C++ Pass A String
You can write your function to take a const std::string&
:
void print(const std::string& input)
{
cout << input << endl;
}
or a const char*
:
void print(const char* input)
{
cout << input << endl;
}
Both ways allow you to call it like this:
print("Hello World!\n"); // A temporary is made
std::string someString = //...
print(someString); // No temporary is made
The second version does require c_str()
to be called for std::string
s:
print("Hello World!\n"); // No temporary is made
std::string someString = //...
print(someString.c_str()); // No temporary is made
You should be able to call print("yo!") since there is a constructor for std::string which takes a const char*. These single argument constructors define implicit conversions from their aguments to their class type (unless the constructor is declared explicit which is not the case for std::string). Have you actually tried to compile this code?
void print(std::string input)
{
cout << input << endl;
}
int main()
{
print("yo");
}
It compiles fine for me in GCC. However, if you declared print like this void print(std::string& input)
then it would fail to compile since you can't bind a non-const reference to a temporary (the string would be a temporary constructed from "yo")
Well, std::string
is a class, const char *
is a pointer. Those are two different things. It's easy to get from string
to a pointer (since it typically contains one that it can just return), but for the other way, you need to create an object of type std::string
.
My recommendation: Functions that take constant strings and don't modify them should always take const char *
as an argument. That way, they will always work - with string literals as well as with std::string
(via an implicit c_str()
).