如何解释C程序的输出

Question

Code:

int main()
{
    unsigned int a = 0xfffffff7;
    char *b = (char *)&a;
    printf("%08x",*b);
}

the output is: fffffff7 . My machine is little-endian. Of course I know *b equals 0xf7 , but I don't know why the output of printf() is like this.

Answer 1

Since your system is small-endian, a is stored in memory as F7 FF FF FF .

b points to the first byte of a . (F7)

*b evaluates to a char . (F7)

*b is promoted to an int in order to pass it as a parameter, since it's of type char (which usually defaults to signed char ) it is sign-extended to become FFFFFFF7 .

Answer 2

 +-----------------------+
 |  F7   <--b=(char *) &a|
 |  FF                   |
 |  FF                   |
 |  FF                   |
 |                       |
 +-----------------------+


 printf("%08x",*b);

//means :

1. *b asking the value b pointer to (F7)

2. %08x is asking for hex, when printing a char as an integer type it is widened to an int before printing. (FFFFFF7 now)