将uint8_t [4]转换为uint32_t,然后在C中再次转换

Question

我写了这个测试将a转换uint8[4]为a uint32_t,但是如何将输出转换u32回原来的uint8[4]？我认为它可以通过位移以相同的方式完成,但我不确定如何...

uint32_t u32(uint8_t b[4]){
  uint32_t u;
  u = b[0];
  u = (u  << 8) + b[1];
  u = (u  << 8) + b[2];
  u = (u  << 8) + b[3];
  return u;
}

void p(uint32_t value){
        printf("\nuint32: %u\n",value);
}

int main(){
 uint8_t b[4];
 char a[4] = "test";
 char tmp[32];

 int i;
 for(i=0;i<4;i++){
   b[i] = a[i];
   sprintf(tmp,"%s%u",tmp,b[i]);
 }

 printf("uint8: ");

 for(i=0;i<4;i++)
   printf("%u",b[i]);

uint32_t t2 = u32(b);
 p(t2);
 return 0;
}

Answer 1

我试着做一点点不同并使用已知的功能而不是自己实现.我的主要观点是教你字节在内存中,你给它们的解释取决于上下文.请注意,有一些应该注意的字节顺序问题.此外,最好的教训是使用逐位操作.

#include <stdint.h>
#include <stdio.h>
#include <arpa/inet.h>
int main()
{
        uint8_t a[4]={0x1,0x2,0x3,0x4};
        uint32_t b = *((uint32_t*) a);
        /* turning an array to unit32 */
        printf("0x%x\n",htonl(b));
        /* turn an uint32 to array */
        uint32_t c = htonl(b);
        uint8_t d[4] = {0};
        printf("0x%x\n",c);
        for (int i=0; i<4 ;++i)
                d[i] = ((uint8_t*)&c)[3-i];
        for (int i=0; i<4 ;++i)
                printf("0x%x\n",d[i]);
        return 0;
}

输出是:

0x1020304
0x1020304
0x1
0x2
0x3
0x4

拆解它[你可以使用lldb或gdb加载程序]:

root# otool -tv a.out
a.out:
(__TEXT,__text) section
_main:
0000000100000e50    pushq   %rbp
0000000100000e51    movq    %rsp, %rbp
0000000100000e54    subq    $0x30, %rsp
0000000100000e58    movl    $0x0, 0xfffffffffffffffc(%rbp)
0000000100000e5f    movl    0x12b(%rip), %eax
0000000100000e65    movl    %eax, 0xfffffffffffffff8(%rbp)
0000000100000e68    movl    0xfffffffffffffff8(%rbp), %eax
0000000100000e6b    movl    %eax, 0xfffffffffffffff4(%rbp)
0000000100000e6e    movl    0xfffffffffffffff4(%rbp), %edi
0000000100000e71    callq   0x100000f50
0000000100000e76    leaq    0x117(%rip), %rdi
0000000100000e7d    movl    %eax, %esi
0000000100000e7f    movb    $0x0, %al
0000000100000e81    callq   0x100000f66
0000000100000e86    movl    0xfffffffffffffff4(%rbp), %edi
0000000100000e89    movl    %eax, 0xffffffffffffffe0(%rbp)
0000000100000e8c    callq   0x100000f50
0000000100000e91    movl    $0x0, %esi
0000000100000e96    movabsq $0x4, %rdx
0000000100000ea0    leaq    0xffffffffffffffec(%rbp), %rcx
0000000100000ea4    movl    %eax, 0xfffffffffffffff0(%rbp)
0000000100000ea7    movq    %rcx, %rdi
0000000100000eaa    callq   0x100000f60
0000000100000eaf    leaq    0xde(%rip), %rdi
0000000100000eb6    movl    0xfffffffffffffff0(%rbp), %esi
0000000100000eb9    movb    $0x0, %al
0000000100000ebb    callq   0x100000f66
0000000100000ec0    movl    $0x0, 0xffffffffffffffe8(%rbp)
0000000100000ec7    movl    %eax, 0xffffffffffffffdc(%rbp)
0000000100000eca    cmpl    $0x4, 0xffffffffffffffe8(%rbp)
0000000100000ed1    jge 0x100000efe
0000000100000ed7    movl    $0x3, %eax
0000000100000edc    subl    0xffffffffffffffe8(%rbp), %eax
0000000100000edf    movslq  %eax, %rcx
0000000100000ee2    movb    0xfffffffffffffff0(%rbp,%rcx), %dl
0000000100000ee6    movslq  0xffffffffffffffe8(%rbp), %rcx
0000000100000eea    movb    %dl, 0xffffffffffffffec(%rbp,%rcx)
0000000100000eee    movl    0xffffffffffffffe8(%rbp), %eax
0000000100000ef1    addl    $0x1, %eax
0000000100000ef6    movl    %eax, 0xffffffffffffffe8(%rbp)
0000000100000ef9    jmpq    0x100000eca
0000000100000efe    movl    $0x0, 0xffffffffffffffe4(%rbp)
0000000100000f05    cmpl    $0x4, 0xffffffffffffffe4(%rbp)
0000000100000f0c    jge 0x100000f3c
0000000100000f12    leaq    0x7b(%rip), %rdi
0000000100000f19    movslq  0xffffffffffffffe4(%rbp), %rax
0000000100000f1d    movzbl  0xffffffffffffffec(%rbp,%rax), %esi
0000000100000f22    movb    $0x0, %al
0000000100000f24    callq   0x100000f66
0000000100000f29    movl    %eax, 0xffffffffffffffd8(%rbp)
0000000100000f2c    movl    0xffffffffffffffe4(%rbp), %eax
0000000100000f2f    addl    $0x1, %eax
0000000100000f34    movl    %eax, 0xffffffffffffffe4(%rbp)
0000000100000f37    jmpq    0x100000f05
0000000100000f3c    movl    $0x0, %eax
0000000100000f41    addq    $0x30, %rsp
0000000100000f45    popq    %rbp
0000000100000f46    ret
0000000100000f47    nopw    (%rax,%rax)
__OSSwapInt32:
0000000100000f50    pushq   %rbp
0000000100000f51    movq    %rsp, %rbp
0000000100000f54    movl    %edi, 0xfffffffffffffffc(%rbp)
0000000100000f57    movl    0xfffffffffffffffc(%rbp), %edi
0000000100000f5a    bswapl  %edi
0000000100000f5c    movl    %edi, %eax
0000000100000f5e    popq    %rbp
0000000100000f5f    ret

Answer 2

您的u32()函数是阅读大端数字的与主机端无关的代码的一个很好的练习。
它当然可以轻松反转，只需从结尾开始并转到函数的开头，反汇编您的uint32_t：

void u8from32 (uint8_t b[4], uint32_t u32)
    b[3] = (uint8_t)u32;
    b[2] = (uint8_t)(u32>>=8);
    b[1] = (uint8_t)(u32>>=8);
    b[0] = (uint8_t)(u32>>=8);
}