Unexplained inability to compare string in C++ within templates

Question

I'm getting the standard

error C2678: binary '==' : no operator found which takes a left-hand operand of type 'std::string' (or there is no acceptable conversion)

error. However, I have included string, and there is no apparent way for me to explain this.

Here is my code.

#include <string>
#include <vector>
#include <type_traits>

// [...]
template <typename T>
class Graph {
private:
    // [...]
    struct Vertex {
        T name;
        // [...]
    };
    std::vector<Vertex> verts;
public:
    // [...]
    template <typename P>
    int vertex(P item) {
        if (std::is_same<P,T>::value) {
            for (unsigned int i = 0; i < verts.size(); i++){
                if (verts[i].name == item) {
                    return i;
                }
            }
        } else if (std::is_same<P,int>::value) {
            return item;
        }
        return -1;
    }
}

The compiler is VS2012, the platform is Windows 8.1 64bit.

If P is not a string, your code will fail. C++ is not a dynamic language - your code is evaluated at compile time. You make provisions for the case when P is an int - indicating you expect that possibility. But in that case, your code will fail to compile, because you're still comparing it with a string. — Charles Salvia
– Charles Salvia, Commented May 18, 2014 at 0:14
I was under the impression from This SO Question that C++ can switch branches at compile time based on inference. — PythonNut
– PythonNut, Commented May 18, 2014 at 0:20
@PythonNut, if makes a decision at runtime. It's not conditional compilation. The condition, in this case, is a compile-time value, but it's still used during runtime. — Qaz
– Qaz, Commented May 18, 2014 at 0:33
So is the code in the linked question invalid? Basically P, my method is completely wrong? I was hoping std could pull off some magic with macros or something... oh well. — PythonNut
– PythonNut, Commented May 18, 2014 at 0:35
No, C++ will not switch branches with runtime if. You need to use meta-programming facilities like std::enable_if to do what you're trying to accomplish — Charles Salvia
– Charles Salvia, Commented May 18, 2014 at 1:11

Dimitrios Bouzas · Accepted Answer · 2014-05-18 00:41:55Z

2

Making an educated guess, in if (verts[i].name == item) it seems that verts[i].name has been instantiated as a std::string while item hasn't been instantiated as a std::string. Thus, the compiler rightfully complains that there's no overloaded operator to compare a std::string against X.

Update:

Another guess is that X is being instantiated as an int or a double because if it was instantiated as a std::string the compiler would complain about attempting to convert a std::string to an int by returning it in Graph::vertex() member function.

edited May 18, 2014 at 0:41

answered May 18, 2014 at 0:15

Dimitrios Bouzas

43.1k11 gold badges103 silver badges177 bronze badges

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

3 Comments

Charles Salvia Over a year ago

verts[i] is of type Vertex, but verts[i].name is likely of type std::string

PythonNut Over a year ago

Very sorry, edited Question to specify the structure of Vertex.

PythonNut Over a year ago

I'm not sure this argument still applies... as both the LHS and RHS are of type std::string.

rici · Accepted Answer · 2014-05-18 01:28:10Z

1

Seems like you just need two vertex methods, like this (no template):

int vertex(const T& item) {
  for (unsigned int i = 0; i < verts.size(); i++){
    if (verts[i].name == item) {
      return i;
    }
  }
}

int vertex(int item) {
  return item;
}

That won't work if T turns out to be int, but that doesn't seem likely to work with your original code either.

The above also fails to implement the default vector(SomeOtherType) case as returning a default -1. It will just fail to compile, which to me seems reasonable. YMMV.

answered May 18, 2014 at 1:28

rici

243k30 gold badges263 silver badges364 bronze badges

1 Comment

PythonNut Over a year ago

I turns out this and some template parameter inference does exactly what I had in mind originally. Thanks.

Collectives™ on Stack Overflow

Unexplained inability to compare string in C++ within templates

2 Answers 2

Update:

3 Comments

1 Comment

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Update:

3 Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Linked

Related