每行打印一個單詞

Question

我需要編寫一個程序，每行打印其輸入一個單詞。 這是到目前為止我得到的：

#include <stdio.h>
main(){
    int c;
    while ((c = getchar()) != EOF){
        if (c != ' ' || c!='\n' || c!='\t')
            printf("%c", c);
        else 
            printf("\n");
    }
}

邏輯很簡單。 我檢查輸入是否不是換行符，制表符或空格，並在這種情況下將其打印出來，否則將打印換行符。

當我運行它時，會得到如下結果：

input-->  This is
output--> This is

它打印整個內容。 這里出什么問題了？

Answer 1

if (c != ' ' || c!='\\n' || c!='\\t')這將永遠不會為假。

也許您的意思是： if (c != ' ' && c!='\\n' && c!='\\t')

Answer 2

按照上面的注釋，也不要使用printf嘗試putchar，也應該使用&&代替||。

這是我的代碼-

#include<stdio.h>

main()
{
    int c, nw;                  /* nw for word & c for character*/

    while ( ( c = getchar() ) != EOF ){

    if ( c != ' ' && c != '\n' && c != '\t')
    nw = c;
    else {
        nw = '\n';
    }

    putchar (nw);

    }

}

該代碼將為您提供所需的輸出

Answer 3

您可以使用string.h庫中的strtok函數，該函數可以通過提供定界符將輸入切成多個單詞。

這是一段注釋完美的代碼，可以滿足您的需求

#include <stdio.h> 
#include <string.h> 

int main(int argc, char *argv[])
{
    char line[1000]=""; // the line that you will enter in the input
    printf("Input the line:\n>>");
    scanf("%[^\n]",line); // read the line till the you hit enter button
    char *p=strtok(line," !#$%&'()*+,-./'"); // cut the line into words 
    // delimiter here are punctuation characters  (blank)!#$%&'()*+,-./'

    printf("\nThese are the words written in the line :\n");
    printf("----------------------------------------\n");
    while (p!=NULL) // a loop to extract the words one by one 
    {
        printf("%s\n",p); // print each word
        p=strtok(NULL," !#$%&'()*+,-./'"); // repeat till p is null 
    }

    return 0;
}

如果執行上面的代碼，我們將得到

Input the line:
>>hello every body how are you !

These are the words written in the line :
----------------------------------------
hello
every
body
how
are
you

Answer 4

suggest the code implement a state machine, 
where there are two states, in-a-word and not-in-a-word.  
Also, there are numerous other characters that could be read 
(I.E. ',' '.' '?' etc) that need to be check for. 

the general logic:

state = not-in-a-word
output '\n'

get first char
loop until eof
    if char is in range a...z or in range A...Z
    then 
        output char
        state = in-a-word
    else if state == in-a-word
    then
         output '\n'
         state = not-in-a-word
    else
         do nothing
    end if
    get next char
end loop

output '\n'

Answer 5

我認為簡單的解決方案是

#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;
}