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Asked: 2026-05-31T11:47:14+00:00

While trying to figure out how to round a float like 1.255 to the

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While trying to figure out how to round a float like 1.255 to the nearest hundredth, I found something interesting. I’m using gcc 4.4.5 on Debian 6. 

int   x = (1.255 * 100) + 0.5;   //  gives me back 125 instead of 126. 
float y = (1.255 * 100) + 0.5;   //  gives me back 126.000000.

Why is is that when I save to an int I get back 125 and not 126 ? In fedora when I save the above expression to an int I get back 126. Is this a gcc bug in debian ? Any help would be greatly appreciated.
Thanks.

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1 Answer

  1. Editorial Team

    Although this looks like a “typical” floating-point question, it’s more complicated than that.

    This one involves a combination of 3 things:

    Let’s break this down:

    Floating-point literals are of type double by default. Hence 1.255 is of type double.

    Thus the expression:

    (1.255 * 100) + 0.5

    is done using type double.

    But because binary floating-point can’t represent 1.255 exactly, the expression evaluates to:

    (1.255 * 100) + 0.5 = 125.99999999999999000000

    and is of type double.

    Since this is less than 126, storing it to an integer will result in 125.
    Storing it to float will round it to the nearest float, which results in 126.

    int    x = (1.255 * 100.) + 0.5;
float  y = (1.255 * 100.) + 0.5;
double z = (1.255 * 100.) + 0.5;

cout << fixed;
cout << x << endl;
cout << setprecision(20) << y << endl;
cout << setprecision(20) << z << endl;

    Output:

    125
126.00000000000000000000
125.99999999999999000000
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