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Category:IntergraphQ:
After using * operator for refference, what's next?
I'm studying pointers and found out that if you put * operator after the name of a variable, it means that it is a reference to that variable.
Now I have a question about when I use that operator.
I.E:
int main() {
int *p = 0;
*p = 1;
}
Is that the same as doing:
int main() {
int *p = 0;
p = &p;
}
I.E:
*p = 1;
And what's that 0?
In the first case, I set the value of *p, right?
And in the second case, I set the value of p, right?
A:
First off, the behaviour of a pointer and the value of its expression is determined by the language standard, not by how you interpret the expression. But if you're asking about the interpretation of the expression, then that is determined by the language standard.
In C++, pointer arithmetic and assigning of pointers behave in a consistent way. That is, you are allowed to perform pointer arithmetic and assign pointers as long as you have the relevant type. For example, int * p = 0; p = p + 1; is allowed, but int * p = 0; p = &p; is not.
However, C (and in C++, a language that is based on C) does not require that the two different cases be equivalent. In C, pointer arithmetic is slightly different than it is in C++: it involves subtraction, and it can't be combined with pointer assignment, because it's not a context where you're allowed to modify the expression.
So, in the first case,
int *p = 0;
*p = 1;