*a++ = *b++ (what does it mean, how it works) - c

Question

What would be the values after performing this operation?
#include <stdio.h>
int main() {
int *a = 0;
int *b = 3;
*a++ = *b++;
printf("%d", a);
printf("%d", b);
return 0;
}
The code above gives me a segmentation fault.

Answer 1

*a++ = *b++ (what does it mean, how it works)
*a++ = *b++;
means
*(a++) = *(b++);
x++ increments x and returns the original value. So the following is equivalent:
*a = *b; // Copy the `int` to which `b` points into the `int` to which `a` points.
a = a + 1; // Make `a` point to the following `int`.
b = b + 1; // Make `b` point to the following `int`.
Before: After:
a a
+----------+ +----------+ +----------+ +----------+
| ---------->| x | | ------+ | p |
+----------+ +----------+ +----------+ | +----------+
| y | +--->| y |
+----------+ +----------+
| | | |
b b
+----------+ +----------+ +----------+ +----------+
| ---------->| p | | ------+ | p |
+----------+ +----------+ +----------+ | +----------+
| q | +--->| q |
+----------+ +----------+
| | | |
The code above gives me a segmentation fault.
You assigned garbage to a and b. 0 as a pointer is the NULL pointer, and 3 isn't a valid pointer.

Answer 2

Given
#include <stdio.h>
int main() {
int *a = 0;
int *b = 3;
*a++ = *b++;
printf("%d", a);
printf("%d", b);
return 0;
}
the printed values can not be predicted as the code invokes undefined behavior in multiple ways.
First, both *a and *b invoke undefined behavior by dereferencing invalid pointers - a is initialized to a null pointer value, and b is initialized to point to address 3, which is almost certainly invalid also.
Second,printf("%d", a); invokes undefined behavior by trying to print an int * variable with the %d format specifier for int. The proper code would be
printf("%p", ( void * ) a);
It's not clear what the currently-posted code is supposed to do.