C数组计数(初学者)

问题描述

我目前正在阅读Kernighan&的"The C Programming Language" Richie和我正在努力弄清生产线的功能.我想我只是有点愚蠢，还不太了解他们的解释.

I'm currently reading 'The C Programming Language' by Kernighan & Richie and I'm struggling to work out what a line does. I think I'm just being a little stupid, and not quite understanding their explanation.

++ndigit[c-'0'];

我不得不稍稍更改程序，就像以前ndigit给我的垃圾值一样，所以我只是将数组实例化为0，而不是使用for循环遍历它，并以这种方式更改值.

I've had to change the program ever so slightly as ndigit was previously giving me garbage values, so I just instantiate the array to 0, instead of traversing it with a for loop, and changing the values that way.

#include <stdio.h>

main()
{
    int c, i, nwhite, nother;
    int ndigit[10]= {0};

    nwhite = nother = 0;

    while ((c = getchar()) != EOF)
        if (c >= '0' && c <= '9')
            ++ndigit[c-'0'];
        else if (c == ' ' || c == '\n' || c == '\t')
            ++nwhite;
        else
            ++nother;

    printf("digits =");
    for (i = 0; i < 10; i++)
        printf (" %d", ndigit[i]);
    printf (", white space = %d, other = %d\n", nwhite, nother);
}

使用程序作为输入，我们将其打印到控制台-

Using the program as its input, we get this printed to the console -

digits = 7 2 0 0 0 0 0 0 0 1, white space = 104, other = 291

我了解7 2 0 0 0 0 0 0 0 1是对输入中出现的单个数字出现0次出现7次，出现1次出现两次等的计数.)

I understand that 7 2 0 0 0 0 0 0 0 1 is a count of how many times the single numbers appear in the input 0 appears 7 times, 1 appears twice etc.)

但是，...[c-'0'];的工作方式"如何?

But, how does the ...[c-'0']; 'work'?

推荐答案

您问以下表达式的工作原理

You asked how the below expression works

 c-'0'

从ASCII码0中减去所输入字符的ASCII码，它定义了数组中必须存储计数的位置.

The ASCII code of the entered charecter is subtracted from the ASCII code of 0 and it defines the position in the array where the count has to be stored .

假设您从键盘输入 1 ，其中 1 的ASCII码为49，而 0 的ASCII码为48.因此

Suppose you enter 1 from the keyboard ASCII code for 1 is 49 and ASCII code for 0 is 48.hence

   49-48 =1

，计数将存储在数组索引位置1中.

and the count will be stored in the array index location 1 .

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