C++ If in function not woking

Question

Im completely new to C++ and Im trying to make my own "cmd". Still getting error: "|11|error: return-statement with a value, in function returning 'void' [-fpermissive]|"... Please help :(

Code here:

#include <iostream>
#include <string>

using namespace std;

void cmd(int command)
{
    if (command == login)
    {
        cout << "Prikaz login neni k dispozici!";
        return 0;
    }
    else
    {
        cout << "Prikaz ";
        cout << command;
        cout << " neni znamy!";
        return 0;
    }
}

int main()
{
    cout << "Zadejte prikaz: ";
    cin << command;
    cmd(command);
    return 0;
}

Answer 1

To expand on Philipp Braun's answer, the exact issue was that if a function is declared void, it can't have an actual return value. [Specifically, it can only have a return value of type void, which basically means it can return either nothing or the result of another function that itself returns nothing.]

// Valid.  Function is void, and returns nothing.
void foo(int arg, int u, int ments) {
    doSomethingWith(arg, u, ments);
    return;
}

// Valid.  Returns (with no return value) upon reaching end of function.
void bar(int arg, int u, int ments) {
    doSomethingWith(arg, u, ments);
}

// Valid.  Returns the result of foo(), which itself returns nothing.
void bat(int arg, int u, int ments) {
    doSomethingWith(arg, u, ments);
    return foo(arg, u, ments);
}

// Invalid.  3 is int, not void.
void baz(int arg, int u, int ments) {
    doSomethingWith(arg, u, ments);
    return 3;
}

// Valid again.  Function is int, and returns int.
int quz(int arg, int u, int ments) {
    doSomethingWith(arg, u, ments);
    return 3;
}

// And formally invalid again.  Returns (with no return value) at end of function, instead of returning an int.
// Causes undefined behaviour, which can be... well, anything, really.
int qaz(int arg, int u, int ments) {
    doSomethingWith(arg, u, ments);
}

In a function declaration and/or definition, the first type, shown before the name, is the function's return type, and indicates what the function returns. If it's void, the function isn't allowed to have a return value; any return statements must either be used without an expression, or with an expression of type void.

The return statement takes the following form:

return /* expression */;

In this case, expression can be any valid expression (such as in return 1 + 1; or return functionName();), or can be omitted for functions with a return type of void. Note that it doesn't have to be omitted for functions with a return type of void; the following is a weird, but perfectly valid program:

#include <iostream>

void iReturnVoid() {
    std::cout << "I return void." << std::endl;
}

void andSoDoI() {
    std::cout << "I must inform you that ";
    return iReturnVoid();
}

int main() {
    andSoDoI();
}

In short, the type of expression has to match the type specified at the beginning of the function.

If a return statement isn't explicitly provided, then the function returns upon executing its final statement; this is treated as if the function ends with return 0; for main(), or as if the function ends with return; for any other function.

Also note that if any value-returning function returns without returning a value, whether explicitly by encountering and executing return; or implicitly by reaching the end of the function, then it results in undefined behaviour. (main() is immune to this happening implicitly, due to its implicit return statement returning 0, but can have trouble with this if it encounters an explicit return statement without a value.

---

I know this may be a tad complex for your first day programming, but it basically boils down to this:

1. The return statement should have the same type as the function itself. If the function is void, it returns void. If the function is int, it returns an int. If the function is char*, it returns a pointer to a char. And so on.
2. If the program reaches the end of main() without encountering a return statement, it implicitly inserts a return 0; statement to indicate that the program completed successfully.
3. If the program reaches the end of any other function without encountering a return statement, it implicitly inserts an empty return statement, specifically "return;". This is fine for void functions, but can cause problems for functions with any other return type.

Answer 2

No idea what you are exactly planning to do, but the following program should work. You simply did not declare your integer variables properly.

#include <iostream>
#include <string>

using namespace std;

int cmd(string command)
{
    if (command == "login")
    {
        cout << "Prikaz login neni k dispozici!";
        return 0;
    }
    else
    {
        cout << "Prikaz ";
        cout << command;
        cout << " neni znamy!";
        return 0;
    }
}

int main()
{
    string command; /* Define String */
    cout << "Zadejte prikaz: ";
    cin >> command; /* Input */
    cmd(command);
    return 0;
}