无法理解 C 中的字数统计 [关闭]

Question

[已解决] 我最近问了一个关于一些我认为我非常理解的代码的问题。

但几天后，当我回顾修订时，新的问题来自同一段代码（来自 Brian. W. Kernighnan 的 THE C PROGRAMMING LANGUAGE 第二版（ISBN-13：978-8131704943））。

代码如下：

#include <stdio.h>

#define IN  1
#define OUT 0

main()
{
    int nw, c, state;
    state = OUT;
    nw = 0;
    while ((c = getchar()) != EOF)
    {
        if (c == ' ' || c == '\n' || c == '\t')
            state = OUT;
        else if (state == OUT)
        {
            state = IN;
            ++nw;
        }
    }
    printf("%d", nw);
}

所以我做了一些随机测试，发现了我无法解释或理解的东西：

1. 为什么我有的时候代码不起作用

   state = OUT 在nw = 0 之后

当我通常运行此代码时，它给了我正确的编号。单词，但是当我交换state = OUT and nw = 0 的顺序时，它总是返回等于 0 的答案，为什么会这样？ 我知道顺序在 C 中很重要，但为什么只有这个特定的顺序？？

2.if-else语句的简要含义是什么？

3.Acc。到书国家变种。被定义为定义 getchar 是否在一个单词中，但我不明白 state var 的确切含义。这样做？？

4.另外作者使用等价和相等是什么？ 我注意到他在制定条件时使用等价，否则使用等价，对吗？？

谢谢...

Answer 1

这是您发布的一些 cmets 的代码：

#include <stdio.h>
#define IN  1 // in the code below, whenever you see IN, it's replaced by 1
#define OUT 0 // same as IN, but for 0

int main(void) {
  int nw, c, state; // nw is the number of the words
  // c stands for the character we get from input and state is the state that are
  // currently.

  // init state as OUT
  state = OUT;
  // init counter to 0
  nw = 0;

  // while user doesn't give EOF
  while ((c = getchar()) != EOF) {

    // we found whitespace, newline or tab
    if (c == ' ' || c == '\n' || c == '\t')
      state = OUT;                            // put state as OUT,
    // so that we do not count them as words
    else if (state == OUT) {                  // enter here only if state is OUT
      // Now, we see that we found a letter/number, which means that a word
      // was typed by the user.

      // set state as IN, so that we remember that we are eating the characters of
      // the word given (in the next loops)
      state = IN;

      // increase the counter of the words
      ++nw;
    }
  }

  // print the number of words received by the user
  printf("%d", nw);
  return 0;
}

现在，让我们看一个示例输入：

sam
dad

当然输出（你应该从上面的 cmets 猜到）是：

2

让我们逐步运行（有趣的）代码输入sam：

// input: "sam" (without the quotes)

while ((c = getchar()) != EOF) { // eat first character, i.e. 's'

  /****** 1st execution of the loop ********************/

  // 's' is not going into this if
  if (c == ' ' || c == '\n' || c == '\t')
    state = OUT;
  // first time we execute the loop so state is OUT,
  // thus we enter the loop
  else if (state == OUT) {

    // set state as IN
    state = IN;

    // increase the counter of the words
    ++nw;
  }

  /****************************************************/

  /****** 2nd execution of the loop ********************/
  // while loop's getchar() gives as 'a'

  // 'a' is not going into this if
  if (c == ' ' || c == '\n' || c == '\t')
    state = OUT;
  // state is IN, so we don't go into this loop
  else if (state == OUT) {
    state = IN;
    ++nw;
  }

  /****************************************************/


  /****** 3rd execution of the loop ********************/
  // same as 2nd, but for 'm'

  /****************************************************/


  /****** 4rth execution of the loop ********************/
  // while loop's getchar() gives as '\n'

  // '\n' is going into this if (the c == '\n' is true
  if (c == ' ' || c == '\n' || c == '\t')
    state = OUT;                // state sets to OUT

  // we are not going into this if else, since we already entered the above if!
  else if (state == OUT) {
    state = IN;
    ++nw;
  }

  /****************************************************/


  /****** 5th execution of the loop ********************/
  // while loop's getchar() gives as 'd' (first letter of dad)

  // 'd' is not going into this if
  if (c == ' ' || c == '\n' || c == '\t')
    state = OUT;
  // state is OUT, so we go into this loop
  else if (state == OUT) {
    // set state as IN
    state = IN;

    // increase counter
    ++nw;
  }

  /****************************************************/

  // and so on :)
}

首先确保您理解这一点，然后回答您自己提出的问题。 :)

另外，我们称之为代码，而不是脚本。不要急于进入脚本。