why this call operator has Compilation Error?

Question

when compiling with mingw32-g++, there is error: no matching function for call to 'for_each(int [9], int*, main()::Output)', but can do well in vs2005?

#include <iostream>
#include <algorithm>

int main() {

  struct Output {
      void  operator () (int v) {
           std::cout << v << std::endl; 
       }
  };

  int a[] = {1, 2, 3, 4, 5, 6, 7, 8, 9};
  std::for_each(a, a + sizeof(a)/sizeof(a[0]), Output());

  return 0;
}

@kain64b - I am not able to understand this code... can anyone describe a bit more on this... — kapilddit, Aug 13 '12 at 06:46
I'm kind of wondering why `main` is returning 1 after no apparent error. — chris, Aug 13 '12 at 07:05

AnT stands with Russia · Accepted Answer · 2012-08-13T16:10:29.897

5

In pre-C++11 version of the language types used as template arguments were required to have linkage. Local class declarations in C++ have no linkage, which is why they can't be used as template arguments in C++98/C++03. In those versions of the language you have to declare your class in namespace scope.

The linkage requirement was removed in C++11. Your code is valid from C++11 point of view. Apparently, you are compiling in pre-C++11 mode.

edited Aug 13 '12 at 16:10

answered Aug 13 '12 at 06:44

AnT stands with Russia

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Unless you happen to have a C++11 conformat compiler (or one that doesn't know the rules - VS2005). – Bo Persson Aug 13 '12 at 06:50
@Flexo: Why would it be a problem? Argument deduction mechanism will make `for_each` to receive its last argument *by value*. No problem whatsoever. – AnT stands with Russia Aug 13 '12 at 06:55

score 3 · Answer 2 · edited May 23 '17 at 12:12

3

You have to declare the struct outside of main. See this question for an explanation.

edited May 23 '17 at 12:12

Community

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answered Aug 13 '12 at 06:43

Björn Pollex

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why this call operator has Compilation Error?

2 Answers2