Thread: largest integer in c

It is not the size of the integer that is the problem, it's the size of a stack-allocated array - the variable arr is about 2MB large, which probably blows the stack.

Either use a global/static variable, or use dynyamic memory allocation.

And to answer your actual question, the largest integer value is about 2147 million.

--
Mats

It's 0xFFFFFFFF. But the problem isn't that. You are creating a local variable on the stack with that value, which is too big for the stack, hence the crashing.

Originally Posted by mpi_beginner

Hi,

I was trying to execute the following code which was given in an instruction.
However, the program crashes every time I try to execute it.

Code:

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[])
{
  int nb = 4194304;
  
  int arr[nb];
  int i=0;
  for(i=0; i<nb; i++)
           arr[i] = i;
  
  system("PAUSE");	
  return 0;
}

What is the largest int value that can be represented?

it depends on the size of the stack - use maloc for dynamic array allocation - at least you can check the failure to allocate array and exit without crash

Originally Posted by Elysia

It's 0xFFFFFFFF.

Unsigned yes. int is signed, so it is 0x7FFFFFFFF.

--
Mats

Oh, and I misread the number of digits, your variable is about 16MB, which is certainly beyond the scope of normal stack sizes.

--
Mats

Originally Posted by mpi_beginner

Thanks for the info. It works now with dynamic allocation.

Just a question out of interest: Isnt it more than 2MB?

Yes, I realized that and corrected it about 20 minutes ago in the last post before yours. It's a bit more than 16MB.

--
Mats

Originally Posted by matsp

Unsigned yes. int is signed, so it is 0x7FFFFFFFF.

--
Mats

True, but I always think of that 0xFFFFFFFF is the biggest hex number you can stuff into an integer. Obviously, if signed, it will just represent -1, but still...

BTW: the minimum and maximum values specific data types can store are defined as constants in limits.h. Some of these constants are:

• INT_MAX: max of (signed) int
• INT_MIN: min of (signed) int
• UINT_MAX: max of unsigned int
• etc.

Another way to figure it out: signed values stored in two's complement representation have a range of -2^(n-1) to +2^(n-1)-1, where n is the number of bits in the number. If you use sizeof, it will tell you the number of bytes in a number; multiply this by 8 (or CHAR_BIT in limits.h if you want to be picky) and you get the number of bits, which you can use in those formulae.

Even if the stack was big enough, that's not standard C syntax. You are not allowed to declare an array using a variable to indicate the size. Your compiler apparently allows it, but it's wrong.

Originally Posted by brewbuck

You are not allowed to declare an array using a variable to indicate the size.

True for C90, false for C99. The current standard allows variable length arrays. You can make gcc emit a warning which states this explicitly:

Code:

int main(int argc)
{
	int x[argc];
	return 0;
}

Tell gcc to compile with "-std=c89 -pedantic" and it will reply:

Code:

warning: ISO C90 forbids variable length array ‘x’

Greets,
Philip