Array increment with pointers while leaving pointer address unchanged

Question

I am studying for a midterm and one of the questions is Which of the following statements will increment a value in the array, and leave the pointer address unchanged? Circle 0 or more.

int array[10];
int * ptr = array;

1) *ptr++;

2)(*ptr)++;

3)*++ptr;

4)++*ptr;

I have seen 1 and 2 used before and I believe it's just getting the de-referenced values without changing the pointer. But I was surprised to find 3 and 4 is actually valid and I am confused as to how to understand or even read it? Is it the same thing? I believe the answer is that all 4 of them are valid.

Knowing the precedence of operators will help you get started. — Captain Obvlious
– Captain Obvlious, Commented Oct 21, 2013 at 1:26

Charlie · Accepted Answer · 2013-10-21 01:34:51Z

2

http://en.cppreference.com/w/cpp/language/operator_precedence will be helpful to you. Read through it and put the parenthesis in to the examples based on the precedence of the operators and everything will hopefully make some sense.

The first would become *(ptr++) for instance.

answered Oct 21, 2013 at 1:34

Charlie

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resigned · Accepted Answer · 2013-10-28 22:19:35Z

Every C++ expression yields a value (main effect). There can also be side effects (as in the examples above) which occur before (e.g. pre-increment in examples 3 and 4) or after (e.g. post-increment in examples 1 and 2) the main effect. So in your examples:

1) side effect occurs last:

main effect: dereference ptr to get array[0]

side effect: post increment ptr by 4 bytes (on a 32 bit machine)

2) side effect occurs last:

main effect: dereference ptr to get array[0]

side effect: post increment *ptr (=array[0]) by 1

3) side effect occurs first:

side effect: pre increment ptr by 4 bytes (on a 32 bit machine)

main effect: dereference *ptr to get array[0]

4) side effect occurs first:

side effect: pre increment *ptr (= array[0]) by 1

main effect: dereference *ptr to get array[0]

Notice the main effect is obtained by ignoring the side effect operator (++) and is therefore the same in all cases. The main effect is often combined with assignment as in:

int y = *++ptr;    // ptr is incrented to point at array[1] and then y becomes equal to *ptr (= array[1])

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