Difference in Call by Reference Vs Call by pointer argument todo pre and postincrement

Question

I know how to do call by reference and call by pointer operations. But I am confused about doing both of them in pre and post-increment operation. Here the code snippet.

Call by reference with pointer arguments

void fun2(int *x) // It works like int *x=&x;
{
    ++*x;  // *x++ does not provide lvalue, so will not work here [ I think So ]
}

Call by reference with reference arguments

void fun3(int& x)
{
    ++x;  // x++; [both works fine but I can't understand why?]
}

Here's the Driver code

int main()
{
    int x=10;
    cout<<x<<"\n"; //10
    fun2(&x);
    cout<<x<<"\n"; //11
    fun3(x);
    cout<<x<<"\n"; //12
    return 0;
}

Why in fun2() *x++ gives the output 10 instead of giving 11 but ++*x works fine and in fun3() both x++ and ++x works fine?

Sorry while asking here I mistyped it. No, while writing code I write it fine void fun3(int &x). But don't know why call-by-pointer *x++ not changing the value of x — Sumon Dey, Jan 12 '20 at 04:58
See the linked duplicate. It is about C (I am still trying to find a good C++ duplicate.) but it applies exactly the same to C++. — walnut, Jan 12 '20 at 05:00
See this question https://stackoverflow.com/questions/18481740/pointer-expressions-ptr-ptr-and-ptr — samnaction, Jan 12 '20 at 05:02
Does this answer your question? [Pointer expressions: \*ptr++, \*++ptr and ++\*ptr](https://stackoverflow.com/questions/18481740/pointer-expressions-ptr-ptr-and-ptr) — samnaction, Jan 12 '20 at 05:07

score 0 · Accepted Answer · answered Jan 12 '20 at 05:18

In fun2(), *x++ does not do what you think it does.

Per operator precedence, the ++ postfix increment operator has a higher precedence than the * dereference operator. As such, the statement *x++ is treated as *(x++), incrementing the pointer before then dereferencing it. The value being pointed at is untouched.

For what you are expecting, you need to use (*x)++ instead, dereferencing the pointer and then incrementing the value being pointed at.

You don't have the same problem with ++*x because the ++ prefix increment operator has the same precedence as the * dereference operator but is evaluated first, so ++*x is treated as ++(*x).

In func3(), there is no pointer to dereference. x is a reference to the caller's int (x in main()), and so both increment operators are being invoked directly on that int.

Difference in Call by Reference Vs Call by pointer argument todo pre and postincrement

1 Answers1