Code detecting both string and const char array

Question

I'm running these two lines of code to add to a map later on:

o.add("-i", 800,             "int option");  
o.add("-w", "'switch on' option (no third parameter)");

To add them i'm using my two add functions defined as:

template<class T>
void add(std::string name, T value, std::string desc);
template<class T>
void add(std::string name, std::string desc);

The first one works fine, and returns the values I want, but if I add the second one, I get the error:

error: no matching function for call to ‘Opt::add(const char [3], const char [40])’

My question is why it's using my strings in the first one properly, and my strings in the second are being thought of as const char arrays.

Thank you in advance.

Qaz · Accepted Answer · 2012-10-24 21:55:58Z

1

Since you're not using the template argument in your second overload, remove it:

template<class T>
void add(std::string name, T value, std::string desc);

void add(std::string name, std::string desc);

A working sample can be found here.

answered Oct 24, 2012 at 21:55

Qaz

62.3k14 gold badges149 silver badges210 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

Classtronaut Over a year ago

Thanks alot, for some reason I thought I was using T later on in my function, but it turns out I'm not. For education's sake though, why exactly would that cause my strings to be detected as char arrays?

user1610015 Over a year ago

Because they are char arrays. The compiler saw two overloads of add, one with three parameters, and one with a non-automatically deducible template parameter. Since the call didn't match either one of the overloads, it showed you the actual types with which you were trying to call the function. And in C++, unlike in C# and Java, string literals have char array type, not std::string type, although they are automatically convertible to std::string, because std::string has a non-explicit constructor that takes a char array.

Qaz Over a year ago

@Classtronaut, Yes, it shows you the type you pass in. It can still be converted to a different type, as happens in the first example.

user1610015 · Accepted Answer · 2012-10-24 21:59:59Z

1

The error message is weird, but to use the second overload you need to explicitly specify the template argument (since there's no way to automatically deduce it):

o.add<T>("-w", "'switch on' option (no third parameter)");

Or, if you don't actually need the template parameter in this case, just make it a non-template method.

answered Oct 24, 2012 at 21:59

user1610015

6,6982 gold badges17 silver badges19 bronze badges

Collectives™ on Stack Overflow

Code detecting both string and const char array

2 Answers 2

3 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

3 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related