c - C prime 生成器中 realloc 的看似未定义的行为(程序已发布，可运行)

Question

我编写了一个程序来查找质数，并使用 realloc() 来更改“*primes” block 的大小。当我输入大于 6 的任何位数时，我发现前两个值似乎是未定义的行为，我无法理解为什么会发生这种情况(后面的数字都是正确的)。请停一下。

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

typedef int int64_t;

void *safe_calloc(size_t,size_t);
void *safe_realloc(void*,size_t);

int main()
{
    printf("Prime Generator\nPlease enter the number of usable bits: ");
    int bits =0, maxNum=1, size = 6, elems = 3;

    scanf("%d",&bits);

    if(bits<1)
        exit(0);

    int i=0;
    for(; i < bits; i++)
        maxNum*=2;

    int *primes = safe_calloc(size*sizeof(int),sizeof(int));
    int *temp = safe_calloc(size*sizeof(int),sizeof(int));

    primes[0] = 1;
    primes[1] = 2;
    primes[2] = 3;

    int n = 3,j;
    for(; n < maxNum; n+=2){

        for(j = 2; j < elems; j++){

            if(n%primes[j] == 0)
                break;
            else if(j == elems-1){

            primes[elems++] = n;
            if((size-elems) < 2){

                for(i = 0; i < elems; i++)
                    printf("%d\n",primes[i]);
                printf("\n");

                size += 8;  // add 8 spaces to the prime storage array
                temp = safe_realloc(primes,size*sizeof(int));

                for(i = 0; i < elems; i++)
                    printf("%d\n",primes[i]);
                printf("\n");

                memmove(temp,primes,(size-8)*sizeof(int)); // copy over data to new array, just to be sure
                primes = temp;

                for(i = 0; i < elems; i++)
                    printf("%d\n",primes[i]);
                printf("\n");
                }
            }
        }
    }
    if(bits == 1){
        printf("1");
    }
    else{
        for(i = 0; i < elems; i++)
            printf("%d\n",primes[i]);
        printf("\n");
    }
    free(temp);
    free(primes);

    return 0;
}
void *safe_calloc(size_t length,size_t dataSize){
    int *tmp;
    if ((tmp = calloc(length,dataSize)) == NULL) {
        printf("ERROR: calloc failed");
        exit(0);
    }
    return tmp;
}
void *safe_realloc(void* ptr, size_t arraySize){
    int *tmp;
    if ((tmp = realloc(ptr,arraySize)) == NULL) {
        printf("ERROR: realloc failed");
        exit(0);
    }
    return tmp;
}

Answer 1

调用后不要使用你放入realloc 的指针——只使用返回的指针。您也不需要复制/memmove:

The contents of the object shall remain unchanged up to the lesser of the new and old sizes. If the new size of the memory object would require movement of the object, the space for the previous instantiation of the object is freed.