References as function arguments?

Question

I have a trouble with references. Consider this code:

void pseudo_increase(int a){a++;}  
int main(){  
    int a = 0;
    //..
    pseudo_increase(a);
    //..
}

Here, the value of variable a will not increase as a clone or copy of it is passed and not variable itself.
Now let us consider an another example:

void true_increase(int& a){a++;}
int main(){  
    int a = 0;
    //..
    true_increase(a);
    //..
}

Here it is said value of a will increase - but why?

When true_increase(a) is called, a copy of a will be passed. It will be a different variable. Hence &a will be different from true address of a. So how is the value of a increased?

Correct me where I am wrong.

Answer 1

Consider the following example:

int a = 1;
int &b = a;
b = 2; // this will set a to 2
printf("a = %d\n", a); //output: a = 2

Here b can be treated like an alias for a. Whatever you assign to b, will be assigned to a as well (because b is a reference to a). Passing a parameter by reference is no different:

void foo(int &b)
{
   b = 2;
}

int main()
{
    int a = 1;
    foo(a);
    printf("a = %d\n", a); //output: a = 2
    return 0;
}

Answer 2

When true_increase(a) is called , copy of 'a' will be passed

That's where you're wrong. A reference to a will be made. That's what the & is for next to the parameter type. Any operation that happens to a reference is applied to the referent.