Post-increment in a while loop

Question

The following code confused me a bit:

char * strcpy(char * p, const char * q) {
  while (*p++=*q++);
  //return
}

This is a stripped down implementation of strcpy function. From this code, we see that pointer p and q are incremented then dereferenced and q is assigned to p until the \\0 char has been reached.

I would like someone to explain the first iteration of the while loop.

Answer 1

Because the ++ is after the variables, they aren't incremented until after the expression is evaluated. That's why it's the post-increment operator; the pre-increment is prefixed ( ++p ). *++p would write to the second spot, *p++ writes to the first.

Answer 2

p++ is post-incrementing the pointer p . So the current value of p is operated upon by the deference operator * before p is incremented.

Your reasoning would've been correct if the while loop was written as follows:

while (*++p=*++q);

In this case the increment would happen before dereferencing.

Answer 3

No, the increment happens after the assignment.

If it were *(++p) , the pointer p would be incremented and after that assigned.

Answer 4

The expressions x++ and ++x have both a result (value) and a side effect .

The result of the expression x++ is the current value of x . The side effect is that the contents of x are incremented by 1.

The result of the expression ++x is the current value of x plus 1. The side effect is the same as above.

Note that the side effect doesn't have to be applied immediately after the expression is evaluated; it only has to be applied before the next sequence point. For example, given the code

x = 1;
y = 2;
z = ++x + y++;

there's no guarantee that the contents of x will be modified before the expression y++ is evaluated, or even before the result of ++x + y++ is assigned to z (neither the = nor + operators introduce a sequence point). The expression ++x evaluates to 2, but it's possible that the variable x may not contain the value 2 until after z has been assigned.

It's important to remember that the behavior of expressions like x++ + x++ is explicitly undefined by the language standard; there's no (good) way to predict what the result of the expression will be, or what value x will contain after it's been evaluated.

Postfix operators have a higher precedence than unary operators, so expressions like *p++ are parsed as *(p++) (ie, you're applying the * operator to the result of the expression p++ ). Again, the result of the expression p++ is the current value of p , so while (*p++=*q++); doesn't skip the first element.

Note that the operand to the autoincrement/decrement operators must be an lvalue (essentially, an expression that refers to a memory location such that the memory can be read or modified). The result of the expression x++ or ++x is not an lvalue, so you can't write things like ++x++ or (x++)++ or ++(++x) . You could write something like ++(*p++) ( p++ is not an lvalue, but *p++ is), although that would probably get you slapped by anyone reading your code.

Answer 5

The right hand side of the expression (*q++) will be evaluated prior to *p++, and both will only be incremented after the assignment takes place.

Read the statement right to left and remember post-increment (q++ instead of ++q) happens after everything else in the line is resolved.

*q --> dereference q
=  --> assign the value
*p --> to p

increment both.

Do this until qp taking q's element = 0 which is when it reaches the null terminator.

Answer 6

This is a stripped implementation of strcpy function. From this code, we see that pointer p and q are increamented, than dereferenced and q is assigned to p until \\0 char has been reached.

It happens the other way around. The value at *p is set to *q , then both pointers are incremented.

When you have int foo = bar++ the increment happens after foo has been set. To have it happen first you would do int foo = ++bar

Answer 7

The value of q++ is q
The value of ++q is q+1

Answer 8

The while loop's condition is performing post-increment. Equivalently:

while (true) {
    char* old_p = p;
    const char* old_q = q;

    ++p; // or p++;
    ++q; // or q++;

    *old_p = *old_q;
    if (*old_p == '\0')
        break;
}