Although this question might be answered somewhere and I could not find it.
Below written first statement work whereas second does not? WHY?
int main() {
int x = 1, y = 2;
int *p = &++x; // First
std::cout << "*p = " << *p << std::endl;
// int *q = &x++; // Second
// std::cout << "*q = " << *p << std::endl;
}
In this declaration
int *p = &++x;
there are used two unary operators: pre-increment ++ and taking of address. Unary operators are executed from right to left. So at first the variable x is incremented and its address is assigned to the pointer p, The result of the pre-increment operator is lvalue of the incremented object.
So for example such an expression like this
++++x;
is correct.
In this declaration
int *p = &x++;
there are used the postfix operator post-increment ++ and the unary operator of taking address. Postfix operators have a higher priority relative to unary operators. SO at first the post-increment is executed. Its result is a temporary object that has the value of the variable x before incrementing. Then the operator of taking of address is executed.
However you may not take an address of a temporary object. So for this declaration the compiler will issue an error.
Opposite to the pre-increment operator such an expression like this
x++++;
is invalid.
From the C++ 17 Standard (5.3.2 Increment and decrement)
1 The operand of prefix ++ is modified by adding 1, or set to true if it is bool (this use is deprecated). The operand shall be a modifiable lvalue. The type of the operand shall be an arithmetic type or a pointer to a completely-defined object type. The result is the updated operand; it is an lvalue, and it is a bit-field if the operand is a bit-field....
And (5.2.6 Increment and decrement)
1 The value of a postfix ++ expression is the value of its operand. [ Note: the value obtained is a copy of the original value — end note ]...
In C the both operations yield a value. So in C you also may not write
++++x;