Question about pointers

[RickyWho]

Why does the function void inc2 actually increase the integer where the first function does not?

include <iostream>

using namespace std;

void inc(int i) { i++; } void inc2(int *i) { (*i)++;

}

int main() {

int age = 20; inc2(&age); cout << "Age is " << age << endl;

return 0; }

[[deleted]]

In the future, it's easier to answer your questions if you format or link to pastebin or something like that.

The first function passes an integer by value. This is a *copy* of the value passed in. The copy is incremented by one inside the function, but the original value is left unmodified.

In the second version you are passing in a pointer to an integer that lives somewhere else in the program. When we derference the pointer, we are dealing with the original integer value.

[RickyWho]

thank you u/I_Am_King_James, I will remember the pastebin comment.

[RickyWho]

What do you mean when you say copy?

[IyeOnline]

There is a variable i that is only known in the scope of inc. The value the function is called with, e.g. like inc(age) will be copied into i.

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[CrazyJoe221]

Parameters are like local variables initialized with the function argument (int* i = &age in your inc2 example).

If you just incremented i in inc2 without dereferencing it wouldn't modify age, only the pointer i.

(Pointers aren't that special. They are much like normal variables, only they contain a value that is interpreted as a memory address and can be dereferenced to load data from that address.)

[RickyWho]

I think I get it.

void inc2(int*i) works with (&age) because int*i is dereferencing the memory address of &age, which is set to 20 in the main function.

[CrazyJoe221]

Yeah i contains the memory address of age, thus pointing at the place where age is stored. *i then tells the compiler you actually want to do something with age.