FAQ: Code Challenge: C++ Functions - Average

This community-built FAQ covers the “Average” exercise from the lesson “Code Challenge: C++ Functions”.

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I would like to ask about the fuctions. Okay if;
“void something(std::string 1, std::string 2)” why does, “double something(double 1, double 2)”.

What’s the differences?
Is void is only for std::string and double is the same as void but only for double, so what does bool is for and what is the “void” for integer?
Is it something like this? I’m so confused.

2 Likes

I got this:

#include

// Define average() here:
double average(double num1, double num2)
{
return (num1+num2)/2;
}

int main() {

std::cout << average(42.0, 24.0) << “\n”;
std::cout << average(1.0, 2.0) << “\n”;

}

namespace “std” has no member “cout”
You are missing an include.

double average(double num1, double num2)
{
    return (num1 + num2) / 2;
}

int main()
{

    std::cout << average(42.0, 24.0) << "\n";
    std::cout << average(1.0, 2.0) << "\n";
}

I have a question reg the average exercise. The code i wrote was this. what is the difference from it to the correct answer?

My code:
#include

// Define average() here:

void average(double num1, double num2){
double average_var;
average_var = (num1 + num2) / 2;
return average_var;
}

int main() {
std::cout << average(42.0, 24.0) << “\n”;
std::cout << average(1.0, 2.0) << “\n”;
}

ANSWER FROM EXERCISE:
#include

// Define average() here:
double average(double num1, double num2) {
return (num1 + num2) / 2;
}

int main() {
std::cout << average(42.0, 24.0) << “\n”;
std::cout << average(1.0, 2.0) << “\n”;
}

When you format your code

#include

// Define average() here:

void average(double num1, double num2)
{
    double average_var;
    average_var = (num1 + num2) / 2;
    return average_var;
}

int main()
{
    std::cout << average(42.0, 24.0) << “\n”;
    std::cout << average(1.0, 2.0) << “\n”;
}

ANSWER FROM EXERCISE :
#include

    // Define average() here:
    double
    average(double num1, double num2)
{
    return (num1 + num2) / 2;
}

int main()
{
    std::cout << average(42.0, 24.0) << “\n”;
    std::cout << average(1.0, 2.0) << “\n”;
}

You can see that double from the answer is a double function. Your double function is a void. Void returns nothing, yet you are attempting to return average_var.

It will be easier to read it like this and compare where you went wrong.

#include

// Define average() here:

void average(double num1, double num2)
{
    double average_var;
    average_var = (num1 + num2) / 2;
    return average_var;
}

double average(double num1, double num2)
{
    return (num1 + num2) / 2;
}

int main()
{
    std::cout << average(42.0, 24.0) << “\n”;
    std::cout << average(1.0, 2.0) << “\n”;
}