How can I print maximum value of an unsigned integer?

Question

I want to print the maximum value of the unsigned integer which is of 4 bytes.

#include \"stdafx.h\"
#include \"conio.h\"

int _tmain(int argc, _TCHAR* argv[         


        

Answer 1

Use %u as the format string to print unsigned int, %lu for unsigned long, and %hu for unsigned short.

Answer 2

Curious and easy way to do this is just

printf("%lu",-1);

it prints whatever the biggest unsigned in your computer:)

Answer 3

Here is the code:

#include <stdio.h>

int main(void) {
    unsigned int a = 0;
    printf("%u", --a);
    return 0;
}

Output:

4294967295

How it works is that 0 is the minimum value for unsigned int and when you further decrease that value by 1 it wraps around and moves to the highest value.

Answer 4

There is the macro defined in <limits.h>: UINT_MAX.

Answer 5

The %d format treats its argument as a signed int. Use %u instead.

But a better way to get the maximum value of type unsigned int is to use the UINT_MAX macro. You'll need

#include <limits.h>

to make it visible.

You can also compute the maximum value of an unsigned type by converting the value -1 to the type.

#include <limits.h>
#include <stdio.h>
int main(void) {
    unsigned int max = -1;
    printf("UINT_MAX = %u = 0x%x\n", UINT_MAX, UINT_MAX);
    printf("max      = %u = 0x%x\n", max, max);
    return 0;
}

Note that the UINT_MAX isn't necessarily 0xffffffff. It is if unsigned int happens to be 32 bits, but it could be as small as 16 bits; it's 64 bits on a few systems.

Answer 6

Use %u as the printf format string.