log10, log10f, log10l
From cppreference.com
| Defined in header
<math.h>
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| float log10f( float arg );
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(1) | (since C99) |
| double log10( double arg );
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(2) | |
| long double log10l( long double arg );
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(3) | (since C99) |
| Defined in header
<tgmath.h>
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| #define log10( arg )
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(4) | (since C99) |
1-3) Computes the common (base-10) logarithm of
arg.
4) Type-generic macro: If
arg has type long double, log10l is called. Otherwise, if arg has integer type or the type double, log10 is called. Otherwise, log10f is called.
Contents |
[edit] Parameters
| arg | - | floating point value |
[edit] Return value
If no errors occur, the common (base-10) logarithm of arg (log
10(arg) or lg(arg)) is returned.
If a domain error occurs, an implementation-defined value is returned (NaN where supported).
If a pole error occurs, -HUGE_VAL, -HUGE_VALF, or -HUGE_VALL is returned.
[edit] Error handling
Errors are reported as specified in math_errhandling.
Domain error occurs if arg is less than zero.
Pole error may occur if arg is zero.
If the implementation supports IEEE floating-point arithmetic (IEC 60559),
- If the argument is ±0, -∞ is returned and FE_DIVBYZERO is raised.
- If the argument is 1, +0 is returned
- If the argument is negative, NaN is returned and FE_INVALID is raised.
- If the argument is +∞, +∞ is returned
- If the argument is NaN, NaN is returned
[edit] Example
Run this code
#include <stdio.h> #include <math.h> #include <float.h> #include <errno.h> #include <fenv.h> #pragma STDC FENV_ACCESS ON int main(void) { printf("log10(1000) = %f\n", log10(1000)); printf("log10(0.001) = %f\n", log10(0.001)); printf("base-5 logarithm of 125 = %f\n", log10(125)/log10(5)); // special values printf("log10(1) = %f\n", log10(1)); printf("log10(+Inf) = %f\n", log10(INFINITY)); //error handling errno = 0; feclearexcept(FE_ALL_EXCEPT); printf("log10(0) = %f\n", log10(0)); if(errno == ERANGE) perror(" errno == ERANGE"); if(fetestexcept(FE_DIVBYZERO)) puts(" FE_DIVBYZERO raised"); }
Possible output:
log10(1000) = 3.000000
log10(0.001) = -3.000000
base-5 logarithm of 125 = 3.000000
log10(1) = 0.000000
log10(+Inf) = inf
log10(0) = -inf
errno == ERANGE: Numerical result out of range
FE_DIVBYZERO raised[edit] See also
| (C99)(C99)
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computes natural (base-e) logarithm (ln(x)) (function) |
| (C99)(C99)(C99)
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computes base-2 logarithm (log2(x)) (function) |
| (C99)(C99)(C99)
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computes natural (base-e) logarithm of 1 plus the given number (ln(1+x)) (function) |
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C++ documentation for log10
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