Thread: A "cout" question

  1. #1
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    A "cout" question

    How do I make it so that this statement shows .8 instead of 0:
    cout << 4/5;

    It keeps displaying 0 instead of .8. Thanks for your help.

  2. #2
    C++ Developer XSquared's Avatar
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    cout<<4.0/5.0;
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    Now how would I do that if it were a variable?

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    If it were a variable, the << operator would know how. If both top/bottom are floats, it will output 0.8. But if they are ints, then you will get the result of integer division, which drops the remainder and the << will output an int, or 0.

  5. #5
    ATH0 quzah's Avatar
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    Type cast it. I believe C++ likes you to use "static_cast". IIRC, it's something like:
    Code:
    int var1 = 4, var2 = 5;
    cout << static_cast<float>(var1) / static_cast<float>(var2);
    Quzah.
    Hope is the first step on the road to disappointment.

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    Here is my line of code:

    cout << "PI= " << 4 * static_cast<float>(c) / static_cast<float>(n) << endl;
    Both of the variables, c & n, were defined as integers. I still can't get the line to print out any decimals. I'm going to fiddle around with it, but maybe you know what I am doing wrong? Thanks for all your help already quzah.

  7. #7
    ATH0 quzah's Avatar
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    Well, if you look at that line, you'll see that 4 isn't a floating point number. Cast it also, or just ad a .0 to it.

    Quzah.
    Hope is the first step on the road to disappointment.

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    Thanks for your help, it worked!

  9. #9
    Registered User hk_mp5kpdw's Avatar
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    Quote Originally Posted by scrub05
    Both of the variables, c & n, were defined as integers. I still can't get the line to print out any decimals.
    Even casting and performing a floating point division operation might result in no decimals being output if the result of that operation results in a value such as X.0. The decimals will not be shown if there isn't anything to show or unless you tell cout to format its output differently:

    Code:
    #include <iostream>
    #include <iomanip>
    
    using namespace std;
    
    int main()
    {
        float f = 5.0f;
        cout << "f is: " << f << endl;
        cout << setiosflags(ios::fixed) << setprecision(4)
             << "f is: " << f << endl;
    
        int c = 8, n = 4;
        float result = 4 * static_cast<float>(c) / static_cast<float>(n);
    
        cout.unsetf(ios::fixed);
        cout << "Result is: " << result << endl;
        cout << "Result is: " << setiosflags(ios::fixed)
             << setprecision(2) << result << endl;
    
        return 0;
    }
    My output:
    Code:
    f is: 5
    f is: 5.0000
    Result is: 8
    Result is: 8.00
    So you see it can work but depending on what your values for c and n are you might not be seeing the decimals.
    Last edited by hk_mp5kpdw; 01-06-2005 at 04:54 PM.
    "Owners of dogs will have noticed that, if you provide them with food and water and shelter and affection, they will think you are god. Whereas owners of cats are compelled to realize that, if you provide them with food and water and shelter and affection, they draw the conclusion that they are gods."
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  10. #10
    Registered User Scribbler's Avatar
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    I'd sooner write a function which prototypes the arguments as Double, as well as returns a Double (regardless what type you passed to it). Then allow C++'s Promotion Rules to do the work for you. Much more elegant IMHO. Here's a sample proggie...

    Code:
    #include <iostream>
    using namespace std;
    
    double divide ( double, double );
    
    int main()
    {
        int x, y;
        
        x = 4;
        y = 5;
        
        cout << "4 / 5 = " << divide ( x, y );
        
        while ( cin.get() != '\n' );
        
        return 0;
        
    }
    
    double divide ( double a, double b )
    {
        return a / b;
        
    }
    Last edited by Scribbler; 01-06-2005 at 06:53 PM.

  11. #11
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    I woudl just use a solution like what XSquared Said; define your int as a float,

    ie- Intstead of INT X call it Float X. That should get you squared away.

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