C 编码 :: 将结构写入内存块

C Coding :: Writing Structs into a Chunk of Memory

我正在做一个必须用标准 C 编码的项目。(不是 C++。)前一段时间,我写了下面的程序,将不同结构的内容写入二进制文件:

#include <stdio.h>
#include <stdlib.h>
    
typedef struct structA{
        int a, b, c;
}AAA;
typedef struct structB{
        int a, b, c, d, e;
}BBB;
    
int main( int argc, char **argv ){

        // Create and open the output file
        FILE *fd = fopen("test.txt", "w");

        AAA* a1 = (AAA*)malloc( sizeof(AAA) );
        AAA* a2 = (AAA*)malloc( sizeof(AAA) );
        BBB* b1 = (BBB*)malloc( sizeof(BBB) );

        a1->a = 1;   a1->b = 2;   a1->c = 3;
        a2->a = 4;   a2->b = 5;   a2->c = 6;
        b1->a = 10;  b1->b = 20;  b1->c = 30;  b1->d = 40;  b1->e = 50;

        // Write all these structs to the file:
        fwrite((char*)&a1, sizeof(AAA), 1, fd);
        fwrite((char*)&a2, sizeof(AAA), 1, fd);
        fwrite((char*)&b1, sizeof(BBB), 1, fd);

        // Close the file
        fclose( fd );

        free( a1 );
        free( a2 );
        free( b1 );

        printf("END OF PROGRAM.\n");
        return 0;
}

上面的代码完美运行……尽管我不能通过查看输出来判断:

me@ubuntu:/home/me# more test.txt
▒$▒&V
me@ubuntu:/home/me#

我有另一个程序可以读取这个文件并从结构中提取所有信息。所以我知道上面的代码正是我想要的。

但是现在,我需要将这些结构写入 一块分配的内存,而不是写入文件。我认为这很容易:

#include <stdio.h>
#include <stdlib.h>
    
typedef struct structA{
        int a, b, c;
}AAA;
typedef struct structB{
        int a, b, c, d, e;
}BBB;
    
int main( int argc, char **argv ){

        u_char* BlockOfMemory = (u_char*) malloc( sizeof(u_char) * 100 );

        AAA* a1 = (AAA*)malloc( sizeof(AAA) );
        AAA* a2 = (AAA*)malloc( sizeof(AAA) );
        BBB* b1 = (BBB*)malloc( sizeof(BBB) );

        a1->a = 1;   a1->b = 2;   a1->c = 3;
        a2->a = 4;   a2->b = 5;   a2->c = 6;
        b1->a = 10;  b1->b = 20;  b1->c = 30;  b1->d = 40;  b1->e = 50;

        // Write all these structs into BlockOfMemory:
        memcpy ( BlockOfMemory, &a1, sizeof( AAA ) );
        memcpy ( (BlockOfMemory+sizeof(AAA)), &a2, sizeof( AAA ) );
        memcpy ( (BlockOfMemory+sizeof(AAA)+sizeof(AAA)), &b1, sizeof( BBB ) );

        printf("==>  %hhn\n", BlockOfMemory);

        free( a1 );
        free( a2 );
        free( b1 );
        free( BlockOfMemory );

        printf("END OF PROGRAM.\n");
        return 0;
}

成功了吗?我不知道:

me@ubuntu:/home/me# gcc -Wall writeBlock.c
me@ubuntu:/home/me# ./a.out
==>
END OF PROGRAM.
me@ubuntu:/home/me#

这里的objective是内存块必须包含与二进制文件完全相同的信息。我在我的代码编译和运行的奇怪情况下,但考虑到我拥有的工具(VI 和 GCC),我无法验证我的代码是否正确或偏离目标。

有人可以指教吗?另外, memcpy() 会是这里使用的函数吗?谢谢。

编辑:修复了第一个程序时我错误地添加了第二个“free( b1 );”因为剪切粘贴错误。

正如已经指出的,您正在将指针 a1 的地址传递给 memcpy,a2b1 也是如此。应该是

memcpy ( BlockOfMemory, a1, sizeof( AAA ) );
memcpy ( (BlockOfMemory+sizeof(AAA)), a2, sizeof( AAA ) );
memcpy ( (BlockOfMemory+sizeof(AAA)+sizeof(AAA)), b1, sizeof( BBB ) );

如果你想打印 BlockOfMemory 的内容,你需要转换回 AAA*BBB* 并用指针运算移动,就像这样

unsigned char* tmp = BlockOfMemory;
AAA* x = (AAA*)tmp;
printf("==>  %d %d %d\n", x->a, x->b, x->c);

tmp += sizeof(AAA);
x = (AAA*)tmp;
printf("==>  %d %d %d\n", x->a, x->b, x->c);

tmp += sizeof(AAA);
BBB* y = (BBB*)tmp;
printf("==>  %d %d %d %d %d\n", y->a, y->b, y->c, y->d, y->e);