Question

我想每行打印一个单词，但输出不打印每个单词的最后一个字符。任何人都可以帮助找到错误？

#include <stdio.h>  
#define IN 1  /* Inside a text */  
#define OUT 0  /* Outside a text */ 
main()  
{  
    int c,nw,state;  
    state=OUT;  
    while((c=getchar())!=EOF) 
    {  
       if(c==' '||c=='\n'||c=='\t')  
       {  
            state=OUT;  
            printf("\n");                      
       }  

       else if(state==OUT)  
       {  
            putchar(c);   /* To print the first character */
            state=IN;  
            ++nw;  
            c=getchar();  
            putchar(c);  /* To print the other characters of the word*/
       }    
    }  
}

使用上面的代码，不会打印每个单词的最后一个字符。

Answer 1

我认为可以在不维持州的情况下解决。欢迎提出建议

#include <stdio.h>

int main(void) {
    // your code goes here
    int c;
    while((c=getchar())!=EOF)
    {
        if(c==' ' || c=='\t' || c=='\b')
        {
            printf("\n");
            while(c==' ' || c=='\t' || c=='\b')
            c=getchar();
        }
        if(c!=EOF)
        putchar(c);
    }
    return 0;
}

Answer 2

要修复的示例

#include <stdio.h>  

#define IN  1 /* Inside a text */  
#define OUT 0 /* Outside a text */ 

int main(void){
    int c, nw=0, state=OUT;

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

Answer 3

我认为这是解决方案

#include <stdio.h>

main()
{
    int c, inspace;
    inspace = 0;


    while ((c=getchar())!= EOF) 
    {
        if (c == ' ' || c == '\t' || c == '\n')
        {
            if (inspace == 0)
            {
                putchar('\n');
                inspace = 1;
            }
        }

        else 
        {
            putchar(c);
            inspace = 0;
        }

    }
}

Answer 4

for( ; (c = getchar()) != EOF; )
{
    flag = 0;
    if(c == '\n' || c == '\t' || c == ' ')
        { state = OUT;

          }
    else if(state == OUT){
         flag = 1;
         state = IN;
        }
    if(flag == 1 && count!=0)
       putchar('\n');

    if(state == IN)
        putchar(c);

    count++;

 }

Answer 5

你可以试试这个。

function addTimeout(fn, timeout) {
    if (timeout === undefined) timeout = 1000; // default timeout
    return function(...args) {
        return new Promise(function(resolve, reject) {
            const tid = setTimeout(reject, timeout,
                new Error("promise timed out"));
            fn(...args)
                .then(function(...args) {
                    clearTimeout(tid);
                    resolve(...args);
                })
                .catch(function(...args) {
                    clearTimeout(tid);
                    reject(...args);
                });
        });
    };
}

Answer 6

#include<stdio.h>

 int main(){
   int c;
    while((c=getchar())!=EOF)      
       if(c=='\t'|| c=='\n' || c==' ')
             putchar('\n');
                else
                   putchar(c);

          return 0;
  }

Answer 7

每行打印输入一个单词的程序

OP的代码肯定与c=getchar(); putchar(c);的以下内容有关，因为它打印c是字母，空格，'\n'或EOF。< / p>

else if(state==OUT) {  
  putchar(c);
  state=IN;  
  ++nw;  
  c=getchar();  
  putchar(c);  // Error to print without testing `c`
}

OP state想法后只需要进行一些更改。

#include <stdio.h>
#include <ctype.h>
#define IN 1  /* Inside text */
#define OUT 0  /* Outside text */

int main(void) {
  unsigned long long wc = 0;  // no need for a narrow `int`, how about something wider?
  int state = OUT;
  int ch;
  while ((ch = getchar()) != EOF) {
    if (state == OUT) {
      if (!isspace(ch)) {
        wc++;        // new word begun
        state = IN;
        putchar(ch);
      }
    } else /* state == IN */ {
      if (isspace(ch)) {   // add test
        putchar('\n');
        state = OUT;
      } else {
        putchar(ch);
      }
    }
  }
  if (state == IN) { // add test in case last word not followed by a white-space
    putchar('\n');
  }
  printf("Word count:%llu\n", wc);
  return 0;
}

Answer 8

这是@BLUEPIXY的一个版本，仅在添加字符时才打印字符，然后在换行符上将单词分开，并在换行符的末尾填充换行符。

#include <stdio.h>
#define IN 1  /*Inside of a text*/
#define OUT 0 /*outside of a text*/
#define SPACE ' '
#define NEWLINE '\n'
#define TAB '\t'

int main()
{
  int nw = 0, c, state = OUT;

  while ((c = getchar()) != EOF)
  {
    if (c == SPACE || c == NEWLINE || c == TAB)
    {
      /* Pads the last character(a new line) with another new line */
      if (c == NEWLINE)
      {
        putchar('\n');
      }
      if (state == IN)
      {
        state = OUT;
        putchar('\n');
      }
    }
    else
    {
      if (state == OUT)
      {
        state = IN;
        ++nw;
      }
      putchar(c);
    }
  }
  printf("\nTotal word count: %d\n", nw);
  return 0;
}

Answer 9

其中一条评论说，如果没有#defines或标志，这是可以解决的。确实有可能。而且它使代码非常短。这是我的解决方案：

int main()
{
    int c;

    while ((c = getchar()) != EOF)
    {
        if (c != ' ')
        {
            putchar(c);
        }
        else
        {
            printf("\n");
        }
    }
}

Answer 10

使用state，但仅呼叫getchar。

#include <stdio.h>

#define IN 1
#define OUT 0

int main() {
    int c;
    int state;

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

Answer 11

您还可以：

#include <stdio.h>

#define IN  1
#define OUT 0

main()
{
    int state;
    int c;
    state = OUT;
    while ((c = getchar()) != EOF) {
        if ((c != '\n') && (c != ' ') && (c != '\t')) {
            putchar(c);
            state = IN;
        }
        else if (state == IN) {
            putchar('\n');
            state = OUT;

        }
    }
}

一个程序，每行打印一个输入的单词

11 个答案: