I want to know why strcmp() returns different values if used more than once in the same function. Below is the program. The first case I am aware of why it prints -6. But in the second case, why does it print -1?
#include<stdio.h>
#include<string.h>
int main()
{
char a[10] = "aa";
char b[10] = "ag";
printf("%d\n",strcmp(a, b));
printf("%d\n",strcmp("aa","ag"));
return 0;
}
And the output it produces is below
[sxxxx@bhlingxxx test]$ gcc -Wall t51.c
[sxxxx@bhlingxxx test]$ ./a.out
-6
-1
Why is the output of second strcmp() -1? Is it the Compiler who plays here? If so What is the exact optimization it does?
The C standard says the following regarding the return value of strcmp:
Section 7.24.4.2p3:
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
So as long as the result fits that description it is compliant with the C standard. That means the compiler can perform optimizations to fit that definition.
If we look at the assembly code:
.loc 1 7 0
leaq -32(%rbp), %rdx
leaq -48(%rbp), %rax
movq %rdx, %rsi
movq %rax, %rdi
call strcmp
movl %eax, %esi
movl $.LC0, %edi
movl $0, %eax
call printf
.loc 1 8 0
movl $-1, %esi # result of strcmp is precomputed!
movl $.LC0, %edi
movl $0, %eax
call printf
In the first case, arrays are passed to strcmp to a call to strcmp and a call to printf are generated. In the second case however, string constants are passed to both. The compiler sees this and generates the result itself, optimizing out the actual call to strcmp, and passes the hardcoded value -1 to printf.