Can this get to the end of the function and not return anything? - c++

Question

I am making a c++ text based game and while making a feature for eating food I made this function names eating. I am getting a warning saying its possible to reach the end with no return value. How is it possible to get to the end and not return something?
int inventory::eat(std::string eating)
{
if (!consumables.empty())
{
for (int i = 0; i < consumables.size(); i++)
{
if (consumables[i].name == eating)
{
return consumables[i].effect;
}
else
{
return 404;
}
}
}
else
{
return 505;
}
}

Answer 1

Presumably the compiler is unable to tell that if consumables.empty() == false, then consumables.size() > 0.
I'd probably rewrite it as this, but I'm worried about the if/else inside your for loop.
int inventory::eat(std::string eating) {
for (int i = 0; i < consumables.size(); i++) {
if (consumables[i].name == eating) {
return consumables[i].effect;
} else {
return 404;
}
}
return 505;
}

Answer 2

Assuming no other thread modifies consumables while eat() is running, then no, return will not be skipped. But the compiler doesn't know that. It doesn't knower that !empty() and size() > 0 mean the same thing for a container. They are just two separate method calls.
That said, I would suggest writing the code more like this, which is easier to read, and avoids the warning:
int inventory::eat(const std::string &eating)
{
if (consumables.empty())
return 505;
for (int i = 0; i < consumables.size(); ++i)
{
if (consumables[i].name == eating)
return consumables[i].effect;
}
return 404;
}
Or, if you are using C++11 and consumables supports iterators:
#include <algorithm>
int inventory::eat(const std::string &eating)
{
if (consumables.empty())
return 505;
auto found = std::find_if(
consumables.begin(), consumables.end(),
// replace 'consumable' below with your actual type name as needed...
[&](const consumable &c){ return c.name == eating; }
);
if (found != consumables.end())
return found->effect;
return 404;
}