LinuxQuestions.org - code substitutions

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- - code substitutions (https://www.linuxquestions.org/questions/programming-9/code-substitutions-891356/)

code substitutions

GNU g++ seems to try to automatically match STL templates to patterns
in your code. E.g. if you code a swap looking function it substitutes
std::swap, is this so or am I inadvertently specifying a template ?

Avoid specifying "using namespace std" in your code!

Also, you can always reference functions using the :: operator. Here's an example that you can play with:

Code:

#include <algorithm>

#include <iostream>



template <typename T>

void swap(T& x, T& y)

{

  std::cout << "I'm on strike!" << std::endl;

}



int main()

{

  int x = 10;

  int y = 20;



  std::cout << "x = " << x << ", y = " << y << std::endl;



  std::swap(x, y);  // calls the C++ library.



  std::cout << "x = " << x << ", y = " << y << std::endl;



  ::swap(x, y);  // calls the swap() function defined above.



  std::cout << "x = " << x << ", y = " << y << std::endl;

}

I(Loarn) was trying to get the compiler to use my code and not what it considered a match
from the template supply, What finally dawned on me, I have been away for awhile, was that
is was not necessarily using the template but since I had #included <algorithm> and was
compiling with the -E flag, it was including the entire algorithm header file in the
assembler stage of the compile.
Thanks for your assistance, please ignore the misspelling I was schooled in the progressive
era.