[Solved] Confused as to why increment function doesn’t actually increment variable [closed]

Question

in this code snipped you declared a function with an argument passed to it by value:

void increment(int value) {
    value++;
}

What it really does, it increments this value … inside the function, and that is it. It does not return the value nor updates any other variable.

There are several ways you can address the problem.

int increment(int value) {
   value++;
   return value;
}

int main() {
   int a = 5;
   a = increment(a);
   ...
}

In the above example new value of a will be incremented.

you can use pointers to update a variable which was declared outside the scope of the function. You can think of pointers as addresses of the variable.

void increment(int *pointer) {
   (*pointer)++;
}

int main() {
   int a = 5;
   a = increment(&a);
   ...
}

In the above example you updated value of the variable a by using its address (pointer).

you can use references. They are just proxys to a variable and in general behave as pointers. The difference is that allow a syntax appropriate for a proxy:

void increment(int &proxy) {
   proxy++;
}

int main() {
   int a = 5;
   a = increment(a);
   ...
}

In this case proxy is just a way to reference the variable outside the function scope and ++ will update the value of variable a

Accepted Answer