How i understand this.

Question

1 void myFunc(int *x) {
2 *x = 100;
3 }

4
5 int main() {
6 int var = 20;
7 myFunc(&var);
8 cout << var;
9 }
10 // Outputs 100
in line 7 the address of var is passed to line 1 parameter which implies that int *x is assigned the value of var.
but in line 2 *x is given another value that is 100. so *x has two values now. that is 100 and address of var.
plis make correction if i m wrong

Answer 1

I don't blame you for being confused - this is a complicated concept! Think of it this way - & and * work together. & fetches the address of a variable. * points to that variable when you give it an address you got with &.

You got lines 7 and 1 correct - var's address is being passed to myFunc(), and stored in a pointer that can use that address. It's like you told myFunc() "var is here! You can use it and change it!"

When you assign 100 to *x, what you're really doing is assigning 100 to var, which *x is pointing at. And when myFunc() ends, var will still be 100.

The only way to reassign a pointer itself is to give it a new address to point at with &.

Answer 2

https://code.sololearn.com/cIxFf27nurVi ; it may help just check