c - 如何将字符串赋值给 char* 指针？

Question

#include <stdio.h>

struct Analysis {
    int lnlen;   
    int arr[2]; 
    char* name;
};

int main()
{
    struct Analysis ana_space[2];
    char *ptr = (void*) &ana_space;

    ana_space[0].lnlen = 0;
    ana_space[0].arr[0] = 1;
    ana_space[0].arr[1] = 2;
    ana_space[0].name = "Peter";

    printf("\n%d\n", *ptr);     // print 0;

    *ptr = 10;                  // how to use memcpy here;

    printf("\n%d\n", *ptr);     // print 10;

    ptr = ptr + sizeof(int);    // advance pointer by int;

    printf("\n%d\n", *ptr);     // print 1;  

    ptr = ptr + 2*sizeof(int);  // advance pointer by 2 ints;

    printf("\n%s\n", *ptr);     // print "Peter"; --------------not work

    //*ptr = "Jim";             // how to assign new name "Jim" into that memory;
    return 0;
}

输出:

0

10

1

(null)

我想使用 char * 作为指针来遍历内存地址以获取一些数据并将值存储到内存中。

对于 int 和 int array，它工作正常，但不适用于字符串。

如何打印字符串并将新的字符串值存入内存？

Answer 1

您的方法不可移植。使用offsetof会更好以确保您可以可靠地指向 struct 成员的地址。

int main()
{
    struct Analysis ana_space[2];
    char *ptr = (void*) &ana_space;

    size_t offset1 = offsetof(struct Analysis, arr);
    size_t offset2 = offsetof(struct Analysis, name);

    ana_space[0].lnlen = 0;
    ana_space[0].arr[0] = 1;
    ana_space[0].arr[1] = 2;
    ana_space[0].name = "Peter";

    // advance pointer to point to arr.
    ptr = ptr + offset1;

    // advance pointer to point to name
    ptr = ptr + (offset2-offset1);

    // Cast the pointer appropriately before dereferencing.
    printf("\n%s\n", *(char**)ptr); 

    // how to assign new name "Jim" into that memory;
    *(char**)ptr = "Jim";
    printf("\n%s\n", *(char**)ptr);

    return 0;
}

您的使用:

printf("\n%d\n", *ptr);     // print 0;

*ptr = 10;                  // how to use memcpy here;

printf("\n%d\n", *ptr);     // print 10;

并且预期的输出是有缺陷的。它只适用于小端系统。我建议使用:

printf("\n%d\n", *(int*)ptr);

*(int*)ptr = 10;

printf("\n%d\n", *(int*)ptr);