Why std::log10(16.0) / std::log10(2.0) != std::log2(16.0)?

Asked May 24 '18 at 14:16

Active May 24 '18 at 14:16

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I just realized that in C++ std::log2((float)16) returns 4 but replacing with std::log10((float)16) / std::log10(2.0) returns 4.0000001902931812. I am storing my answer in a double variable.

I am not sure why ?

The replacement works nicely in windows calculator and matlab.

asked May 24 '18 at 14:16

Anirban

Why do you expect it to be equal? Numerical inaccuracy are understandable. – user202729 May 24 '18 at 14:17
2

https://stackoverflow.com/questions/588004/is-floating-point-math-broken – 0___________ May 24 '18 at 14:17
`std::log10((float)16)` why do you cast to `float` if you want a `double` in the end? – Holt May 24 '18 at 14:20
`std::log10(2.0)` is not equal to the log10 of 2, just near it. Rest of the calculation propagates that approximation. Same for `std::log10((float)16)`. – chux - Reinstate Monica May 24 '18 at 14:22
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If you want a `double` literal, include a decimal (`10.`). If you want a `float` use the `f` suffix (`10.f`). – François Andrieux May 24 '18 at 14:33

Why std::log10(16.0) / std::log10(2.0) != std::log2(16.0)?

0 Answers0