如何在一个元素为零的数组中查找元素数 [英] How to find number of elements in an array which has one of its element as zero

问题描述

我已经声明了一个包含30个元素的数组：-

I have declared an array of 30 elements as:-

unsigned char some_array[30];

我将一些元素存储在数组中，例如：-

I store some elements in the array, for eg:-

{0xFB, 0x07, 0x0B, 0x02, 0x00, 0x0C, 0x03, 0x08, 0x0F, 0x0F, 0x07, 0x0A, 0x09, 0x06, 0x0D}

我需要找到此数组中的元素数。

I need to find the number of elements in this array.

sizeof (some_array) / sizeof (some_array[0]); 

产生30，这是数组可以存储的元素总数。

yields 30, which is the total number of elements the array can store.

strlen(some_array);

收益4，即数组中0x00元素的索引。

yields 4, which is the index of 0x00 element in the array.

如何确定数组中存储的元素数量？

How can i determine the number of elements stored in the array ??

推荐答案

实际值，您可以在整个数组上执行 memset ，这将设置一个值，您确定该值不会属于初始化的值范围，例如0xff。然后执行 for 循环并计数。

Before assigning the actual values, you can do a memset on the entire array which would set a value you are sure won't be part of your initialized range of values, example 0xff. Then do a for loop and count.

int cnt = 0;
memset (some_array, 0xff, 30);

for (i = 0; i < 30; i++) {
    if (some_array[i] != 0xff) {
        cnt++;
    }
}

printf ("%d\n", cnt);

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