无法根据用户的字符串动态分配内存； realloc：下一个大小无效

如何解决无法根据用户的字符串动态分配内存； realloc：下一个大小无效

我正在尝试最小化以下程序的内存分配。我试图动态分配内存的整个思路是

1. 为30个字节的用户字符串输入动态分配内存
2. 如果用户的输入大于分配的内存，请调用realloc（）函数以使字符串适合101个字节
3. 如果用户的输入大于100个字节，请调用exit（）并终止程序。

当我实现realloc（）函数以检查用户输入是否适合动态分配的char数组（与calloc一致）时，我的程序失败，并显示以下输出：

realloc(): invalid next size
Aborted (core dumped)

我不确定该如何解决。

代码：

#include <stdio.h>  //gets() was removed from this library
#include <stdlib.h> //for calloc()
#include <string.h> //for strlen()
#include <unistd.h> //for sleep(2)
int len = 0;
struct item
{
    //i cant make sure pointer points to nothing because structures are only a templates.
    //i cant do operations here!!!!!!!
    char *itemName;
    int qty;
    float price;
    float amount;
};

int readItem(struct item *pProduct)

{
    printf("hey,what is the product name: ");
    gets(pProduct->itemName);

    //checking whether user's input fits into the allocated pointer size
    len = strlen(pProduct->itemName);
    printf("%d\n\n",len);

    if (len > 30)
    {
        //error here
        pProduct->itemName = (char *)realloc(pProduct->itemName,101);

        printf("\nyour input is too big. re enter it\n");
        gets(pProduct->itemName);

        if (pProduct->itemName == NULL)
        {
            puts("memory is not avaible");
            exit(1);
        }
        len = strlen(pProduct->itemName);
        printf("%d\n\n",len);
        if (len > 100)
        {
            puts("memory is not avaible");
            exit(1);
        }
    }
    printf("\nEnter the product quantity: ");
    scanf("%d",&pProduct->qty);

    printf("\nEnter the product price: ");
    scanf("%f",&pProduct->price);

    //calculating the amount
    pProduct->amount = (float)pProduct->qty * pProduct->price;

    return 0;
}

void printItem(struct item *pProduct)
{
    puts("\n\nthe database:");
    sleep(1);

    printf("----------------------\n");
    printf("Name: \"%s\"\n",(*pProduct).itemName);
    printf("Price: %.2f\n",pProduct->price);
    printf("Quantity: %d\n",pProduct->qty);
    printf("Amount: %.2f\n",pProduct->amount);
    printf("----------------------\n");
}

int main()
{

    //assigning variable product and piinter pProduct to a tag name item,struct item product,*pProduct = NULL;
    //storing an address to a pointer pProduct
    //making a structure to a pointer
    pProduct = &product;

    //(char *) is casting char to a pointer type

    pProduct->itemName = (char *)calloc(30,sizeof(char));

    readItem(pProduct);
    printItem(pProduct);

    free(pProduct->itemName);
    pProduct->itemName = NULL;
    pProduct = NULL;

    return 0;
}

我知道REALLOC确实会造成麻烦。更准确地说，如果我输入34个字符，realloc似乎可以工作，但是如果将其加倍至68，尽管realloc中提到了大小，它仍会引发错误

我对调用栈的了解为零，但这对某人可能有用：

    libc.so.6!__GI_raise(int sig) (/build/glibc-YYA7BZ/glibc-2.31/sysdeps/unix/sysv/linux/raise.c:50)
     
    libc.so.6!__GI_abort() (/build/glibc-YYA7BZ/glibc-2.31/stdlib/abort.c:79)
     
    libc.so.6!__libc_message(enum __libc_message_action action,const char * fmt) (/build/glibc-YYA7BZ/glibc-2.31/sysdeps/posix/libc_fatal.c:155)
     
    libc.so.6!malloc_printerr(const char * str) (/build/glibc-YYA7BZ/glibc-2.31/malloc/malloc.c:5347)

    libc.so.6!_int_realloc(mstate av,mchunkptr oldp,size_t oldsize,size_t nb) (/build/glibc-YYA7BZ/glibc-2.31/malloc/malloc.c:4564)
     
    libc.so.6!__GI___libc_realloc(size_t bytes,void * oldmem) (/build/glibc-YYA7BZ/glibc-2.31/malloc/malloc.c:3226)
     
    libc.so.6!realloc_hook_ini(void * ptr,size_t sz,const void * caller) (/build/glibc-YYA7BZ/glibc-2.31/malloc/hooks.c:41)
     
    readItem(struct item * pProduct) (/home/max/My stuff/programming/Udemy/section 14,reading and wrinting files/test.c:22)
     
    main() (/home/max/My stuff/programming/Udemy/section 14,reading and wrinting files/test.c:72)

解决方法

建议；

分配一个长缓冲区，比可能需要的更长。然后通过调用fgets()

获得数据

char buffer[1024];
fgets( buffer,sizof( buffer ),stdin );

-或-

使用getline()函数读取数据

char *buffer = NULL;
size_t byteCount;
getline( &line,&byteCount,stdin );

注意：为了使图示简单，我没有进行上面的错误检查。