How to read from input until newline is found usin

2020-01-28 05:42发布

问题:

I was asked to do a work in C when I'm supposed to read from input until there's a space and then until the user presses enter. If I do this:

scanf("%2000s %2000s", a, b);

It will follow the 1st rule but not the 2nd.
If I write:

I am smart

What I get is equivalent to:
a = "I";
b = "am";
But It should be:
a = "I";
b = "am smart";

I already tried:

scanf("%2000s %2000[^\n]\n", a, b);

and

scanf("%2000s %2000[^\0]\0", a, b);

In the 1st one, it waits for the user to press Ctrl+D (to send EOF) and that's not what I want. In the 2nd one, it won't compile. According to the compiler:

warning: no closing ‘]’ for ‘%[’ format

Any good way to solve this?

回答1:

scanf (and cousins) have one slightly strange characteristic: any white space in the format string (outside of a scanset) matches an arbitrary amount of white space in the input. As it happens, at least in the default "C" locale, a new-line is classified as white space.

This means the trailing '\n' is trying to match not only a new-line, but any succeeding white-space as well. It won't be considered matched until you signal the end of the input, or else enter some non-white space character.

To deal with this, you typically want to do something like this:

scanf("%2000s %2000[^\n]%c", a, b, c);

if (c=='\n')
    // we read the whole line
else
    // the rest of the line was more than 2000 characters long. `c` contains a 
    // character from the input, and there's potentially more after that as well.


回答2:

scanf("%2000s %2000[^\n]", a, b);


回答3:

use getchar and a while that look like this

while(x = getchar())
{   
    if(x == '\n'||x == '\0')
       do what you need when space or return is detected
    else
        mystring.append(x)
}

Sorry if I wrote a pseudo-code but I don't work with C language from a while.



回答4:

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

int main(void)
{
  int i = 0;
  char *a = (char *) malloc(sizeof(char) * 1024);
  while (1) {
    scanf("%c", &a[i]);
    if (a[i] == '\n') {
      break;
    }
    else {
      i++;
    }
  }
  a[i] = '\0';
  i = 0;
  printf("\n");
  while (a[i] != '\0') {
    printf("%c", a[i]);
    i++;
  }
  free(a);
  getch();
  return 0;
}


回答5:

I am too late, but you can try this approach as well.

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

int main() {
    int i=0, j=0, arr[100];
    char temp;
    while(scanf("%d%c", &arr[i], &temp)){
        i++;
        if(temp=='\n'){
            break;
        }
    }
    for(j=0; j<i; j++) {
        printf("%d ", arr[j]);
    }

    return 0;
}


回答6:

Sounds like a homework problem. scanf() is the wrong function to use for the problem. I'd recommend getchar() or getch().

Note: I'm purposefully not solving the problem since this seems like homework, instead just pointing you in the right direction.



回答7:

#include <stdio.h>
int main()
{
    char a[5],b[10];
    scanf("%2000s %2000[^\n]s",a,b);
    printf("a=%s b=%s",a,b);
}

Just write s in place of \n :)



标签: c format scanf