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.
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.
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.
No, the increment happens after the assignment.
If it were *(++p), the pointer p would be incremented and after that assigned.
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++) (i.e., 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.
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 q p taking q's element = 0 which is when it reaches the null terminator.
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
The value of q++ is q
The value of ++q is q+1
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;
}