有人可以解释一下这个 C 程序的输出吗？

Question

Here is the code:

int a=256;
char *x= (char *)&a;
*x++ = 1;
*x =x[0]++;
printf("a=%d\n", a);

The output is:

a=257

Answer 1

int a=256;

Initialized integer variable with value 256. On little-endian machine memory layout is:

00 01 00 00 ...

char *x= (char *)&a;

x is pointer to least significant byte of a (on little endian machine).

00 01 00 00 ...
 ^ -- x points here

*x++ = 1;

Set byte where x points to 1, then move x to next byte. Memory is:

01 01 00 00 ...
    ^-- x points here

*x =x[0]++;

Unspecified behaviour, *x and x[0] are equal ( x[0] is *(x+0) ) . Post-increment is ignored in your implementation.

UPD: actually it is not ignored but overwritten by assignment. x[0]++ increases second byte:

01 02 00 00
    ^ -- x

and then value taken before increment ( 01 ) placed to the same place by *x=

01 01 00 00

Answer 2

I'm going to take the posted code one line at a time.

int a=256;

I presume this is straightforward enough.

char *x= (char *)&a;

This sets a char pointer to point to just one byte of the multi-byte int value. This is a low-level, machine-dependent, and not necessarily meaningful operation. (Also x is a poor, unidiomatic name for a pointer variable.)

*x++ = 1;

This both sets the byte pointed to by x , and increments x to point to the next byte. In general that's a sensible operation — for example, it's how we often fill characters into a text buffer one at a time. In this context, though, it's borderline meaningless, because it's rare to move along the bytes of an int variable one at a time, setting or altering them.

*x =x[0]++;

And this line is completely meaningless. I'm not going to try to explain what it does, because I literally can't.

printf("a=%d\n", a);

Obviously this prints the value of a , although after what poor a has been through, it's hard to say what kind of bloody mess might be left of its bits and bytes.