Question

I’d like to refactor some old C code of mine, and I was curious if I can replace all ptr++ with ptr += 1 where ptris some pointer, without changing any behavior. Here’s an example of what I mean, from K&R Section 5.3:

/* strlen: return length of string s*/
int strlen(char *s)
{
    int n;
    for (n = 0; *s != '\0'; s++)
        n++;
    return n;
}

When I replace the s++ with s += 1, I get the same results, but I’m wondering if this will be the case for all types. I also did a test for ints:

int size = 10;
int *int_array = (int*)calloc(size, sizeof(int));
for (int i = 0; i < size; i++)
    int_array[i] = i;

for (int i = 0; i < size; i++) {
    printf("*int_array = %d\n", i, *int_array);
    int_array++;
}

If I replace the line int_array++; with int_array += 1;, I get the same result.

After thinking about this some more, I realize there could be a problem if the value is used in an expression. Would it be safer I just moved the increment onto another line like so:

int a = 5;
int b = a++;

would become:

int a = 5;
int b = a;
a += 1;

Conclusion

What I thought could be a problem, incrementing pointers of different types, is not a problem. See @bdonlan’s response for the reason why.

This doesn’t mean that you can replace all x++ with x += 1 and expect the same behavior. You can, however, replace ++x with (x += 1) safely, since they are equivalent.

Answer 1

a += 1 is equivalent to ++a (C99 §6.5.3.1/2). In a line like int b = a++; this means it is not equivalent to a++; a++ would return the old value of a, while a += 1 returns the new value.

Note that if you don’t use the result of a++ (ie, you have a statement containing just a++;), then they are effectively identical.

Also, note that _all pointer arithmetic is done in increments of the pointed-to type’s size (§6.5.6/8). This means that:

ptr = ptr + x;

is equivalent to:

ptr = (ptr_type *)( (char *)ptr + x * sizeof(*ptr) );

This is the same whether you use +, ++, +=, or [] (p[x] is exactly equivalent to *(p + x); you can even do things like 4["Hello"] because of this).