simple ques, dont no wat these int are doing?

Hi, im very new to c++ and am trying to learn by actually studying code etc.
I took this simple code from this tutorial website.

Code:

#include <iostream>
  
int
multiply(int x, int y)
{
return x*y;
}
int divide(int x, int
y)
{ return x/y;}
int add(int x, int y) 
{
return
x+y;
}
int subtract(int x, int y)
{ return x-y;}

using namespace std;

int
main(){ char op='c'; int x, y;
while(op!='e') {

cout<<"What operation would you like to perform: add(+),subtract(-), divide(/),multiply(*), [e]xit?"; 
cin>>op;

switch(op)
{
case '+':
cin>>x;
cin>>y;

cout<<x<<"+"<<y<<"="<<add(x,
y)<<endl;
break;
case '-':
cin>>x;
cin>>y;

cout<<x<<"-"<<y<<"="<<subtract(x,
y)<<endl;
break;
case '/':
cin>>x;
cin>>y;

cout<<x<<"/"<<y<<"="<<divide(x,
y)<<endl;
break;
case '*':
cin>>x;
cin>>y;

cout<<x<<"*"<<y<<"="<<multiply(x,
y)<<endl;
break;
case 'e':
return 0;
default:

cout<<"Sorry, try again"<<endl;
} }
return
0;}

now, i know what nearly all of this is doing. except why the intergers are there at the start.
Later ion in the program it use the case senario which i understand, then it goes on to tell the computer what to do in each case, i honestly dont no why they are there at the start.

im sorry for this proabably being a really low question but i dnt no lol

which ints are you confused about? could you be more specific because there are a lot at the top of the program, lol

are you confused about the function calls to multiply and divide?

Code:

int multiply(int x, int y)
{
return x*y;
}

int divide(int x, int y)
{ 
return x/y;
}
int add(int x, int y) 
{
return x+y;
}
int subtract(int x, int y)
{ 
return x-y;
}

thats what I am getting from your question... and what the means is when you call the function it gives you back an integer. That way is I said

Code:

int i = add( 3, 4);

it would return the integer 7.
The ints inside the ()'s mean that you are telling the program that you are always sending it integers.


Now I may have misread what you mean, so please let me know if I did misread it.

A method declaration is always like this

Code:

return type "name of function" ( as many as arguments as you like ) {

  return -whatever type you said it should return ;
}

so basically what

Code:

int multiply(int x, int y)
{
return x*y;
}

Means is that the return type of the function is an integer , the name of the function by what you call it from withing your main program or any other function or whatsoever is then multiply, between ( ) are the arguments

so the function multiply takes two arguments two integers called x and y , then as i said in the beginning it returns an int, since x and y are both int´s we can easily say return x*y;

You could also write it like this

Code:

int multiply(int x, int y)
{
int result;
result = x*y
return result;
}

Thats basically all about it ... I hope that answers your question ( didnt quite understand where you see any problem or what you dont understand ).

::edit::

Didnt see dpro already answered your question actually ( its getting late , better go to sleep )

yeah, i c now
where i saw

cout<<x<<"-"<<y...etc

i thought it was doing the work, , now ive just ran through it i no its just writing it on the screen

sorry about that

Originally Posted by jonezy

yeah, i c now
where i saw

cout<<x<<"-"<<y...etc

i thought it was doing the work, , now ive just ran through it i no its just writing it on the screen

sorry about that

Yeppers, any time you see anything in quotes, it will always be treated as a string and won't be parsed as a keyword/operator/anything else.

Also, with the way cin is overloaded using the >> operator, you can actually do this:

Code:

cin >> x >> y >> z >> w;  //This allows the input for all 4 in one line.