C 中 64 位变量的按位移位 >= 32 操作

Question

I'm trying to make shift operation that is longer than 32 for 64 bit variable. Can someone tell what is messed up with my code.

Processor architechture is AMD64 and development environments are Visual studio Community 2015 and DevC++

My code looks like this:

#include <stdio.h>
#include <stdint.h>
int main(int argc, char *argv[])
{
    uint64_t x = ~0;
    printf("x is %8x \n", x);
    x = ~((uint64_t)(1) << 31);
    printf("x is %8x \n", x);
    x = ~((uint64_t)(1) << 32);
    printf("x is %8x \n", x);
    return 1;
}

Output is:

x is ffffffff
x is 7fffffff
x is ffffffff

I've been thinking on this for whole morning now.. I'm really a beginner with C on bigger than embedded 8-bit architechtures :)

-Codester

Answer 1

• uint64_t x = ~0;

  This code won't work if int is 32 bits. If you want to set a uint64_t to "all ones" in a portable manner, you need to do uint64_t x = ~(uint64_t)0; .

• %8x

  This is not the correct format specifier for uint64_t . You should use PRIx64 from inttypes.h . Example:

   #include <inttypes.h> printf("x is %8" PRIx64 "\\n", x);

Answer 2

Adding extra llX to printf format solved my problem:

#include <stdio.h>
#include <stdint.h>
int main(int argc, char *argv[])
{
    uint64_t x = ~0;
    printf("x is %8llX \n", x);
    x = ~((uint64_t)(1) << 31);
    printf("x is %8llX \n", x);
    x = ~((uint64_t)(1) << 32);
    printf("x is %8llX \n", x);
    return 1;
}

So many times answer is too simple :P