需要有关此memcpy和数组代码的帮助 [英] Need help with this memcpy and array code

问题描述

我想在主函数中合并2个数组.但是由于某些原因，即使我可以使array3 [0]和array3 [1]具有正确的值，但是当它们复制到数据包中时，它们的值是不同的.知道为什么吗?
感谢您的帮助.

I want to merge the 2 arrays in the main function. But for some reason, even though I can get array3[0] and array3[1] to have the right values, when they are copied into the packet, their values are different. Any idea why ?
Thanks for your help.

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

unsigned char *function1() 
{
    unsigned char array2[] = { 0x4a,0xb2 };
    return (array2 );
}

main()
{
   unsigned char temp[] = { 0xaa, 0x0b, 0x03,0x04,0x05,0x06,0x07,0x08,0x09 };
   unsigned char x[2];
   unsigned char *packet;
   int pkt_len;
   pkt_len = sizeof(temp) + sizeof(x);
   packet = (unsigned char*) malloc ( pkt_len +1);

   memset( packet, 0x00, pkt_len +1);
   unsigned char *pointer1 = malloc ( sizeof(temp) + 1);

   memset( pointer1, 0x00, sizeof(temp) +1);
   memcpy (pointer1, temp, sizeof(temp) );

   memcpy (packet, pointer1, sizeof(temp) );
   printf("\nPacket before copy is 0x%x\n", packet[8]);

   unsigned char *array2 = malloc ( sizeof (x) + 1)  ;
   array2 = (char *)function1();
   printf("\nArray2 is 0x%x\n", array2[0]);
   memcpy (packet + sizeof(temp), array2, sizeof(x) );
   printf("After copy, Packet contents are 0x%x\n", packet[9]);
}

推荐答案

首先:您确定要从function1()返回局部变量的地址吗?函数返回时，没有保证array2将指向有意义的内存块.

其次:您确定您的程序是有效的C吗?我不再是C方面的专家，但是在C89中完全声明，在不声明语句之后声明变量. array2和pointer1是我正在查看的变量.

第三:您正在进行大量的复制和内存设置，因此很难知道您实际上打算做什么.例如:

First off: Are you sure you want to return an address of a local variable from function1()? There''s no guarentee that array2 will point to a meaningful block of memory when the function returns.

Secondly: Are you sure your program is valid C? I''m not an expert on C anymore but declaring variables after non declaring statements was a complete no-no in C89. array2 and pointer1 are the variables I''m looking at in particular.

Thirdly: You''ve got loads of redundant copying and memsetting going on so it''s a bit hard to see what you''re actually intending to do. For example:

memset( pointer1, 0x00, sizeof(temp) +1);  
memcpy (pointer1, temp, sizeof(temp) );

可以替换为:

could be replaced with:

memcpy ( pointer1, temp, sizeof temp );
pointer1[ sizeof temp] = 0x00;

第四:在C语言中，您无需强制转换malloc的返回值. void *与任何指针类型兼容.

第五:每当有经验的C程序员看到sizeof(x)时，他们便开始寻找称为x的类型.取一个对象的大小时，不要使用花括号.

第六:指针1的要点是什么?似乎完全是多余的.

如果您只想动态创建一个内存块并将其他两个数组复制到其中，则可以用大约三行代码完成:
-分配内存
-将第一个数组复制到分配的内存中
-将第二个数组复制到第一个数组之后的已分配内存中

干杯，

灰

Fourthly: In C you don''t need to cast the return value of malloc. void * is compatible with any pointer type.

Fifthly: Whenever an experienced C programmer sees sizeof( x ) they start looking for a type called x. Leave off the braces when taking the size of an object.

Sixth: What''s the point of pointer1? It seems totally redundant.

If all you want to is dynamically create a lump of memory and copy two other arrays into it you can do it in about three lines:
- allocate the memory
- copy the first array into the allocated memory
- copy the second array into the allocated memory after the first array

Cheers,

Ash

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