I am writing a program. And in the main function "int main()" i call to a function called, lets say, "int X()". Inside "int X()" I would like to call to another function "void Y()". Any ideas how to do this? I tried, inside the X() function, doing "Y();" and "void Y();" but to no prevail. Any tips on getting this to work? if at all possible?
ex.
#include<iostream>
int X()
{
Y();
}
void Y()
{
std::cout << "Hello";
}
int main()
{
X();
system("pause");
return 0;
}
You must declare Y() before using it:
void Y();
int X()
{Y();}
You must define or declare your functions before you use them. For example:
void Y(); //this is just a declaration, you need to implement this later in the code.
int X(){
//...
Y();
//...
return someIntValue; //you will get warned if function supposed to return something does not do it.
}
OR
void Y(){
//code that Y is supposed to do...
}
int X(){
//...
Y();
//...
}
When you call the function you do not write its type anymore (to call functon Y you write: Y(arguments); and not void Y(arguments);). You write the type only when declaring or defining the function.
When the compiler reaches:
int X()
{
Y();
}
it doesn't know what Y is. You need to declare Y before X by inverting their declarations:
void Y()
{
std::cout << "Hello";
}
int X()
{
Y();
}
int main()
{
X();
system("pause");
return 0;
}
You should also provide a return value for X, otherwise a warning will pop up.
And please, don't follow the suggestion of using using namespace std;. The way you writing std::cout is just fine.
And here is the working example.
You need to declare the Y function before the X function uses it.
Write this line before the definition of X:
void Y();
You need to have cout << X(); X() it's just for void function!
X() even if is void and the problem is that Y is defined only after X). 2. There already is an accepted answer that solves the problem.
using namespace std;. You suck. Please stop giving advice to people.