Thread: problem with c=(f-32)*(5/9)

  1. #1
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    problem with c=(f-32)*(5/9)

    hello, my code works if it is written like i have posted it.
    But when i put two brackets at the red line ( c=(f-32)*(5/9); ) my c variable is always 0,000 .
    Why is that so ?

    Code:
    int
    Code:
    main()
    
    {
    float f,c;
    printf("give temp at Fahrenheit: \n\n");
    scanf("%f",&f);
    c=(f-32)*5/9;
    printf("the temperature in celsius is: %f \n\n",c);
    return 0;
    }

  2. #2
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    Because (5/9) is 0. And 0 times anything gives 0.

    EDIT

    To be more specific: the expression you are probably looking for is: (f-32)*((float)5/9)
    Last edited by Absurd; 04-02-2016 at 03:54 AM.

  3. #3
    misoturbutc Hodor's Avatar
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    Actually I think I'd prefer (f - 32) * 5.0 / 9

  4. #4
    Tweaking master Aslaville's Avatar
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    Quote Originally Posted by Hodor View Post
    Actually I think I'd prefer (f - 32) * 5.0 / 9
    Personally, I'd prefer (f - 32) * 5.f / 9

    It's even more hip

  5. #5
    Lurking whiteflags's Avatar
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    Personally I try to write everything out all the time.

    c = (f - 32.0f) / 1.8f;

  6. #6
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    Ok.! my problem was just hilarious! thank you very much!!! 8-)

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