如何在特定的内存位置创建数组？ c

发布于 2025-01-21 18:14:50 字数 627 浏览 3 评论 0原文

char string1[] = "dog";
char *string2 = string1 + strlen(string1) + 1;
printf("address of string1: %p\n", string1);
printf("address of string2: %p\n", string2);
string2 = malloc(sizeof(char)*4);
string2[0] ='c';string2[1] ='a';string2[2] ='t';string2[3] ='\0';
printf("address of string2: %p\n", string2);

字符串的地址1：0000000061FE14
字符串的地址2：000000000061FE18
字符串的地址2：00000000000000AD4D50

我想在内存位置0x61fe18，

即使我设法在“ dog \ 0”之后，我想在内存位置0x61fe1。成功获得String2的所需内存位置2，Malloc函数重新分配了该位置。我希望“ CAT”数组从0x61Fe18开始，但是Malloc将其重新分配到0xad4d50

是否有办法将数组分配到所需的起始位置？

原文

char string1[] = "dog";
char *string2 = string1 + strlen(string1) + 1;
printf("address of string1: %p\n", string1);
printf("address of string2: %p\n", string2);
string2 = malloc(sizeof(char)*4);
string2[0] ='c';string2[1] ='a';string2[2] ='t';string2[3] ='\0';
printf("address of string2: %p\n", string2);

address of string1: 000000000061FE14
address of string2: 000000000061FE18
address of string2: 0000000000AD4D50

I want to make a 'cat' string at a memory location 0x61FE18, right after "dog\0"

Even though I've managed to successfully get the desired memory location of string2, the malloc function reassigns the location. I want the "cat" array to start at 0x61FE18 but malloc reassigns it to 0xAD4D50

Is there a way to allocate array to a desired starting location?

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

韶华倾负 2025-01-28 18:14:51

你的意思是这样吗？

unsigned char *data = malloc(1000);
if (data) {
    strcpy((char*)data, "cat"); // `data` points to "cat" followed by garbage
    strcpy((char*)data + 4, "dog"); // `data` points to "cat\0dog" garbage
    (int*)(data+24) = 0x01020304; // `data` points to "cat\0dog\0XXXXXXXXXXXXXXXX\1\2\3\4"
                                  // depending on endianness and sizeof int
    free(data);
}

you mean something like this?

unsigned char *data = malloc(1000);
if (data) {
    strcpy((char*)data, "cat"); // `data` points to "cat" followed by garbage
    strcpy((char*)data + 4, "dog"); // `data` points to "cat\0dog" garbage
    (int*)(data+24) = 0x01020304; // `data` points to "cat\0dog\0XXXXXXXXXXXXXXXX\1\2\3\4"
                                  // depending on endianness and sizeof int
    free(data);
}

回复收藏 0 原文