using const

This is a discussion on using const within the C++ Programming forums, part of the General Programming Boards category; Hello, I have some code I am trying to work out. My reason is that the passing to g1(s) is ...

  1. #1
    UK2
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    Sep 2003
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    112

    using const

    Hello,

    I have some code I am trying to work out.

    My reason is that the passing to g1(s) is not allowed is that the function g1 could change the value of s.

    However, g1 doesn't change the value it is only printing the value. However, as the function f1 cannot know what function g1 will do with the value it will ALWAYS complain of a compile error.

    Am I right with my thinking?

    Many thanks,

    Code:
    void g1(std::string& s)
    {
         std::cout << "S: " << s << std::endl;
    }
     
     void f1(const std::string& s)
     {
       g1(s);          // Compile-time Error since s is const
     
       std::string localCopy = s;
       g1(localCopy);  // OK since localCopy is not const
     }

  2. #2
    C++ Witch laserlight's Avatar
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    However, g1 doesn't change the value it is only printing the value. However, as the function f1 cannot know what function g1 will do with the value it will ALWAYS complain of a compile error.

    Am I right with my thinking?
    As far as I understand it, yes. Of course, "cannot" may be too strong, since this is a matter of language design.

    Another way of looking at it is that f1 promises its caller that it will not change what s references. g1 does not make such a promise, so f1 cannot pass s to g1 without risking that its own promise may be broken.
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  3. #3
    and the hat of sweating
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    Make the function take a const string then:
    Code:
    void g1( const std::string& s )
    "I am probably the laziest programmer on the planet, a fact with which anyone who has ever seen my code will agree." - esbo, 11/15/2008

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  4. #4
    UK2
    Join Date
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    Another way of looking at it is that f1 promises its caller that it will not change what s references. g1 does not make such a promise, so f1 cannot pass s to g1 without risking that its own promise may be broken.
    Great explanation. Easy to understood.

    Thanks

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