Logarithm base 2 in C language

Question

Please, can someone help me with how to calculate a simple log2 in C? I tried with this code, but it does not work:

printf( "log( %f ) = %f\n", x, log2(x) );

Possible duplicate of [How to write log base(2) in c/c++](http://stackoverflow.com/questions/3064926/how-to-write-log-base2-in-c-c) — chus, Dec 01 '16 at 00:25

score 5 · Answer 1 · answered Nov 30 '16 at 23:54

5

#include <stdio.h>
#include <math.h>

int main() {
    double x = 42.0;

    printf( "log( %f ) = %f\n", x, log2(x) );

    return 0;
}

OUTPUT

% ./a.out
log( 42.000000 ) = 5.392317
%

answered Nov 30 '16 at 23:54

cdlane

40,441
5
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81

So why does OP's code not work? – Right leg Dec 01 '16 at 02:18
1

@Rightleg Probably missing `` which means that the return value incorrectly defaults to `int`. – Raymond Chen Dec 01 '16 at 02:32
1

He also might need to compile with `-lm`, depending which version of `gcc` he is using. – RoadRunner Dec 01 '16 at 03:03

RoadRunner · Accepted Answer · 2016-12-01T02:08:53.507

You can also create a helper function which converts to any log base you wish:

Something like this:

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

double 
my_log(double x, int base) { 
    return log(x) / log(base); 
} 

int 
main(void) {
    double x = 42.0;

    printf("log(%f) = %f\n", x, my_log(x, 2));

    return 0;
}

Compiled with:

gcc -Wall -o logprog logprog.c -lm

Output:

log(42.000000) = 5.392317

Logarithm base 2 in C language

2 Answers2