将 lambda 表达式传递给模板调用函数时没有匹配的函数编译错误？

Question

Code:

#include <iostream>

template <class FunctorType>
void caller(const FunctorType& func) {
  func();
}

int main() {
  double data[5] = {5., 0., 0., 0., 0.};
  auto peek_data = [data]() { std::cout << data[0] << std::endl; };
  auto change_data = [data]() mutable { data[0] = 4.2; };

  caller(peek_data);    // This works
  caller(change_data);  // This doesn't
  return 0;
}

If I compile this with clang++ -std=c++11 mutable_lambda.cpp , I got error: no matching function for call to object of type 'const (lambda at mutable_lambda.cpp:8:22)' .

Question: Why passing the second lambda expression with mutable copy capture failed to compile? Thanks in advance!

Answer 1

A mutable lambda has a non- const operator() . You are trying to call this non- const operator() through a const reference. That doesn't work for the same reason that calling any non- const non-static member function doesn't work through a const reference.

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If you want to allow caller to modify the passed function object (through a non- const operator() call) if it is passed as non- const argument, then take it by forwarding reference instead of const reference:

template <class FunctorType>
void caller(FunctorType&& func) {
  func(); // maybe std::forward<FunctorType>(func)();
}

---

Also note that change_data does not change the data in main . It changes a copy of it stored inside the lambda. If you want it to change the data in main , you need to capture data by-reference and then the lambda itself also doesn't need to be mutable anymore:

auto change_data = [&data]() { data[0] = 4.2; };