Question

我需要帮助才能清楚地理解此代码，请帮忙。我无法弄清楚这个程序如何跟踪响应数组中给出的数量。我不明白for循环中的内容，特别是这行++频率[回复[answer]];

    #include<stdio.h>
    #define RESPONSE_SIZE 40
    #define FREQUENCY_SIZE 11

    int main(void)
    {
        int answer; /* counter to loop through 40 responses */
        int rating; /* counter to loop through frequencies 1-10 */

        /* initialize frequency counters to 0 */

        int frequency[FREQUENCY_SIZE] = {0};

        /* place the survey responses in the responses array */

        int responses[RESPONSE_SIZE] = {1,2,6,4,8,5,9,7,8,10,1,6,3,8,6,10,3,8,2,7,6,5,7,6,8,6,7,5,6,6,5,6,7,5,6,4,8,6,8,10};

        /* for each answer, select value of an element of array responses
        and use that value as subscript in array frequency to determine element to increment */

        for(answer = 0 ; answer < RESPONSE_SIZE; answer++){
            ++frequency[responses[answer]];
        }

        printf("%s%17s\n", "Rating", "Frequency");

        /* output the frequencies in a tabular format */
        for(rating = 1; rating < FREQUENCY_SIZE; rating++){
            printf("%6d%17d\n", rating, frequency[rating]);
        }

        return 0;
    }

Answer 1

for(answer = 0 ; answer < RESPONSE_SIZE; answer++){
    ++frequency[responses[answer]];     //  <---
}

上面的循环只计算一个数字出现在数组responses中的次数，并存储在数组frequency中该数字的索引处。这一行在第一个循环中执行 -

 ++frequency[responses[answer]];

因此，它会增加数组responses[answer]的索引frequency的值。

假设responses[answer]的值为1，则数组1的索引frequency的值会递增。

第二个for循环仅用于输出，如上所述。

Answer 2

++frequency[responses[answer]]是一种密集的写作方式

int r = response[answer];
frequency[r] = frequency[r] + 1;

警告frequency[r]仅评估一次。

因此，如果answer等于0，则responses[answer]等于1，因此我们将1添加到frequency[1]。

修改

下表显示了frequency通过循环发生的情况（旧值=＆gt;新值）：

answer response[answer] frequency[response[answer]] ------ ---------------- --------------------------- 0 1 frequency[1]: 0 => 1 1 2 frequency[2]: 0 => 1 2 6 frequency[6]: 0 => 1 3 4 frequency[4]: 0 => 1 ... ... ... 10 1 frequency[1]: 1 => 2

等

需要帮助理解这个C代码（数组）

2 个答案: