Passing std::set to method in C++

Question

Given the following prerequisits:

shared.h

struct A {
    int x;
    int y;
}

typedef set<A> setOfA;

implementation1.cpp

#include "shared.h"
void Implementation1::someFunction()
{
    //...
    setOfA setinstance;
    //...
    Implementation2* i = new Implementation2();
    i->functionF(setinstance);
}

implementation2.h

#include "shared.h"
void Implementation2::functionF(setOfA&);

EDIT: now this should be clearer...

I want to pass a setOfA to another function of a different class - everything is compiling just fine. I get the following linker problem:

undefined reference to 'implementation::functionF(std::set<A, std::less<A>, std::allocator<A> >&)'

Just to get it right - the implementation could not be found, right? This cannot be a typedef problem, because everything compiles just fine... What am I missing here?

Have you implemented `functionF`? Are you including the source file that contains it when you build the program? — Mike Seymour, Dec 21 '11 at 17:36

Bill Lynch · Answer 1 · 2011-12-21T18:44:47.430

5

The linker is unable to find the definition of void functionF(setOfA&);

Somewhere, you need:

void Implementation2::functionF(setOfA&) {
    ...
}

Just to elaborate on this slightly, shouldn't you have an implementation2.cpp file that implements the above code?

edited Dec 21 '11 at 18:44

answered Dec 21 '11 at 17:32

Bill Lynch

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score 2 · Answer 2 · edited May 23 '17 at 12:11

2

You didn't define a comparator for A, so a set of them can't exist.

Define a function:

bool operator<(A& const lhs, A& const rhs)

and ensure that it implements a strict weak ordering; in this case, probably:

bool operator<(A& const lhs, A& const rhs) {
    if (lhs.x != rhs.x)
       return (lhs.x < rhs.x);

    return (lhs.y < rhs.y);
}

edited May 23 '17 at 12:11

Community

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answered Dec 21 '11 at 17:30

Lightness Races in Orbit

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You're right, but shouldn't this cause a compiler error instead of a linker error? – R. Martinho Fernandes Dec 21 '11 at 17:32
@R.MartinhoFernandes: Um, maybe. Hmm. – Lightness Races in Orbit Dec 21 '11 at 17:33

score 1 · Answer 3 · edited Dec 21 '11 at 18:48

1

My major problem was that I was working in the wrong namespace. I agree to close this thread, because it's probably just helping me with my specific problem - on the other hand the hint to implement the operator< gave me the clue. For the sake of completeness, here's my operator< implementation.

bool operator<(const A& e) const {
if (x != e.x)
  return x < e.x;
else
  return y < e.y;
}

edited Dec 21 '11 at 18:48

Edward Strange

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answered Dec 21 '11 at 18:01

Eric

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Good, but prefer a non-member function. – Lightness Races in Orbit Dec 21 '11 at 18:08

Passing std::set to method in C++

3 Answers3