Thread: Limited output

  1. #1
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    Limited output

    Say I have a double wich has a value of 4.567
    How can let the output, using cout, be 4.57 or 4.6 ?

  2. #2
    Code Goddess Prelude's Avatar
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    Code:
    #include <iostream>
    #include <iomanip>
    
    int main()
    {
      double d =  4.567;
    
      std::cout.precision ( 3 );
      std::cout<< d <<std::endl;
    
      std::cout.precision ( 2 );
      std::cout<< d <<std::endl;
    }
    -Prelude
    My best code is written with the delete key.

  3. #3
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    And how to make it output a 7 as 7.0?

  4. #4
    Code Goddess Prelude's Avatar
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    >And how to make it output a 7 as 7.0?
    By using the std::fixed manipulator for cout. Or you could go here
    Code:
    #include <iostream>
    #include <iomanip>
    
    int main()
    {
      double d =  7;
    
      std::cout.precision ( 1 );
      std::cout<< std::fixed << d <<std::endl;
    }
    -Prelude
    My best code is written with the delete key.

  5. #5
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    I got: "fixed; undeclared identifier"

  6. #6
    Code Goddess Prelude's Avatar
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    >I got: "fixed; undeclared identifier"
    Then you didn't use it correctly, review the code I gave you or do this:
    Code:
    #include <iostream>
    #include <iomanip>
    
    int main()
    {
      double d =  7;
    
      std::cout.precision ( 1 );
      std::cout.setf ( std::ios_base::fixed, std::ios_base::floatfield );
      std::cout<< d <<std::endl;
    }
    -Prelude
    My best code is written with the delete key.

  7. #7
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    I want to use it in a function in a class, that makes any difference?

  8. #8
    Code Goddess Prelude's Avatar
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    >I want to use it in a function in a class, that makes any difference?
    No, not really. It would be pretty dumb to allow output formatting in global functions but not member functions. Perhaps if you posted the code that is giving you problems I could tell you what isn't working like you would expect.

    -Prelude
    My best code is written with the delete key.

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