Why ++(*p) changes the pointer value?

Question

If I have this code:

int A[5] = { 2, 1, 3, 55 };
int *p = A;
cout << ++(*p);

the result is 3 and the value of the first position of A is 3 also, why?

I mean, by hierarchy of operators () is more hierarchical than ++, then we would need operate *p first:

++(*p) => ++(2) => 3 

with any change in A vector?

Answer 1

*p is not just "2", it's an lvalue, i.e. this "2" has a well-defined location. The value at this location is modified by the ++ operator - by definition of the ++ operator.

If you don't want to modify the value, use + 1 instead: *p + 1.

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In C/C++, lvalue is a value with a defined location in memory. This value can be changed - by an assignment, incremented, decrement. For example,

int x = 0;
x = 1; // ok, x is an lvalue, assignment changes the value from 0 to 1

int *p = &x;
*p = 2; // ok, *p is an lvalue, assignment changes the value from 1 to 2

In contrast, an rvalue is a value without a defined location - for example, a result of an arithmetic operation. This value can't be assigned, incremented or decremented (it doesn't mean it can't be used in a larger expression).

For example,

int x = 0, y = 1;
(x + y) = 3; // compilation error, (x + y) is an rvalue

2++; // compilation error, 2 is an rvalue

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Here's a pretty simple article explaining lvalues / rvalues: https://eli.thegreenplace.net/2011/12/15/understanding-lvalues-and-rvalues-in-c-and-c

Answer 2

The ++(*p) is the same as ++p[0] and ++A[0] All change the first element of the array.

Why ++(*p) changes the pointer value

It does not. The pointer value is value kept in the p. It is called "address" or "reference". It does not change.