为什么realloc会在重复调用时失败,因为使用malloc分配大块有效？

Question

我试图从stdin读入(从文件传递值).我正在从字符串中读取每个字符并将其存储到动态分配的字符串指针中.需要时我会重新分配内存.我想尽可能多地填充字符.虽然我可以将其限制为100,000个字符.但是在一些迭代之后realloc失败了.但是如果我在malloc的第一次初始化期间指定一个大块的大块,比如1048567,我就能完全读取该字符串.为什么是这样？

以下是我的计划:

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

int display_mem_alloc_error();

enum {
    CHUNK_SIZE = 31    //31 fails. But 1048567 passes.
};

int display_mem_alloc_error() {
    fprintf(stderr, "\nError allocating memory");
    exit(1);
}

int main(int argc, char **argv) {
    int numStr;                  //number of input strings
    int curSize = CHUNK_SIZE;    //currently allocated chunk size
    int i = 0;                   //counter
    int len = 0;                 //length of the current string
    int c;                       //will contain a character
    char *str = NULL;            //will contain the input string
    char *str_cp = NULL;         //will point to str
    char *str_tmp = NULL;        //used for realloc

    str = malloc(sizeof(*str) * CHUNK_SIZE);
    if (str == NULL) {
        display_mem_alloc_error();
    }    
    str_cp = str;   //store the reference to the allocated memory

    scanf("%d\n", &numStr);   //get the number of input strings
    while (i != numStr) {
        if (i >= 1) {   //reset
            str = str_cp;
            len = 0;
            curSize = CHUNK_SIZE;
        }
        c = getchar();
        while (c != '\n' && c != '\r') {
            *str = (char *) c;
            //printf("\nlen: %d -> *str: %c", len, *str);
            str = str + 1;
            len = len + 1;
            *str = '\0';
            c = getchar();
            if (curSize / len == 1) {
                curSize = curSize + CHUNK_SIZE;
                //printf("\nlen: %d", len);
                printf("\n%d \n", curSize);    //NB: If I comment this then the program simply exits. No message is displayed.
                str_tmp = realloc(str_cp, sizeof(*str_cp) * curSize);
                if (str_tmp == NULL) {
                    display_mem_alloc_error();
                }
                //printf("\nstr_tmp: %d", str_tmp);
                //printf("\nstr: %d", str);
                //printf("\nstr_cp: %d\n", str_cp);
                str_cp = str_tmp;
                str_tmp = NULL;
            }
        }
        i = i + 1;
        printf("\nlen: %d", len);
        //printf("\nEntered string: %s\n", str_cp);
    }
    str = str_cp;
    free(str_cp);
    free(str);
    str_cp = NULL;
    str = NULL;
    return 0;
}

谢谢.

Answer 1

当你 realloc

str_tmp = realloc(str_cp, sizeof(*str_cp) * curSize);
if (str_tmp == NULL) {
    display_mem_alloc_error();
}
//printf("\nstr_tmp: %d", str_tmp);
//printf("\nstr: %d", str);
//printf("\nstr_cp: %d\n", str_cp);
str_cp = str_tmp;
str_tmp = NULL;

你让我们str_cp指向新的内存块,但str仍然指向旧的,现在的freed块.因此,当您访问str下一次迭代中指向的内容时,将调用未定义的行为.

您需要保存str相对于的偏移量str_cp,并且在重新分配之后,让我们str指向旧块的旧偏移量.

并且*str = (char *) c;是错误的,尽管它在功能上等同于正确的非零机会*str = c;.