Why float variable saves value by cutting digits after point in a weird way?

Question

I have this simple code line:

float val = 123456.123456;

when i print this val or look in scope, it stores value 123456.13

Ok, it's fine, it can't store all those digits after point just in 4 bytes, but why does it make 13 after the point? Shouldn't it be 12?

(using vc++ 2010 express on win32)

thanks to everyone for the answers. i got it now. i just though it should drop overlimited digits after point, instead of rounding it. — Kosmo零, Feb 17 '12 at 13:30

score 10 · Answer 1 · answered Feb 17 '12 at 13:24

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In binary,123456.123456 is 11110001001000000.000111111001... (infinite). It rounds to 11110001001000000.001, or 123456.125. That rounds to 123456.13 when printed.

answered Feb 17 '12 at 13:24

Rick Regan

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LihO · Answer 2 · 2012-02-17T13:27:35.933

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The value stored in val is equal to 123456.125. You are getting .13 because you are rounding it:

float val = 123456.123456;
printf("%.4f %.2f\n", val, val);

output: 123456.1250 123456.13

You should use double in this case to avoid truncation. Compiler should also warn you: "warning C4305: 'initializing' : truncation from 'double' to 'float'".

edited Feb 17 '12 at 13:27

answered Feb 17 '12 at 13:22

LihO

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Sergey Kalinichenko · Accepted Answer · 2012-02-17T14:31:47.197

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When represented as a float, your number has an exponent of 16 (i.e. the value is its mantisse times 2^16, or 65536). The mantisse then becomes

123456.123456 / 65536 = 1.8837909462890625

In order to fit in a 32-bit float, the mantisse is truncated to 23 bits, so now it becomes 1.883791. When multiplied back by 65536, it becomes 123456.125.

Note the 5 in the third position after the decimal point: the output routine of C++ that you used rounds it up, making your final number look like 123456.13.

EDIT Explanation of the rounding: (Rick Regan's comment)

The rounding occurs first in binary (to 24 bits), in decimal to binary conversion, and then to decimal, in printf. The stored value is 1.1110001001000000001 x 2^16 = 1.8837909698486328125 x 2^16 = 123456.125. It prints as 123456.13, but only because Visual C++ uses "round half away from zero" rounding.

Rick has an outstanding article on the subject, too.

If you would like to play with other numbers and their float representations, here is a very useful IEEE-754 calculator.

edited Feb 17 '12 at 14:31

answered Feb 17 '12 at 13:26

Sergey Kalinichenko

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Could you please explain bit more.. I could not able to understand particularly this line "When represented as a float, your number has an exponent of 16". Why it is needed? – Rasmi Ranjan Nayak Feb 17 '12 at 13:29
@RasmiRanjanNayak 32-bit float is represented as a 23 bit mantisse, a seven-bit binary exponent, and a sign bit. Logically, you divide your original number by two and increment the exponent until the only remaining digit in front of the decimal point is `1`. For `123456.123456`, 16 divisions are required. – Sergey Kalinichenko Feb 17 '12 at 13:35
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@dasblinkenlight At best this description is misleading. The rounding occurs first in binary (to 24 bits), in decimal to binary conversion, and then to decimal, in printf. The stored value is 1.1110001001000000001 x 2^16 = 1.8837909698486328125 x 2^16 = 123456.125. It prints as 123456.13, but only because Visual C++ uses "round half away from zero" rounding (see my article http://www.exploringbinary.com/inconsistent-rounding-of-printed-floating-point-numbers/.) – Rick Regan Feb 17 '12 at 14:19
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@RickRegan Wow, nice article! Thanks for the reference, I added your comment to the answer to make it more complete. Thank you very much for your help! – Sergey Kalinichenko Feb 17 '12 at 14:33

score 2 · Answer 4 · edited May 23 '17 at 12:20

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Try printing the value of std::numeric_limits<float>::digits10. This is roughly speaking how much precision in base 10 a float has. You're trying to exceed it, so you're experiencing a loss of precision (meaning that digits beyond the significant ones are not really meaningful).

See e.g. What is the meaning of numeric_limits<double>::digits10

edited May 23 '17 at 12:20

Community

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answered Feb 17 '12 at 13:15

John Zwinck

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score 2 · Answer 5 · answered Feb 17 '12 at 13:17

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It's totally compiler dependent. Check it in GCC. It should be xxx.12

answered Feb 17 '12 at 13:17

Rasmi Ranjan Nayak

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If the format used internally is IEEE-754 it shouldn't be compiler-dependent at all. – Matteo Italia Feb 17 '12 at 13:18
I think we are giving "compiler-dependent" two different meanings; what I - and others - understood is that you are saying that "it's normal that it varies from compiler to compiler". Are you saying instead that it's a *compiler-specific bug*? – Matteo Italia Feb 17 '12 at 13:22

Why float variable saves value by cutting digits after point in a weird way?

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