Post-increment on a dereferenced pointer?

Question

Trying to understand the behaviour of pointers in C, I was a little surprised by the following (example code below):

#include <stdio.h>  void add_one_v1(int *our_var_ptr) {     *our_var_ptr = *our_var_ptr +1; }  void add_one_v2(int *our_var_ptr) {     *our_var_ptr++; }  int main() {     int testvar;      testvar = 63;     add_one_v1(&(testvar));         /* Try first version of the function */     printf("%d\n", testvar);        /* Prints out 64                     */     printf("@ %p\n\n", &(testvar));      testvar = 63;     add_one_v2(&(testvar));         /* Try first version of the function */     printf("%d\n", testvar);        /* Prints 63 ?                       */     printf("@ %p\n", &(testvar));   /* Address remains identical         */ } 

Output:

64 @ 0xbf84c6b0  63 @ 0xbf84c6b0 

What exactly does the *our_var_ptr++ statement in the second function (add_one_v2) do since it's clearly not the same as *our_var_ptr = *our_var_ptr +1?

Answer 1

This is one of those little gotcha's that make C and C++ so much fun. If you want to bend your brain, figure out this one:

while (*dst++ = *src++) ; 

It's a string copy. The pointers keep getting incremented until a character with a value of zero is copied. Once you know why this trick works, you'll never forget how ++ works on pointers again.

P.S. You can always override the operator order with parentheses. The following will increment the value pointed at, rather than the pointer itself:

(*our_var_ptr)++; 

Answer 2

Due to operator precedence rules and the fact that ++ is a postfix operator, add_one_v2() does dereference the pointer, but the ++ is actually being applied to the pointer itself. However, remember that C always uses pass-by-value: add_one_v2() is incrementing its local copy of the pointer, which will have no effect whatsoever on the value stored at that address.

As a test, replace add_one_v2() with these bits of code and see how the output is affected:

void add_one_v2(int *our_var_ptr) {     (*our_var_ptr)++;  // Now stores 64 }  void add_one_v2(int *our_var_ptr) {     *(our_var_ptr++);  // Increments the pointer, but this is a local                        // copy of the pointer, so it doesn't do anything. }