realloc()：int数组的下一个大小无效

Question

我试图找出作为参数的整数下的素数。

它的工作方式是，有一个循环，从上一个素数+2开始，再到参数平方下的素数。

正如您所看到的，这是递归的，其中的基本情况是当参数小于6时，它返回1。有一个以{3}开头的素数的全局列表，当找到一个新的素数时会得到更多的内存。

不管怎么说这是密码。我放了一些printfs进行调试。

#include <stdio.h>
#include <math.h>
#include <stdlib.h>
int number_of_primes = 1;
int s(int number);
int *primes;
int main(int argc, char *argv[])
{
    primes = (int *)malloc(sizeof(int));
    primes[0] = 3;
    printf("\nNumber of primes under %i is %i\n", atoi(argv[1]), s(atoi(argv[1])) + 1);
    free(primes);
}

int s(int number)
{
    printf("the number is %i\n", number);
    if (number < 5)
    {
        return 1;
    }
    int limit = s((int)sqrt(number)) + 1;
    printf("limit is %i\n", limit);
    // For every odd number after the last prime to number
    for (int odd = primes[limit - 2]; odd < number + 1; odd = odd + 2)
    {
        printf("primes are: ");
        for (int i = 0; i < number_of_primes; i++)
        {
            printf("%i, ", primes[i]);
        }
        printf("\n");
        printf("\ncurrent odd is %i, number is %i, limit is %i\n", odd, number, limit);
        // If it is not a multiple of any of any primes
        for (int prime_index = 0; prime_index < limit; prime_index++)
        {
            printf("prime_index is %i, and primes[prime_index] is %i\n", prime_index, primes[prime_index]);
            if (primes[prime_index])
            {
                if (odd % primes[prime_index] == 0)
                {
                    break;
                }
            }
            if (prime_index == limit - 1)
            {
                printf("new_prime_is %i, number_of_primes is %i, and primes[number_of_primes - 1] is %i\n", odd, number_of_primes, primes[number_of_primes - 1]);
                primes = realloc(primes, sizeof(int));
                primes[number_of_primes] = odd;
                number_of_primes++;
            }
        }
    }
    return number_of_primes;
}

由于某种原因，这在29岁之前是可行的。然后我得到

the number is 29
the number is 5
the number is 2
limit is 2
primes are: 3, 

current odd is 3, number is 5, limit is 2
prime_index is 0, and primes[prime_index] is 3
primes are: 3, 

current odd is 5, number is 5, limit is 2
prime_index is 0, and primes[prime_index] is 3
prime_index is 1, and primes[prime_index] is 0
new_prime_is 5, number_of_primes is 1, and primes[number_of_primes - 1] is 3
limit is 3
primes are: 3, 5, 

current odd is 5, number is 29, limit is 3
prime_index is 0, and primes[prime_index] is 3
prime_index is 1, and primes[prime_index] is 5
primes are: 3, 5, 

current odd is 7, number is 29, limit is 3
prime_index is 0, and primes[prime_index] is 3
prime_index is 1, and primes[prime_index] is 5
prime_index is 2, and primes[prime_index] is 0
new_prime_is 7, number_of_primes is 2, and primes[number_of_primes - 1] is 5
primes are: 3, 5, 7, 

current odd is 9, number is 29, limit is 3
prime_index is 0, and primes[prime_index] is 3
primes are: 3, 5, 7, 

current odd is 11, number is 29, limit is 3
prime_index is 0, and primes[prime_index] is 3
prime_index is 1, and primes[prime_index] is 5
prime_index is 2, and primes[prime_index] is 7
new_prime_is 11, number_of_primes is 3, and primes[number_of_primes - 1] is 7
primes are: 3, 5, 7, 11, 

current odd is 13, number is 29, limit is 3
prime_index is 0, and primes[prime_index] is 3
prime_index is 1, and primes[prime_index] is 5
prime_index is 2, and primes[prime_index] is 7
new_prime_is 13, number_of_primes is 4, and primes[number_of_primes - 1] is 11
primes are: 3, 5, 7, 11, 13, 

current odd is 15, number is 29, limit is 3
prime_index is 0, and primes[prime_index] is 3
primes are: 3, 5, 7, 11, 13, 

current odd is 17, number is 29, limit is 3
prime_index is 0, and primes[prime_index] is 3
prime_index is 1, and primes[prime_index] is 5
prime_index is 2, and primes[prime_index] is 7
new_prime_is 17, number_of_primes is 5, and primes[number_of_primes - 1] is 13
primes are: 3, 5, 7, 11, 13, 17, 

current odd is 19, number is 29, limit is 3
prime_index is 0, and primes[prime_index] is 3
prime_index is 1, and primes[prime_index] is 5
prime_index is 2, and primes[prime_index] is 7
new_prime_is 19, number_of_primes is 6, and primes[number_of_primes - 1] is 17
primes are: 3, 5, 7, 11, 13, 17, 19, 

current odd is 21, number is 29, limit is 3
prime_index is 0, and primes[prime_index] is 3
primes are: 3, 5, 7, 11, 13, 17, 19, 

current odd is 23, number is 29, limit is 3
prime_index is 0, and primes[prime_index] is 3
prime_index is 1, and primes[prime_index] is 5
prime_index is 2, and primes[prime_index] is 7
new_prime_is 23, number_of_primes is 7, and primes[number_of_primes - 1] is 19
primes are: 3, 5, 7, 11, 13, 17, 19, 23, 

current odd is 25, number is 29, limit is 3
prime_index is 0, and primes[prime_index] is 3
prime_index is 1, and primes[prime_index] is 5
primes are: 3, 5, 7, 11, 13, 17, 19, 23, 

current odd is 27, number is 29, limit is 3
prime_index is 0, and primes[prime_index] is 3
primes are: 3, 5, 7, 11, 13, 17, 19, 23, 

current odd is 29, number is 29, limit is 3
prime_index is 0, and primes[prime_index] is 3
prime_index is 1, and primes[prime_index] is 5
prime_index is 2, and primes[prime_index] is 7
new_prime_is 29, number_of_primes is 8, and primes[number_of_primes - 1] is 23
realloc(): invalid next size
Aborted (core dumped)

当我运行英勇时，我会得到一堆无效的读物。

Answer 1

你的内存管理错了。

两者均：

primes = (int *)malloc(sizeof(int));
/* and */
primes = realloc(primes, sizeof(int));

只为一个整数分配内存。在分配的内存之外访问分配的对象，它是一个UB (未定义的行为)，它通过显示错误来显示自己。

要分配(或为size整数分配内存)，您需要：

primes = malloc(size * sizeof(*primes));
/* and */
int *tmp = realloc(primes, size * sizeof(*primes));
if(tmp) primes = tmp;
else {/* memory allocation error handling*/ }

使用sizeof中的对象而不是类型