请帮忙解释一下这个用memcpy的程序
这个程序用了memcpy得到的结果是
0 1072693248 0 1073741824 0 1074266112 0 1074790400 0 1075052544
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书上的解释是“memcpy不知道也不关心数据类型；它只是把一些字节从一个位置复制到另一个位置。复制过程也不进行数据转抽象换。如果使用循环对元素逐个赋值，那么在赋值过程中会将double类型转换成int类型值。此时，对字节按原样进行复制，然后程序把数据作为int类型进行解释。”
我对这段话的理解是memcpy 不能在不同类型的数据间复制。对吗？
#include "stdafx.h"
#include "stdio.h"
#include "string.h"
#include "stdlib.h"
int main(void)
{
int a[10] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
double b[5] = {1.0, 2.0, 3.0, 4.0, 5.0};
memcpy(a, b, 5*sizeof(double));
for(int i=0; i<10; i++)
{
printf("%d ", a[i]);
}
putchar('\n');
return 0;
}
/*
*void * memcpy ( void * destination, const void * source, size_t num );
Copy block of memory
Copies the values of num bytes from the location pointed by source
directly to the memory block pointed by destination.
The underlying type of the objects pointed by both the source and destination
pointers are irrelevant for this function; The result is a binary copy of the data.
The function does not check for any terminating null character in source - it
always copies exactly num bytes.
To avoid overflows, the size of the arrays pointed by both the destination and
source parameters, shall be at least num bytes, and should not overlap
(for overlapping memory blocks, memmove is a safer approach).
*/
#include <stdio.h>
#include <string.h>
void get_byte(void *s, void *d, size_t size)
{
/* -----------------------------
*
[解决办法]
0
[解决办法]
1
[解决办法]
2
[解决办法]
3
[解决办法]
* -----------------------------
*32bit,4byte
*/
asm (
"loop_start:"
"movl (%%esi), %%ebx\n\t"
"bswap %%ebx\n\t"
//获取0
"movl $0, %%edx\n\t"
"movb %%bl, %%dl\n\t"
"movl %%edx, (%%edi)\n\t"
//获取1
"movl $0, %%edx\n\t"
"movb %%bh, %%dl\n\t"
"movl %%edx, 4(%%edi)\n\t"
"bswap %%ebx\n\t"
//获取2
"movl $0, %%edx\n\t"
"movb %%bh, %%dl\n\t"
"movl %%edx, 8(%%edi)\n\t"
//获取3
"movl $0, %%edx\n\t"
"movb %%bl, %%dl\n\t"
"movl %%edx, 12(%%edi)\n\t"
"addl $4, %%esi\n\t"
"addl $16, %%edi\n\t"
"loop loop_start\n\t"
:
:"S"(s),"D"(d),"c"(size));
}
void get_byte_2(int *s, int *d, size_t size)
{
int i, j;
for(i = 0; i < size; ++i)
{
for(j = 0; j < 4; ++j)
{
*(d + 4 * i + j) = (*s >> (8 * j)) & 0xff;
}
}
}
int main(void)
{
int a[10] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, a_byte[40] = {0}, b_byte[40] = {0};
double b[5] = {1.0, 2.0, 3.0, 4.0, 5.0};
int i = 0, j = 0;
void *p = b, *q = p + 1;
//上面给出两中方法，效果一样的，但是做法不一样！选择一个即可
//获取b各变量的字节值，和复制后做比较
get_byte(b, b_byte, 10);//按32位处理，故传入10
//获取a各个变量的字节值
get_byte(a, a_byte, 10);
//测试a的字节和b的字节关系
for(i = 0; i < 40; ++i)
{
if(*(a_byte + i) != *(b_byte + i))
break;
}
if(i == 40)
printf("equal!\n");
else
printf("unequal!\n");
//字节拷贝
memcpy(a, b, 40);
//重新获取a的各字节值
get_byte(a, a_byte, 10);
//再测试ab的关系
for(i = 0; i < 40; ++i)
{
if(*(a_byte + i) != *(b_byte + i))
break;
}
if(i == 40)
printf("equal!\n");
else
printf("unequal!\n");
return 0;
}
