我无法弄清楚下面的代码是如何产生ffffffaa的输出的，请帮助我理解吗？

Question

Hello i am unable to figure out how the below code produce output as ffffffaa please help me understand

#include<stdio.h>
 int main()
{
  int a=0xaaaaaaaa;
  char *p=(char*)&a;
  printf("%x\n",*p);
}  

Answer 1

Variadic functions like printf perform the default argument promotion on its trailing arguments. A char is promoted to an int and when char is signed on the platform a sign extension is performed.

To avoid the sign extension use unsigned char types or cast *p to unsigned char in the printf call of your program.

Answer 2

I'm unable to figure out how the code above can produce ffffffaa because it cannot do that with a "%d" specifier ;-).

You probably mistyped you code sample and it should read:

int main() {
    int a=0xaaaaaaaa;
    char *p=(char*)&a;
    printf("%x\n",*p);
}  

For this code to produce ffffffaa , char must be 8 bit signed, and int must be stored in little endian order or have a size of at most 32 bits.

&a is the address of the first byte of a in memory. Dereferencing this address as a char * loads one byte and sign extends it as an int before passing it to printf as this function is variadic. Extra arguments to variatic functions with types smaller than int are promoted to int and passed as such, and floating point types smaller than double are promoted to double and passed as such. printf receives an int but prints the hexadecimal representation of an unsigned int for the format specifier %x . It's a good thing int and unsigned int are passed the same way ;-)

Since the value 0xaaaaaaaa has the same representation in little endian and big endian order, your code will not depend on endianness as long as int is 32 bits or less. But on a machine with 64 bit int in big endian order (such as some late PowerPCs), the actual memory layout of a would be 00 00 00 00 aa aa aa aa . Your code would then produce 0 output.

You can learn about endianness with this code sample:

#include <stdio.h>    
int main(void) {
    int i, a = 0x12345678;
    unsigned char *p = (unsigned char *)&a;
    printf("Memory layout for %x is:", a);
    for (i = 0; i < (int)sizeof(a); i++)
        printf(" %02x", p[i]);
    printf("\n");
}

Different compilers may produce different output on the same machine if they target a different architecture (32 vs 64 bits).