std::generate algorithm

This is a discussion on std::generate algorithm within the C++ Programming forums, part of the General Programming Boards category; Hi All, I have just been having a mess around with the function call operator and the generate algorithm. When ...

  1. #1
    Registered User subdene's Avatar
    Join Date
    Jan 2002
    Posts
    359

    std::generate algorithm

    Hi All,

    I have just been having a mess around with the function call operator and the generate algorithm. When I first saw the output of this which is:

    0 1 2 3 4 5 6 7 8 9

    Code:
    #include "stdafx.h"
    #include <iostream>
    #include <algorithm>
    #include <iterator>
    
    struct g
    {
            g()
    	{
    		n = 0;
    	}
           
    	int operator()() 
    	{ 
    		return n++; 
    	}
            int n;
    };
    
    int main()
    {
            int a[10];
    	int size = (sizeof(a) / sizeof(a[0]));
            std::generate(a, a + size, g());
            std::copy(a, a+10, std::ostream_iterator<int>(std::cout, " "));
    	getchar();
    }
    My natural instinct was to think that this would be 0 0 0 0 0 0 etc...
    However, it looks like the implementation of the generate algorithm creates only one instance of type g, then uses that same instance for subsequent iterations. Is this correct?

    Thanks
    Be a leader and not a follower.

  2. #2
    The larch
    Join Date
    May 2006
    Posts
    3,573
    It uses the instance you pass to it.
    I might be wrong.

    Thank you, anon. You sure know how to recognize different types of trees from quite a long way away.
    Quoted more than 1000 times (I hope).

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