为什么此uint32_t的值为-2147483648？

Question

Using gcc compiler, why is this uint32_t -2147483648 when I set the MSB? It's unsigned - shouldn't it be positive?

#include <stdio.h>
#include <stdint.h>

#define BIT_SET(a,b) ((a) |= (1<<(b)))

int main()
{
    uint32_t var = 0;
    BIT_SET(var, 31);

    printf("%d\n", var); //prints -2147483648

    return 0;
}

Answer 1

%d is for signed integers. What you want is %u which is for unsigned integers. printf is not smart enough to guess the type of your data.

Source

Answer 2

printf() doesn't know the type of something you pass it and has to trust you to use the correct conversion specifier . But you don't. With %d , printf() interprets your value as an int .

On a platform where unsigned int is the same as uint32_t (very common), you could use %u instead. But beware this isn't portable, eg to platforms where int only has 16 bits. The correct way to print an uint32_t looks like this (you have to include inttypes.h ):

printf("%" PRIu32 "\n", var);

See eg this inttypes.h reference

Answer 3

You are using wrong format specifier here. %d is for int type, ie, signed integer type.

To print exact width variables, you need to use PRIuN macros as format specifiers.

For unsigned type 32 bit, that would be PRIu32 . For an exhaustive list, see chapter §7.8.1, C11 .

Answer 4

Type of var is uint32_t . But you are printing it using %d which is undefined behaviour . Use PRIu32 from <inttypes.h> to print uint32_t :

printf("%"PRIu32"\n",var);