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Why is there sometimes the & operator inside the parameters of a function?
4 ответов
+ 3
void setA(int &a)
{
a = 6;
}
int main()
{
int a = 5;
cout << "a = " << a << "\n";
setA(a);
cout << "a = " << a << "\n";
}
Here is an example. When we pass a through setA, we pass it by reference, because the arguements of the function is &a (the memory address of a). We then set a to 6. And even if we didn't return a from setA, the a in the main evaluates to 6.
If we didn't include the & operator in the arguements, then a = 6 will only be true inside the function, whereas in main, a will still be 5.
Sorry if this is confusing, my explaination is a bit shit.
+ 2
The andersand (&) operator returns the memory address of an object or a variable.
When used passing variables in a function e.g.
int func(int &a, int &b)
this is called pass by reference, and allows you to modify the original variable that was passed through. When using pass by value (without the &), you are creating temporary variables inside that function, which do not effect the original variables.
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What do you mean by original variables?
0
It makes sense. Great explanation! So if I wanted to print 5 without removing the &, all I'd have to do is print *p in the cout?