The title is a bit lengthy, but it's best explained by an example:
Suppose we have the following functions in C++:
void SomeFunction(int num) { //1
}
void SomeFunction(int& num) { //2
}
void SomeFunction(const int& num) { //3
}
void SomeFunction(const int num) { //4
}
All of these are called the same way:
SomeFunction(5);
or
int x = 5;
SomeFunction(x);
When I tried to compile the code, it rightfully says more than one instance of overloaded function "SomeFunction" matches the argument
My question is: Is there a way to tell the compiler which function I meant to call?
I asked my lecturer if it was possible, and she tried something along
SomeFunction< /*some text which I don't remember*/ >(x);
But it didn't work and she asked me to find out and tell her.
I also encounter this post: How to define two functions with the same name and parameters, if one of them has a reference? And it seems that 1 and 2 can't be written together, but what about 3 and 4? Can either one of those be called specifically?
1 and 4 have the same signature, so you'll need to drop one of those.
The other functions cannot be called directly, but you could add a template function that allows you to specify the desired parameter type:
template<class Arg>
void Call(void f(Arg), Arg arg)
{
f(arg);
}
// Driver Program to test above functions
int main()
{
int i;
Call<int>(SomeFunction, 1);
Call<int&>(SomeFunction, i);
Call<const int&>(SomeFunction, 1);
}
Alternatively you could use a function pointer to choose the signature.
int i;
static_cast<void(*)(int)>(&SomeFunction)(1);
static_cast<void(*)(int&)>(&SomeFunction)(i);
static_cast<void(*)(const int&)>(&SomeFunction)(1);
It would be preferrable to avoid this scenario though and only define overloads for either references or the signature void SomeFunction(int)
.
Note:
SomeFunction<some text which I don't remember>(x);
only works for template functions and SomeFunction
is not a template function, so this is not an option here.