C 控制流，内存分配：中止陷阱 6，用字符串中的字符填充数组，嵌套 for 循环

Question

这是一个关于控制流和可能的内存分配的问题。

给定一个 long long 数字，我正在执行以下操作：

将其转换为字符串
以规定的模式遍历此字符串的字符（for 循环的第一层）
对每个选定的字符执行操作（for 循环的第二层）
用处理后的数据填充数组

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

int main(void)
{
    long long n = 12345678; // I am given a number

    char str[8]; // initialize string of length 8
    sprintf(str, "%2lld", n); // convert n to string
    printf("The string is: %s\n", str); // check that n is converted to string

    int arr[4]; // initialize array of length 4

    for (int i = 6; i >= 0; i -= 2) // select every other char in string, starting from second-to-last char
        {
            for (int j = 0; j < 4; j++) // select position of array
            {       
                arr[j] = (str[i] - '0') * 2; // convert char to int, multiply by 2, and assign to array position
                printf("The digit is %c and the product is %d\n", str[i], arr[j]); // announce each entry to the array
            }
        }

    for (int k = 0; k < 4; k++) // print contents of array
    {
        printf("The product at position %d is %d\n", k, arr[k]);
    }
    return 0;
}

这段代码有两个问题：

在 MacOS 终端上运行时生成 Abort trap: 6 错误。
当代码在实践 IDE 环境中成功执行时，会生成以下结果：

The string is: 12345678
The digit is 7 and the product is 14
The digit is 7 and the product is 14
The digit is 7 and the product is 14
The digit is 7 and the product is 14
The digit is 5 and the product is 10
The digit is 5 and the product is 10
The digit is 5 and the product is 10
The digit is 5 and the product is 10
The digit is 3 and the product is 6
The digit is 3 and the product is 6
The digit is 3 and the product is 6
The digit is 3 and the product is 6
The digit is 1 and the product is 2
The digit is 1 and the product is 2
The digit is 1 and the product is 2
The digit is 1 and the product is 2
The product at position 0 is 2
The product at position 1 is 2
The product at position 2 is 2
The product at position 3 is 2

我想要的是解决 Abort trap: 6 错误并得到以下结果：

The string is: 12345678
The digit is 7 and the product is 14
The digit is 5 and the product is 10
The digit is 3 and the product is 6
The digit is 1 and the product is 2
The product at position 0 is 14
The product at position 1 is 10
The product at position 2 is 6
The product at position 3 is 2

我应该在代码中更改什么以实现该目标？

我已经阅读了其他关于 Abort trap: 6 的帖子，但我不知道这段代码是如何在内存中犯错误的 allocation/usage。

Answer 1

您不需要 for 循环来递增 j 作为数组索引，只需在开始时设置为 0 并在每个循环中递增它。

for (int i = 6, j = 0; i >= 0; i -= 2, j++) // select every other char in string, starting from second-to-last char
{
    arr[j] = (str[i] - '0') * 2; // convert char to int, multiply by 2, and assign to array position
    printf("The digit is %c and the product is %d\n", str[i], arr[j]); // announce each entry to the array
}

正如@Keine Lust 在评论中提到的，您必须为空终止字符分配 space。

char str[8];

改为

char str[9];

Answer 2

您的 str 数组不够长：

char str[8]; // initialize string of length 8

要保存一个包含 8 个字符的字符串，您需要一个包含 9 个字节的数组：8 个用于相关字符，一个用于终止空字节。

因为您没有为字符串预留足够的 space，所以您最终写到了数组的末尾。这样做会调用 undefined behavior，在本例中表现为崩溃。

增加数组的大小，以免踩到无效内存：

char str[9];

至于您看到的输出，您在外循环的每次迭代中都覆盖了 arr[j] 的值。不是嵌套循环，而是在 i 和 j.

上迭代的单个循环

C 控制流，内存分配：中止陷阱 6，用字符串中的字符填充数组，嵌套 for 循环

C control flow, memory allocation: Abort trap 6, populate array with characters in string, nested for loops

c

arrays

debugging

for-loop

nested-loops