有关在C'不是说明书“优化？ [英] Regarding optimization for 'not a statment' in c?
问题描述
虽然学习的编译器优化，我在 C 在的Linux 写codeS与 GCC 版本 gcc版本4.4.5（Ubuntu的/ Linaro的4.4.4-14ubuntu5.1）结果
要understant 不是一个声明（NOP）的C.我写了两codeS第一 YC 第二个 XC 和生成自己编译的程序集code 使用 GCC -S 选项。
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.
拳code y.c
desktop:~$ cat y.c
main()
{
int i=0;
}
desktop:~$ gcc -S y.c
desktop:~$
二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
注意：这次在 x.c 最后两行被注释结果。
正如我所期待的。有一个在其生成的汇编codeS没有区别。我比较 x.s 和 y.s 使用差异命令。
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:~$
不过，当我取消最后一个（或说加）在 x.c 两行。
i && i;
i || i;
再编译-S选项x.c和Y.S.比较
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:~$
问：结果
我只是不明白，为什么前pressions 我||我和 I和;＆安培;我并不等同于'不是一个声明？
Question:
I just can not understand why expressions i || i and i && i are not equivalent to 'not a statement'?
为什么编译器传递这两个语句转换为可执行（我们可以用 objdump的拆卸将得到同样的code）。什么是这两位前pression特别。他们不包括 = 操作。
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.
他们是否改变（设置/重置）CPU标志寄存器？我想没有！
Does they change (set/reset) CPU flag-registers ? I guess no!
即使 / 除法运算丢弃，可能会导致被零故障的划分。
Even / division operation discarded that may cause an divided by zero fault.
修改：新增答案
有什么特别的我||我和 I和;＆安培;我前pressions。两者都相当于
不是一条语句。并且可以通过 GCC编译器有一些额外的努力被删除。
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.
要删除此： -o2 和 -03 标志是有用的：结果
这里是我的尝试！
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
多余的线条在RHS是因为文件之间的错位。
The extra lines in at RHS are because of misalignment between files.
我也想补充信息 JAVA 和 C＃编译器放弃这个前pressions不带任何标志。
I also like to add information that JAVA and C# compilers discard this expressions without any flags.
推荐答案
使用 -O2 启用优化，你应该可以看到额外的code消失。
Enable optimizations using -O2, and you should see the extra code disappear.
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