CODEWITHSUNDEEP
c++c++11
kingofthefranks
kingofthefranks

Reputation: 37

Function not returning value, but cout displays it

I've been learning C++ for a bit and tried my hand at making a simple function that returns the area of a room. The return statement doesn't output the value, however using cout I can see the result. Am I missing something here?

#include <iostream>
using namespace std;

int Area(int x, int y);

int main()
{
  int len;
  int wid;
  int area;
  cout << "Hello, enter the length and width of your room." << endl;
  cin >> len >> wid;
  cout << "The area of your room is: ";
  Area(len, wid);
  return 0;
}

int Area(int len, int wid)
{
  int answer = ( len * wid );
  return answer;
}

Upvotes: 2

Views: 1602

Answers (2)

hlt
hlt

Reputation: 6317

return doesn't print anything, and you shouldn't expect it to. All it does is return a value from the function. You can then do whatever you want with that value, including printing it, or assigning it to a variable:

// Does not print:
Area(len, wid);

// Does print:
int result = Area(len, wid);
std::cout << result << "\n";

// Does print:
std::cout << Area(len, wid) << "\n";

Imagine the chaos if every function in a massive codebase suddenly started printing its return value...

Upvotes: 7

mayank
mayank

Reputation: 543

std::cout is used to print data on screen. Functions only return values, so the Area function will return the value which is to be passed in std::ostream::operator<< function to print it. You need to write:

std::cout << Area(len, wid) << "\n";

Upvotes: 8

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