%u). See the NSLog documentation.
As it stands, I'd expect your buffer to overrun, trashing memory as the loop runs and your application to crash.
After reflecting, I believe #FreeAsInBeer is correct and you don't want to iterate through the for loop in this situation and you could probably fix this by using signed ints. However, It seems to me like you would be better off, checking len > sizeof(MSG_INFO) and if this isn't the case handling it differently. Most situations I can think of, I wouldn't want to perform any processing after the for loop, if I'd failed to read sufficient information for a message...
I'm not really sure what is going on here, as the loop should never execute. I've loaded up your code, and it seems that the unsigned part of your int declaration is causing the issues. If you remove both of your unsigned specifiers, your code will execute as it should, without ever entering the loop.
There's a common way to store multiple values in one variable, by using a bitmask. For example, if a user has read, write and execute privileges on an item, that can be converted to a single number by saying read = 4 (2^2), write = 2 (2^1), execute = 1 (2^0) and then add them together to get 7.
I use this technique in several web applications, where I'd usually store the variable into a field and give it a type of MEDIUMINT or whatever, depending on the number of different values.
What I'm interested in, is whether or not there is a practical limit to the number of values you can store like this? For example, if the number was over 64, you couldn't use (64 bit) integers any more. If this was the case, what would you use? How would it affect your program logic (ie: could you still use bitwise comparisons)?
I know that once you start getting really large sets of values, a different method would be the optimal solution, but I'm interested in the boundaries of this method.
Off the top of my head, I'd write a set_bit and get_bit function that could take an array of bytes and a bit offset in the array, and use some bit-twiddling to set/get the appropriate bit in the array. Something like this (in C, but hopefully you get the idea):
// sets the n-th bit in |bytes|. num_bytes is the number of bytes in the array
// result is 0 on success, non-zero on failure (offset out-of-bounds)
int set_bit(char* bytes, unsigned long num_bytes, unsigned long offset)
{
// make sure offset is valid
if(offset < 0 || offset > (num_bytes<<3)-1) { return -1; }
//set the right bit
bytes[offset >> 3] |= (1 << (offset & 0x7));
return 0; //success
}
//gets the n-th bit in |bytes|. num_bytes is the number of bytes in the array
// returns (-1) on error, 0 if bit is "off", positive number if "on"
int get_bit(char* bytes, unsigned long num_bytes, unsigned long offset)
{
// make sure offset is valid
if(offset < 0 || offset > (num_bytes<<3)-1) { return -1; }
//get the right bit
return (bytes[offset >> 3] & (1 << (offset & 0x7));
}
I've used bit masks in filesystem code where the bit mask is many times bigger than a machine word. think of it like an "array of booleans";
(journalling masks in flash memory if you want to know)
many compilers know how to do this for you. Adda bit of OO code to have types that operate senibly and then your code starts looking like it's intent, not some bit-banging.
My 2 cents.
With a 64-bit integer, you can store values up to 2^64-1, 64 is only 2^6. So yes, there is a limit, but if you need more than 64-its worth of flags, I'd be very interested to know what they were all doing :)
How many states so you need to potentially think about? If you have 64 potential states, the number of combinations they can exist in is the full size of a 64-bit integer.
If you need to worry about 128 flags, then a pair of bit vectors would suffice (2^64 * 2).
Addition: in Programming Pearls, there is an extended discussion of using a bit array of length 10^7, implemented in integers (for holding used 800 numbers) - it's very fast, and very appropriate for the task described in that chapter.
Some languages ( I believe perl does, not sure ) permit bitwise arithmetic on strings. Giving you a much greater effective range. ( (strlen * 8bit chars ) combinations )
However, I wouldn't use a single value for superimposition of more than one /type/ of data. The basic r/w/x triplet of 3-bit ints would probably be the upper "practical" limit, not for space efficiency reasons, but for practical development reasons.
( Php uses this system to control its error-messages, and I have already found that its a bit over-the-top when you have to define values where php's constants are not resident and you have to generate the integer by hand, and to be honest, if chmod didn't support the 'ugo+rwx' style syntax I'd never want to use it because i can never remember the magic numbers )
The instant you have to crack open a constants table to debug code you know you've gone too far.
Old thread, but it's worth mentioning that there are cases requiring bloated bit masks, e.g., molecular fingerprints, which are often generated as 1024-bit arrays which we have packed in 32 bigint fields (SQL Server not supporting UInt32). Bit wise operations work fine - until your table starts to grow and you realize the sluggishness of separate function calls. The binary data type would work, were it not for T-SQL's ban on bitwise operators having two binary operands.
For example .NET uses array of integers as an internal storage for their BitArray class.
Practically there's no other way around.
That being said, in SQL you will need more than one column (or use the BLOBS) to store all the states.
You tagged this question SQL, so I think you need to consult with the documentation for your database to find the size of an integer. Then subtract one bit for the sign, just to be safe.
Edit: Your comment says you're using MySQL. The documentation for MySQL 5.0 Numeric Types states that the maximum size of a NUMERIC is 64 or 65 digits. That's 212 bits for 64 digits.
Remember that your language of choice has to be able to work with those digits, so you may be limited to a 64-bit integer anyway.