问题描述
我想在组装中将两个字节char byte1
和char byte2
连接成单个short
。
我该怎么办?使用班次?
我正在使用IA32
解决方法
我只是解决了这个问题,并在有人遇到相同问题的情况下这样做:
concatBytes.s:
.section .data
.global byte1
.global byte2
.section .text
.global concatBytes
concatBytes:
#prologue
pushl %ebp
movl %esp,%ebp
pushl %ebx
#body of the function
movl $0,%eax
movb byte1,%al
movb byte2,%ah
#epilogue
popl %ebx
movl %ebp,%esp
popl %ebp
ret
,
main.c
#include <stdio.h>
#include "concatBytes.h"
char byte1 = '11101010';
char byte2 = '10100001';
int main()
{
short result = 0;
result = concatBytes();
printf("Result = %hd",result);
return 0;
}
concatBytes.h
short concatBytes(void);
concatBytes.c
extern char byte1;
extern char byte2;
short concatBytes(void)
{
return(((short)byte1)^((short)byte2));
}
很明显:
gcc main.c concatBytes.c -o main
main.c:3:14: warning: character constant too long for its type
char byte1 = '11101010';
^
main.c:3:14: warning: overflow in implicit constant conversion [-Woverflow]
main.c:4:14: warning: character constant too long for its type
char byte2 = '10100001';
^
main.c:4:14: warning: overflow in implicit constant conversion [-Woverflow]
这是错误的语法。所以这引出了一个问题,你是说这个意思吗?
#include <stdio.h>
#include "concatBytes.h"
char byte1[] = "11101010";
char byte2[] = "10100001";
int main()
{
short result = 0;
result = concatBytes();
printf("Result = %hd",result);
return 0;
}
或者这个:
#include <stdio.h>
#include "concatBytes.h"
char byte1 = 0xEA;
char byte2 = 0xA1;
int main()
{
short result = 0;
result = concatBytes();
printf("Result = %hd",result);
return 0;
}
假设后者:
0000000000400559 <concatBytes>:
400559: 55 push %rbp
40055a: 48 89 e5 mov %rsp,%rbp
40055d: 0f b6 15 d4 0a 20 00 movzbl 0x200ad4(%rip),%edx # 601038 <byte1>
400564: 0f b6 05 ce 0a 20 00 movzbl 0x200ace(%rip),%eax # 601039 <byte2>
40056b: 31 d0 xor %edx,%eax
40056d: 66 98 cbtw
40056f: 5d pop %rbp
400570: c3 retq
这使您大致了解调用约定,然后只需用“连接”替换该代码的中间部分
如果它是一个字符串,那么您首先需要将每个字符串转换为一个字节,然后进行连接。您可以轻松地找出其中一个...