C++ – What does “warning: not all control paths return a value” mean? (C++)

ccontrolswarnings

The exact warning I get is

warning C4715: 'hand::show' : not all control paths return a value

and hand::show is

std::ostream& hand::show(std::ostream& os) const
{
    if(side == left)
    {
        return os<<display[0]<<display[1]<<display[2]<<display[3]<<display[4];
    }
    if(side == right)
    {
        return os<<display[4]<<display[3]<<display[2]<<display[1]<<display[0];
    }
}

where side is a variable of type orientation

orientation{
    left = -1,
    right = 1
};

What does the warning mean, and what would be the best solution to get rid of it?

Best Answer

Your compiler isn't smart enough to take into account that the only two options for side are left and right, so it thinks it's possible for neither return statement to be executed. When side is neither left nor right, your function doesn't say which value to return.

Related Solutions

C++ – What does the explicit keyword mean

The compiler is allowed to make one implicit conversion to resolve the parameters to a function. What this means is that the compiler can use constructors callable with a single parameter to convert from one type to another in order to get the right type for a parameter.

Here's an example class with a constructor that can be used for implicit conversions:

class Foo
{
public:
  // single parameter constructor, can be used as an implicit conversion
  Foo (int foo) : m_foo (foo) 
  {
  }

  int GetFoo () { return m_foo; }

private:
  int m_foo;
};

Here's a simple function that takes a Foo object:

void DoBar (Foo foo)
{
  int i = foo.GetFoo ();
}

and here's where the DoBar function is called:

int main ()
{
  DoBar (42);
}

The argument is not a Foo object, but an int. However, there exists a constructor for Foo that takes an int so this constructor can be used to convert the parameter to the correct type.

The compiler is allowed to do this once for each parameter.

Prefixing the explicit keyword to the constructor prevents the compiler from using that constructor for implicit conversions. Adding it to the above class will create a compiler error at the function call DoBar (42). It is now necessary to call for conversion explicitly with DoBar (Foo (42))

The reason you might want to do this is to avoid accidental construction that can hide bugs.
Contrived example:

You have a MyString class with a constructor that constructs a string of the given size. You have a function print(const MyString&) (as well as an overload print (char *string)), and you call print(3) (when you actually intended to call print("3")). You expect it to print "3", but it prints an empty string of length 3 instead.

C++ – What does int argc, char *argv[] mean

argv and argc are how command line arguments are passed to main() in C and C++.

argc will be the number of strings pointed to by argv. This will (in practice) be 1 plus the number of arguments, as virtually all implementations will prepend the name of the program to the array.

The variables are named argc (argument count) and argv (argument vector) by convention, but they can be given any valid identifier: int main(int num_args, char** arg_strings) is equally valid.

They can also be omitted entirely, yielding int main(), if you do not intend to process command line arguments.

Try the following program:

#include <iostream>

int main(int argc, char** argv) {
    std::cout << "Have " << argc << " arguments:" << std::endl;
    for (int i = 0; i < argc; ++i) {
        std::cout << argv[i] << std::endl;
    }
}

Running it with ./test a1 b2 c3 will output

Have 4 arguments:
./test
a1
b2
c3

Best Answer

Related Solutions

C++ – What does the explicit keyword mean

C++ – What does int argc, char *argv[] mean

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