错误在gcc中使用数学函数？

Question

When I am passing constant value in log2() as follow

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

int main(int argc, char* argv[]) 
{
int var;
var= log2(16);
printf("%d",var);
return 0;
}

gcc prog.c (NO Error) 4

But when I am passing variable in function log2(var) gives error undefined reference to `log2' I require to link library ie -lm

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

int main(int argc, char* argv[]) 
{ 
int var,i;
i= log2(var);
printf("%d",i);
return 0;
}

Gives error

undefined reference to `log2'

Answer 1

In the first piece of code, the compiler replaces log2(16) with the constant 4 . The compiler usually optimizes constant math this way. That's the reason you don't see an error.

See the generated code for confirmation. This is for your first version:

main:
    pushl   %ebp
    movl    %esp, %ebp
    andl    $-16, %esp
    subl    $32, %esp
    movl    $4, 28(%esp)
    movl    $.LC0, %eax
    movl    28(%esp), %edx
    movl    %edx, 4(%esp)
    movl    %eax, (%esp)
    call    printf
    movl    $0, %eax
    leave
    ret

There is no call to log2. The compiler already replaced it by the constant 4 ( movl $4, 28(%esp) ).

This one is for your second version:

main:
    pushl   %ebp
    movl    %esp, %ebp
    andl    $-16, %esp
    subl    $48, %esp
    fildl   40(%esp)
    fstpl   (%esp)
    call    log2
    fnstcw  30(%esp)
    movzwl  30(%esp), %eax
    movb    $12, %ah
    movw    %ax, 28(%esp)
    fldcw   28(%esp)
    fistpl  44(%esp)
    fldcw   30(%esp)
    movl    $.LC0, %eax
    movl    44(%esp), %edx
    movl    %edx, 4(%esp)
    movl    %eax, (%esp)
    call    printf
    movl    $0, %eax
    leave
    ret

As you can see there is a call log2 in this version. Thats why -lm is needed for the second version.