Thread: Another program (try)

Originally Posted by jasmin89

But somwhere is a problem because i can not compile the code.

What is the compile error? It would come with a line number that tells you where the error was detected. This is not always where the mistake was made, but it can give you a starting point. In this case, the errors do lie where the error message line numbers say, e.g., in main you forgot to call a function, and then you placed what looks like declarations where it looks like you wanted to provide arguments to a function call.

Note that you can simplify your rel function, e.g.,

Code:

double rel(double n1) {
    if (n1 < 0)
        n1 = 0;
    return n1;
}

or even:

Code:

double rel(double n1) {
    return (n1 < 0) ? 0 : n1;
}

You don't need the extra braces and assigning n1 to itself is rather pointless.

Also, the return type of the main function is int, not double.

After you have fixed these problems, you may still find another. That's alright: just keep fixing.

You don't mention the types when you call a function.
printf("%.2f = %.2f", i, rel(i));

1. Make i a double.
2. Notice the use of () on every other function you call.

But when i make i to a double i get this error in line21:

Code:

int main() {
double i = read_data;

printf("%.2f = %.2f", i, rel(i));
}

Error: incompatible types when initializing type 'double' using type 'double' (*)()'

I'm a little surprised that the compiler didn't complain about the conversion from function (pointer) type to int when it correctly complained about the conversion to double. Might be a relic of the old days: are you compiling with compile warnings turned on to a high level?

Anyway, the problem is that you're not calling the function. Call it:

Code:

double i = read_data();

Well, that means you have no clue what the printf is about. You need to read up about it so you can understand it and change it yourself. This is so simple that if we have to hand hold you to fix it, you might as well just quit programming.