Priority operators in C

Question

I found this text (source: https://education.cppinstitute.org/) and I'm trying to understand the second instruction.

Can you answer the question of what distinguishes these two instructions?

 c = *p++;

and

 c = (*p)++;

We can explain: the first assignment is as if the following two disjoint instructions have been performed;

 c = *p;
 p++;

In other words, the character pointed to by p is copied to the c variable; then, p is increased and points to the next element of the array.

The second assignment is performed as follows:

 c = *p;
 string[1]++;

The p pointer is not changed and still points to the second element of the array, and only this element is increased by 1.

What I don't understand is why it is not incremented when the = operator has less priority than the ++ operator.

Answer 1

With respect to this statement expression

c = (*p)++;

, you say

What i dont understand is why [p] is not incremented when the = operator has less priority than the ++ operator.

There is a very simple explanation: p is not incremented as a result of evaluating that expression because it is not the operand of the ++ operator.

That is in part exactly because the = operator has lower precedence: because the precedence of = is so low, the operand of ++ is the expression (*p) rather than the expression c = (*p). Note in particular that p itself is not even plausibly in the running to be the operand in that case, unlike in the variation without parentheses.

Moving on, the expression (*p) designates the thing to which p points, just as *p all alone would do. Context suggests that at that time, that's the same thing designated by string[1]. That is what gets incremented, just as the text says, and its value prior to the increment is the result of the postfix ++ operation.

Answer 2

When the ++ is following a variable, the variable is incremented after it has been used.

So when you have

y = x++;

x is incremented after y gets the value of x.

This is how it works for the -- operator also.

Answer 3

What I don't understand is why it is not incremented when the = operator has less priority than the ++ operator.

The value for example of the expression

x++

is the value of x before incrementing.

So if you'll write

y = x++;

then the variable y gets the value of x before its incrementing.

From the C Standard (6.5.2.4 Postfix increment and decrement operators)

2 The result of the postfix ++ operator is the value of the operand. As a side effect, the value of the operand object is incremented (that is, the value 1 of the appropriate type is added to it). ... The value computation of the result is sequenced before the side effect of updating the stored value of the operand. ...

If instead of the expression

c = (*p)++;

you'll write

c = ++(*p);

then you get the expected by you result. This demonstrates the difference between the postfix increment operator ++ and the prefix (unary) increment operator ++.