log10, log10f, log10l
From cppreference.com
Defined in header <math.h>
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float log10f( float arg ); |
(1) | (since C99) |
double log10( double arg ); |
(2) | |
long double log10l( long double arg ); |
(3) | (since C99) |
Defined in header <tgmath.h>
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#define log10( arg ) |
(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] References
- C11 standard (ISO/IEC 9899:2011):
- 7.12.6.8 The log10 functions (p: 245)
- 7.25 Type-generic math <tgmath.h> (p: 373-375)
- F.10.3.8 The log10 functions (p: 522)
- C99 standard (ISO/IEC 9899:1999):
- 7.12.6.8 The log10 functions (p: 225-226)
- 7.22 Type-generic math <tgmath.h> (p: 335-337)
- F.9.3.8 The log10 functions (p: 459)
- C89/C90 standard (ISO/IEC 9899:1990):
- 4.5.4.5 The log10 function
[edit] See also
(C99)(C99) |
computes natural (base-e) logarithm (ln(x)) (function) |
(C99)(C99)(C99) |
computes base-2 logarithm (log2(x)) (function) |
(C99)(C99)(C99) |
computes natural (base-e) logarithm of 1 plus the given number (ln(1+x)) (function) |
C++ documentation for log10
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