Question

#include <stdlib.h>
#include <stdio.h>

int main()
{
    unsigned long c;
    unsigned long line;
    unsigned long word;
    char ch;
    char lastch = -1;

    c = 0;
    line = 0;
    word = 0;

    while((ch = getchar()) != EOF)
    {
        c ++;
        if (ch == '\n')
        {
            line ++;
        }
        if (ch == ' ' || ch == '\n')
        {
            if (!(lastch == ' ' && ch == ' '))
            {
                word ++;
            }
        }
        lastch = ch;
    }
    printf( "%lu %lu %lu\n", c, word, line );
    return 0;
}

因此，该程序计算标准输入中的字符，行或单词的数量。但其中一个要求是由任何符号（如！， - ，+等）分隔的单词必须被视为2个单词。我如何修改我的代码呢？

Answer 1

创建一个表示单词分隔的字符数组。修改while循环中的第二个if条件，检查数组中是否存在ch，并且该数组中不存在lastch。

修改后的代码：

#include <stdlib.h>
#include <stdio.h>

int main()
{
unsigned long c;
unsigned long line;
unsigned long word;
char ch;
char lastch = -1;
int A[256] = {0};

//Initialize the array indexes which are to be treated as separators.
//Next check the presence of any of these characters in the file.

A[' '] = 1; A['+'] = 1; A['!'] = 1; A['-'] = 1; A['\n'] = 1;
c = 0;
line = 0;
word = 0;

while((ch = getchar()) != EOF)
{
    c ++;
    if (ch == '\n')
    {
        line ++;
    }
    if (A[ch] == 1)
    {
        if (!(A[lastch] == 1 && A[ch] == 1))
        {
            word ++;
        }
    }
    lastch = ch;
}
printf( "%lu %lu %lu\n", c, word, line );
return 0;
}

Answer 2

只需按以下方式使用isalnum（）函数

#include <stdlib.h>
#include <stdio.h>

int main()
{
unsigned long c;
unsigned long line;
unsigned long word;
char ch;
char lastch = -1;

c = 0;
line = 0;
word = 0;

while((ch = getchar()) != EOF)
{
    c ++;
    if(ch=='\n')
      {
        line++;
        continue;
      }
    if (!isalnum(ch))
    {
        word++;
    }
}
printf( "%lu %lu %lu\n", c, word, line );
return 0;
}

将由符号分隔的单词计数为两个单词

2 个答案: