log10, log10f, log10l

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log10
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Macro constants
 
Defined in header <math.h>
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>
#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

#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)
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