problem with if statement

[stmfc]

whats wrong with this function:

int max(int x, int y) {
    return (if (x > y) { return x; } else { return y; });
}

i get an "expected an expression" error

[Meetee]

Use this instead:

int max(int x, int y)
{
    if (x > y)
        return x; 
    else
        return y;
}

Regards

[stmfc]

Use this instead:

int max(int x, int y)
{
    if (x > y)
        return x; 
    else
        return y;
}

Regards

thanks for the reply.
i know that there are many alternative ways to perfom that functionality.
what i want to figure out is the rule which doesnt allow the above code to be considered as legal

[Meetee]

thanks for the reply.
i know that there are many alternative ways to perfom that functionality.
what i want to figure out is the rule which doesnt allow the above code to be considered as legal

Your function has return type int. In your previous code, you have taken return withing return like:

return(if(...) return x; else return y;) 

That is not proper way to return the value.

[dmjpro]

whats wrong with this function:

int max(int x, int y) {
    return (if (x > y) { return x; } else { return y; });
}

i get an "expected an expression" error

Try it.

return x>y?x:y;
 

Kind regards,
Dmjpro.

[weaknessforcats]

whats wrong with this function:


Code: ( text )
int max(int x, int y) {
return (if (x > y) { return x; } else { return y; });
}


i get an "expected an expression" error

The function has an int return type so you have to return an int. That means the code between the parentheses must evaluate to an int. It doesn't.

The key is in the "expected an expression" reported by the compiler. An expression must evaluate to true or false where false is 0 and true is not false.

if statements don't have a value.