Reading in a string with spaces in C

Question

I am trying to read in a string that may or may not include spaces ex. "hello world". By doing the following with a number select menu that is inputted by the user. This is just a small replica of what I am trying to do.

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

int main(void){
  char line[3][80];

  strcpy(line[0],"default line 1\n");
  strcpy(line[1],"default line 2\n");
  strcpy(line[2],"default line 3\n");

  for(int i = 0; i < 3; i++){
    printf("%s", line[i]);
  }

  int option = 0;
  printf("would you like to replace line 1? (1 for yes)\n");
  scanf("%d",&option);
  if(option==1){
   printf("what would you like to replace the line with?\n");
   fgets(line[0],strlen(line[0]),stdin);
  }

  for(int i = 0; i < 3; i++){
    printf("%s", line[i]);
  }
}

Why is it that after I enter 1 to change the line, it prints the statement asking what I want to replace it with and will automatically enter nothing then printing the strings with the first one as empty?

I also have already tried reading the line with sscanf("%[^\\n\\t]s", line[0]); without any luck. Any ideas?

Answer 1

It's because

scanf("%d",&option);

leaves the \\n character in stdin and is consumed by the first call to fgets() . That's why it's best to avoid scanf() in C completely.

You can fix it with:

  scanf("%d",&option);
  getchar(); /* consume the newline */

But I'd suggest using fgets() to read option as well and then you can use strtol() to convert it into an integer.

Note that this statement is not probably what you intended (which limits what you can read into line[0] ).

   fgets(line[0],strlen(line[0]),stdin);

You probably meant to use:

   fgets(line[0],sizeof line[0],stdin);

so that you can read upto the actual size of line[0] .

Please read the C Faq entry as well: http://c-faq.com/stdio/scanfprobs.html

Answer 2

Using fgets() generally seems less error-prone than tangling with scanf() , but if the user enters a string that is as long as or longer than the maximum number of characters specified, any extra characters up to and including the newline remain in the input stream. For this reason I usually write my own version of gets() to get input strings from the user, and if I want numeric input I use strtol() . Here is an example of such a function:

char * s_gets(char *st, int n)
{
    char *ret;
    int ch;

    ret = fgets(st, n, stdin);
    if (ret) {
        while (*st != '\n' && *st != '\0')
            ++st;
        if (*st)
            *st = '\0';
        else {
            while ((ch = getchar()) != '\n' && ch != EOF)
                continue;           // discard extra characters
        }
    }
    return ret;
}

Applied to the OPs problem, I might do something like this:

#include <stdlib.h>                // for strtol()

...

char buf[80];
int option = 0;

printf("would you like to replace line 1? (1 for yes)\n");
s_gets(buf, sizeof(buf));
option = strtol(buf, NULL, 10);

if(option==1){
    printf("what would you like to replace the line with?\n");
    s_gets(line[0],sizeof(line[0]));
}

Answer 3

Your problem is that the '\\n' char is left into stdin and consumed by fgets .

I'd suggest you to always use fgets for read inputs, so

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

int main(void)
{
    char line[3][80];
    char temp[3];

    strcpy(line[0],"default line 1\n");
    strcpy(line[1],"default line 2\n");
    strcpy(line[2],"default line 3\n");

    for(int i = 0; i < 3; i++){
        printf("%s", line[i]);
    }

    int option = 0;
    printf("would you like to replace line 1? (1 for yes)\n");
    fgets(temp,sizeof(temp),stdin);
    option = atoi(temp);

    if(option==1){
        printf("what would you like to replace the line with?\n");
        fgets(line[0],sizeof(line[0]),stdin);
    }

    for(int i = 0; i < 3; i++){
    printf("%s", line[i]);
    }
}