Question

While Learning compiler Optimisation, I write codes in C under Linux with GCC version gcc version 4.4.5 (Ubuntu/Linaro 4.4.4-14ubuntu5.1)
To understant not a statement (nop) in C. I written two codes first y.c second x.c and generate their compiled assembly code using gcc -S option.

Fist code y.c

desktop:~$ cat y.c
main()
{
int i=0;
}
desktop:~$ gcc -S y.c
desktop:~$ 

Second code x.c

desktop:~$ cat x.c
main()
{
int i=0;

/* Loops and if*/
while(0);
for(;0;);
if(0);

/* Arithmetic Operations */
i * i;
i / i;
i - i;
i + i;
i % i;
 +i;
 -i;

/* Comparison operators */
i == i;
i != i;
i < i;
i > i;
i >= i;
i <= i;

/* Bitwise Operations*/
i ^ i;
i | i;
i & i;
 ~i;
i << i;
i >> i;

/* Address Operatins*/ 
 &i;
 *(&i);
 (&i)[i];

/* Ternary conditional operation*/
 i? i : i;

/* Other Operations*/ 
 sizeof(i);
 (char)i;

/* Not-Logical Operation*/ 
 !i; 

/* Logical AND , OR Operators */ 
// i && i;  // Commented && operation 
// i || i;  // Commented || operation
}
desktop:~$ gcc -S x.c

NOTICE: This time Last two lines in x.c are commented.
And as I was expecting. There is no difference in their generated assembly codes. I compared x.s and y.s using diff command.

desktop:~$ diff x.s y.s
1c1
<   .file   "x.c"
---
>   .file   "y.c"
desktop:~$

But When I uncomment last(or say add) two lines in x.c.

i && i;  
i || i; 

Again compile x.c with -S option and compared with y.s.

desktop:~$ tail  x.c  
 sizeof(i);
 (char)i;

/* Not-Logical Operation*/ 
 !i; 

/* Logical AND , OR Operators */ 
i && i;  // unCommented && operation 
i || i;  // unCommented || operation
}
desktop:~$ gcc -S x.c
desktop:~$ diff x.s y.s
1c1
<     .file    "x.c"
---
>     .file    "y.c"
10,21d9
<     movl    -4(%ebp), %eax
<     testl    %eax, %eax
<     je    .L3
<     movl    -4(%ebp), %eax
<     testl    %eax, %eax
< .L3:
<     movl    -4(%ebp), %eax
<     testl    %eax, %eax
<     jne    .L8
<     movl    -4(%ebp), %eax
<     testl    %eax, %eax
< .L8:
desktop:~$ 

Question:
I just can not understand why expressions i || i and i && i are not equivalent to 'not a statement'?

Why compiler transfer this two statements into executable( we can disassemble with objdump will get same code). What is special in this two expression. they do not includes = operation.

Does they change (set/reset) CPU flag-registers ? I guess no!

Even / division operation discarded that may cause an divided by zero fault.

EDIT : Added answer

There is nothing special in i || i and i && i expressions. Both are equivalent to
NOT A STATEMENT. And can be removed by GCC compiler with some extra effort.

To remove this: -o2 and -o3 flags are useful:
Here is my Try!!

desktop:~$ gcc -o2 -S y.c 
desktop:~$ gcc -o2 -S x.c 
desktop:~$ diff x.s y.s -y
    .file   "x.c"                         |     .file   "y.c"
    .text                               .text
    .p2align 4,,15                        <
.globl main                         .globl main
    .type   main, @function                     .type   main, @function
main:                               main:
    pushl   %ebp                            pushl   %ebp
    movl    %esp, %ebp                      movl    %esp, %ebp
    popl    %ebp                          |     subl    $16, %esp
                                  >     movl    $0, -4(%ebp)
                                  >     leave

The extra lines in at RHS are because of misalignment between files.

I also like to add information that JAVA and C# compilers discard this expressions without any flags.

Answer 1

Enable optimizations using -O2, and you should see the extra code disappear.