将字符串作为指针或文字传递时，strcmp（）返回值不一致

Question

I was playing around with strcmp when I noticed this, here is the code:

#include <string.h>
#include <stdio.h>

int main(){

    //passing strings directly
    printf("%d\n", strcmp("ahmad", "fatema"));

    //passing strings as pointers 
    char *a= "ahmad";
    char *b= "fatema";
    printf("%d\n",strcmp(a,b));

    return 0;

}

the output is:

-1
-5

shouldn't strcmp work the same? Why is it that I am given different value when I pass strings as "ahmad" or as char* a = "ahmad" . When you pass values to a function they are allocated in its stack right?

Answer 1

You are most likely seeing the result of a compiler optimization. If we test the code using gcc on godbolt , with -O0 optimization level, we can see for the first case it does not call strcmp :

movl    $-1, %esi   #,
movl    $.LC0, %edi #,
movl    $0, %eax    #,
call    printf  #

Since your are using constants as arguments to strcmp the compiler is able for perform constant folding and call a compiler intrinsic at compile time and generate the -1 then, instead of having to call strcmp at run-time which is implemented in the standard library and will have a different implementation then a likely more simple compile time strcmp .

In the second case it does generate a call to strcmp :

call    strcmp  #
movl    %eax, %esi  # D.2047,
movl    $.LC0, %edi #,
movl    $0, %eax    #,
call    printf  #

This is consistent with the fact that gcc has a builtin for strcmp , which is what gcc will use during constant folding.

If we further test using -O1 optimization level or greater gcc is able to fold both cases and the result will be -1 for both cases:

movl    $-1, %esi   #,
movl    $.LC0, %edi #,
xorl    %eax, %eax  #
call    printf  #
movl    $-1, %esi   #,
movl    $.LC0, %edi #,
xorl    %eax, %eax  #
call    printf  #

With more optimizations options turned on the optimizer is able to determine that a and b point to constants known at compile time as well and can also compute the result of strcmp for this case as well during compile time.

We can confirm that gcc is using builtin function by building with the -fno-builtin flag and observing that a call to strcmp will be generated for all cases.

clang is slightly different in that it does not fold at all using -O0 but will fold at -O1 and above for both.

Note, that any negative result is an entirely conformant, we can see by going to the draft C99 standard section 7.21.4.2 The strcmp function which says ( emphasis mine ):

 int strcmp(const char *s1, const char *s2); 

The strcmp function returns an integer greater than, equal to, or less than zero , accordingly as the string pointed to by s1 is greater than, equal to, or less than the string pointed to by s2.

technosurus points out that strcmp is specified to treat the strings as if they were composed of unsigned char , this is covered in C99 under 7.21.1 which says:

For all functions in this subclause, each character shall be interpreted as if it had the type unsigned char (and therefore every possible object representation is valid and has a different value).

Answer 2

I think you believe that the value returned by strcmp should somehow depend on the input strings passed to it in a way that is not defined by the function specification. This isn't correct. See for instance the POSIX definition:

http://pubs.opengroup.org/onlinepubs/009695399/functions/strcmp.html

Upon completion, strcmp() shall return an integer greater than, equal to, or less than 0, if the string pointed to by s1 is greater than, equal to, or less than the string pointed to by s2, respectively.

This is exactly what you are seeing. The implementation does not need to make any guarantee about the exact return value - only that is less than zero, equal to zero, or greater than zero as appropriate.