C++ Pass A String

Question

Quick probably obvious question.

If I have:

void print(string input)
{
  cout << input << endl;
}

How do I call it like

Answer 1

Make it so that your function accepts a const std::string& instead of by-value. Not only does this avoid the copy and is therefore always preferable when accepting strings into functions, but it also enables the compiler to construct a temporary std::string from the char[] that you're giving it. :)

Answer 2

print(string ("Yo!"));

You need to make a (temporary) std::string object out of it.

Answer 3

Not very technical answer but just letting you know.

For easy stuff like printing you can define a sort of function in your preprocessors like

#define print(x) cout << x << endl

Answer 4

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::strings:

print("Hello World!\n"); // No temporary is made
std::string someString = //...
print(someString.c_str()); // No temporary is made

Answer 5

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()).

Answer 6

Just cast it as a const char *. print((const char *)"Yo!") will work fine.