Question

我是一名崭露头角的C程序员，我编写了这个程序来计算第二版K＆amp; R 1.5.4的单词。我的if语句有问题吗？代码似乎在不应该增加变量时增加，因为它不符合初始测试。

#include <stdio.h>
#define IN 1
#define OUT 0


main()
{
      int word, state, c;

      word = state = OUT;

      while((c = getchar()) != EOF)
      {
               if(c != ' ' || c != '\n' || c != '\t')
               {
                    if(state == OUT)
                    {
                             word++;
                             state = IN;
                    }

                    /*else
                    {
                         state = IN;
                    }*/
               }

               if(c == ' ' || c == '\n' || c == '\t')
               {
                   state = OUT;
               }

               printf("char: %c %x | state: %d | word: %d\n", c, c, state, word); 
      }

      printf("\n//maniac: %d\n", word);

这导致：

>count_word_my.exe
Hello  he   he
char: H 48 | state: 1 | word: 1
char: e 65 | state: 1 | word: 1
char: l 6c | state: 1 | word: 1
char: l 6c | state: 1 | word: 1
char: o 6f | state: 1 | word: 1
char:   20 | state: 0 | word: 1
char:   20 | state: 0 | word: 2
char: h 68 | state: 1 | word: 3
char: e 65 | state: 1 | word: 3
char:   20 | state: 0 | word: 3
char:   20 | state: 0 | word: 4
char:   20 | state: 0 | word: 5
char: h 68 | state: 1 | word: 6
char: e 65 | state: 1 | word: 6
char:
 a | state: 0 | word: 6

char:
 a | state: 0 | word: 7

//maniac: 7

我修改过的K＆amp; R代码：

#include <stdio.h>

#define IN 1 /* inside a word */
#define OUT 0 /* outside a word */

/* count lines, words, and characters in input */
main()
{
      int c, nl, nw, nc, state;

      state = OUT;
      nl = nw = nc = 0;
      while ((c = getchar()) != EOF) {
            ++nc;
            if (c == '\n')
               ++nl;
            if (c == ' ' || c == '\n' || c == '\t')
               state = OUT;
            else if (state == OUT) {
                 state = IN;
                 ++nw;
            }
            printf("char: %c %x | state: %d | word: %d\n", c, c, state, nw);
      }
      printf("%d %d %d\n", nl, nw, nc);
}

K＆amp; R代码导致：

>count_word_test.exe
Hello  he   he
char: H 48 | state: 1 | word: 1
char: e 65 | state: 1 | word: 1
char: l 6c | state: 1 | word: 1
char: l 6c | state: 1 | word: 1
char: o 6f | state: 1 | word: 1
char:   20 | state: 0 | word: 1
char:   20 | state: 0 | word: 1
char: h 68 | state: 1 | word: 2
char: e 65 | state: 1 | word: 2
char:   20 | state: 0 | word: 2
char:   20 | state: 0 | word: 2
char:   20 | state: 0 | word: 2
char: h 68 | state: 1 | word: 3
char: e 65 | state: 1 | word: 3
char:
 a | state: 0 | word: 3

char:
 a | state: 0 | word: 3
2 3 16
^C

当代码在第一个if语句中不满足测试时，如何在'Hello'之后处理第二个空格（0x20）时，我的代码如何递增word / nw？即使它确实达到了第二个if语句，我也认为它会将'state'变量设置为1（IN）。我错过了一些关键的东西。我非常感谢给予的任何帮助。谢谢。

Answer 1

好吧，每个char都会将此评估为真if(c != ' ' || c != '\n' || c != '\t')，因为' '，'\n'和'\t'都没有字符。

应该是：

if(c != ' ' && c != '\n' && c != '\t')

Answer 2

你已经（至少）得到了一个好的答案（我已经投了票）。但请让我详细说明“以防万一”：

if(c == ' ' || c == '\n' || c == '\t') ... 你清楚地理解这一点：如果是“space”或“newline”或“tab”，那么“whitespace == true”。

......但...... if (c != ' ' || c != '\n' || c != '\t') ...说“如果不是空格......或者不是换行符......或者不是标签”

换句话说，'\ n'会评估为“whitespace == false”...因为'\ n'不是空白，而且它不是标签。

你的真正含义是if (c &= ' ' && c != '\n' && c != '\t') ... ......或if (!(c == ' ' || c == '\n' || c == '\t')) ...

换句话说，问题是“C语法”而不是“布尔逻辑”。

这里有两个简短的教程“沉重的理论”......但你可能会喜欢：

K＆amp; R在C中计算1.5.4字

2 个答案: