if statement condition.

[yj1214]

I've noticed that in many programming languages including C++, whenever you put a function as a condition for the if statement, it calls that function.

bool doSomething(){
    cout << "Hello" << endl;
}

int main(){
    if(doSomething()){
        cout << "World!" << endl;
}

this code will print "Hello" to the screen. If you put return true inside the doSomething function, it will print "World!". Does anyone know why this works?

[Albatross]

Because the compiler spliced in a "return false" at the end of doSomething since you didn't specify a return value yourself.

-Albatross

[yj1214]

No, that's not what i'm saying. It was just a thing that I was wondering. (why does if statement calls whatever inside the bracket ex. why does if statement calls doSomething?). But anyway thanks.

[TarikNeaj]

That is exactly what you are saying. And @Albatross answered your question perfectly.

[keskiverto]

You are wondering, why we are not limited to writing either if ( true ) or if ( false ). That would be rather limiting, wouldn't it?

What language syntax has is:
[output]IF ( expression )[/code]
The expression is evaluated during runtime and the result of evaluation is either true or false. That makes sense, doesn't it?

Now, lets take a simple example:
if ( a < b )
That isn't a function call.




... or is it?

Lets reveal a bit more:

bool operator< ( const Foo &, const Foo & ); // function declaration

// code
Foo a = ...;
Foo b = ...;
if ( a < b )

What did look like an innocent "is a less than b" expression, is suddenly a function. Just to be sure, lets write that last line again, but avoiding the operator syntax:
if ( operator< ( a, b ) )
The condition expression contains a function call. The expression can be evaluated only by calling the function.

Does that answer your question?


Then the lazy fellow:

#include <iostream>
using std::cout;

bool foo() {
  cout << "Hello ";
  return false;
}

bool bar() {
  cout << "world ";
  return true;
}

int main() {
  if( foo() && bar() ) {
    cout << "Dolly!\n";
  }
  cout << '\n';
  return 0;
}

[Little Bobby Tables]

can i just take a moment to say thats whats awesome about lisp(s)? you realize everything is a function, such as (+ 3 4)

[Gamer2015]

you realize everything is a function

Isn't that the case for every language?
Or is the awesome thing about lisp that you do realize it?

[Little Bobby Tables]

Isn't that the case for every language?

not neccesarily. for example, in racket:
(+ 3 4)
and in c++:
3 + 4
while the result is exactly the same, operators have a few differences from functions. namely operator precedence. in lisp precedence is very simple, since everything is actually a function. another example is in racket, (while) is a function, and in c++ its a loop. (not sure about (for))

[Albatross]

Also, comparison operators for basic types are very different from function calls at a technical level.

-Albatross