misunderstanding about std::numeric_limits::digits10

Question

According to my understanding (about what I read on the Internet), it seems that std::numeric_limits<double>::digits10 (which for double is equal to 15) represent the number of digits a double can handle, for instance 1.23456789012345 but not 1.234567890123456

On the other hand, double-precision floating-point format range will go until 1.8*10^+308, which seems to represent a number which not hold only on 15 digits.

Where is the incoherence?

The 1.8*10^308 is actually something like 181234567890123000...0000 — SinisterMJ, Oct 29 '12 at 10:48

Maxim Egorushkin · Answer 1 · 2012-10-29T10:54:19.427

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... std::numeric_limits<double>::digits10 (which for double is equal to 15) represent the number of digits a double can handle...

More precisely it's the number of significand digits it can store without loss of precision.

Example in Python:

1e15 == 1e15 + 1
False 

1e16 == 1e16 + 1
True # loss of precision

edited Oct 29 '12 at 10:54

answered Oct 29 '12 at 10:47

Maxim Egorushkin

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score 1 · Answer 2 · answered Oct 29 '12 at 10:47

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std::numeric_limits::digits10: Number of digits (in decimal base) that can be represented without change.

answered Oct 29 '12 at 10:47

Denis Ermolin

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Olaf Dietsche · Answer 3 · 2012-10-29T10:55:58.830

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The one is significant digits. The other is the range of numbers representable. So, you can have:

1.23456789012345*10^308

but not

1.234567890123456*10^308

which shows both significant digits 15 and the range 10^308, which can be represented.

edited Oct 29 '12 at 10:55

answered Oct 29 '12 at 10:49

Olaf Dietsche

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misunderstanding about std::numeric_limits::digits10

3 Answers3