Instereing answer with gcc and clang compiler
Instereing answer with gcc and clang compiler
One day, My companion Tsai ask me a question,“do you know (++i)(++i)(++i) equal answer？”, Then i wrote a program for run.
#include "stdio.h"
int main() {
int d=3;
printf("%d",(++i)+(++i)+(++i));
return 0;;
}
I GOT TWO VALUE FOR ANSWER!!
GCC-7.4.0 COMPILER:
$: gcc 运算符和表达式.c && ./a.out
$:16
CLANG-8.1.0 COMPILER:
$: clang 运算符和表达式.c && ./a.out
$:15
so, I reversed code for analysis.
#reversed assembly(AT&T format)
$: objdump -d a.out > gcc.dmp
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000000000000064a <main>:
64a: 55 push %rbp /赋值？/
64b: 48 89 e5 mov %rsp,%rbp /rsp，rbp为堆栈地址？/
64e: 48 83 ec 10 sub $0x10,%rsp /值为16/
652: c7 45 fc 03 00 00 00 movl $0x3,-0x4(%rbp) /把i=3长字节放在rbp堆栈/
659: 83 45 fc 01 addl $0x1,-0x4(%rbp) /i=i+1放入rbp堆栈/
65d: 83 45 fc 01 addl $0x1,-0x4(%rbp) /i=i+1放入rbp堆栈，此时i=5/
661: 8b 45 fc mov -0x4(%rbp),%eax /把rbp短字节(i=5)堆栈值，移动到寄存器ax/
64a: 55 push %rbp /赋值？/
64b: 48 89 e5 mov %rsp,%rbp /rsp，rbp为堆栈地址？/
64e: 48 83 ec 10 sub $0x10,%rsp /值为16/
652: c7 45 fc 03 00 00 00 movl $0x3,-0x4(%rbp) /把i=3长字节放在rbp堆栈/
659: 83 45 fc 01 addl $0x1,-0x4(%rbp) /i=i+1放入rbp堆栈/
65d: 83 45 fc 01 addl $0x1,-0x4(%rbp) /i=i+1放入rbp堆栈，此时i=5/
661: 8b 45 fc mov -0x4(%rbp),%eax /把rbp短字节(i=5)堆栈值，移动到寄存器ax/
664: 8d 14 00 lea (%rax,%rax,1),%edx /寄存器ax与寄存器ax相加放到寄存器dx/ /dx=10/
667: 83 45 fc 01 addl $0x1,-0x4(%rbp) /i=i+1=6放入rbp堆栈/
66b: 8b 45 fc mov -0x4(%rbp),%eax /把结果i=6移动到寄存器ax/
66e: 01 d0 add %edx,%eax /把寄存器dx与寄存器ax相加，也就是10+6/
670: 89 c6 mov %eax,%esi /把16移动到寄存器si/
672: 48 8d 3d 9b 00 00 00 lea 0x9b(%rip),%rdi # 714 <_IO_stdin_used+0x4> /我也没看懂/
679: b8 00 00 00 00 mov $0x0,%eax /寄存器ax赋值/
67e: e8 9d fe ff ff callq 520 <printf@plt> /调用函数printf/
683: b8 00 00 00 00 mov $0x0,%eax /寄存器ax赋值/
688: c9 leaveq
689: c3 retq
68a: 66 0f 1f 44 00 00 nopw 0x0(%rax,%rax,1)
the self-added operator seem not higher priority,it always self-add, and than register and register add.
Let’s see clang compiler
$: objdump -d a.out > clang.dmp
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0000000000400500 <main>:
400500: 55 push %rbp /赋值?/
400500: 55 push %rbp /赋值?/
400501: 48 89 e5 mov %rsp,%rbp
400504: 48 83 ec 10 sub $0x10,%rsp
400508: 48 bf e4 05 40 00 00 movabs $0x4005e4,%rdi
40050f: 00 00 00
400512: c7 45 fc 00 00 00 00 movl $0x0,-0x4(%rbp)
400519: c7 45 f8 03 00 00 00 movl $0x3,-0x8(%rbp)
400520: 8b 45 f8 mov -0x8(%rbp),%eax
400523: 83 c0 01 add $0x1,%eax /寄存器ax +1，此时寄存器ax的i=4/
400526: 89 45 f8 mov %eax,-0x8(%rbp) /把i=i+1结果给rbp堆栈/
400529: 8b 4d f8 mov -0x8(%rbp),%ecx /把rbp堆栈赋值给寄存器cx，rbp堆栈的值为4/
40052c: 83 c1 01 add $0x1,%ecx /寄存器cx +1，此时寄存器cx的i=5/
40052f: 89 4d f8 mov %ecx,-0x8(%rbp) /寄存器cx值给rbp堆栈，rbp堆栈的值为5/
400532: 01 c8 add %ecx,%eax /寄存器ax与寄存器cx相加，4+5=9，寄存器ax为9/
400534: 8b 4d f8 mov -0x8(%rbp),%ecx /把rbp堆栈的值给寄存器cx，此时寄存器cx的值i=5/
400537: 83 c1 01 add $0x1,%ecx /寄存器cx +1，此时i=6/
40053a: 89 4d f8 mov %ecx,-0x8(%rbp) /寄存器cx值给rbp堆栈，rbp堆栈的值为6/
40053d: 01 c8 add %ecx,%eax /寄存器cx与寄存器ax相加，6+9=15/
40053f: 89 c6 mov %eax,%esi /寄存器ax值赋值到寄存器si/
400541: b0 00 mov $0x0,%al /移动到寄存器al/
400543: e8 b8 fe ff ff callq 400400 <printf@plt>/调用fun函数printf/
400548: 31 c9 xor %ecx,%ecx /异或运算，cx与cx的值相同，结果为0/
40054a: 89 45 f4 mov %eax,-0xc(%rbp) /寄存器ax/
40054d: 89 c8 mov %ecx,%eax /寄存器cx的值给寄存器ax，结果是15/
40054f: 48 83 c4 10 add $0x10,%rsp /rsp堆栈为什么要填入10？/
400553: 5d pop %rbp
400554: c3 retq
400555: 66 2e 0f 1f 84 00 00 nopw %cs:0x0(%rax,%rax,1)
40055c: 00 00 00
40055f: 90 nop
The answer seems also correct, But I confuse what different from gcc and clang compiler…
#Conclusion
Seem gcc build less asm code than clang.
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CC BY 4.0
by the author.