*a++ = *b++（这是什么意思，它是如何工作的）

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.