How to use the return value of a function with parameters inside another function

Question

All these functions are outside the int main():

int func1(int x) {

    int v1 = 6 * x;
    return v1; // the input argument will be 2, so v1 = 12
}

int func2()         


        

Answer 1

EDITED ANWSER

If you need the variable to be a user input, then try this:

int main() {
    int x = 0;
    cin >> x;
    int y = func1(x); // send user input to func1, then to func2
    int e = func2(y); // return result of above to int e
    cout << e << "\n";
}

And then your functions should look like this:

int func1(int x) {
    int v1 = 6 * x;
    return v1; // the input argument will be 2, so v1 = 12
}

int func2(int x){
    int v2 = func1(x) / 4; // It's suppose to be 12 / 4
 }

Answer 2

You get this error because you have to give the function all the arguments it needs. You are not giving an argument to a function so it will not work. sum() doesnt receive any arguments so it has to be called like this

sum()

func1() takes one argument so either change it to a function that takes no arguments

int func1() {

    int v1 = 6 * 2;
    return v1; // the input argument will be 2, so v1 = 12
}

but it shows that you havent thought about what you are doing and its kinda useless

or call it like this without changing the function

int v2 = func1(2) / 4;

EDIT
So you can change func2 to also get one argument and pass it to func1

int func2(int x){
    int v2 = func1(x) / 4;
 }

and in main do this

int x;
cin>>x;
func2(x);

also small thing your func2() should return int but you dont have a return statement

Answer 3

I may be misunderstanding your question, but this works for me:

$ cat test.c
#include <stdio.h>
#include <stdlib.h>

int func1(int x)
{
    int v1 = 6 * x;
    return v1; // the input argument will be 2, so v1 = 12
}

int func2(int x)
{
  int v2 = func1(x) / 4; // It's suppose to be 12 / 4
  return v2;
}

int main( int argc, char *argv[] )
{
  printf( "%i\n", func2( 2 ) );
  return 0;
}

$ gcc test.c

$ ./a.out
3

Answer 4

Could you clarify your question? From my understanding you're doing what you want correctly.

double add_5(double x) {
    return x + 5;
}
double nested_method(double x) {
    double myVariable = add_5(x); //my variable = x + 5
    //other code//
    return myVariable;
}
int main() {
double myInput = 123;
    std::cout << nested_method(myInput) << std::endl; //output should = 128
}

If a method asks for 'n' parameters to call the method you must give it each of those 'n' parameters (of the apropriate type).

ignoring overloads / default values etc

Answer 5

The function needs an argument, so pass it an argument.

It's as simple as that.

Answer 6

Should be:

int func2(){
    int v2 = func1(2) / 4; // It's suppose to be 12 / 4
    return v2;
}

Whenever you call a function, you have to pass the parameters it requires.

You get "too few arguments" because you're not passing the right amount of arguments!