scanf() doesn't accept whitespace

Question

I need a scanf() call to accept whitespace (no tabs or newlines, just ' ' space glyphs).

char buffer[2048];
scanf(" %2048[0-9a-zA-Z ]s", buffer);

This format specifier I got from the answer to this question:

how-do-you-allow-spaces-to-be-entered-using-scanf

While it accepts the first sequence of input just fine, it terminates where the first whitespace character is, with a null character. What's going on? Am I perhaps using the wrong format?

I should say, I'm using scanf() here because safety isn't a concern; I'm the only person who'll ever use this particular program, and the input is rigidly formatted.

@SujithKarivelil - buffer is already a pointer (from the array script), so it's okay — Anne Quinn, Aug 12 '14 at 05:30
The buffer isn't a pointer - it's an array reference, which is [not the same thing](http://c-faq.com/aryptr/aryptr2.html). However, the array reference will [decay to a pointer](http://c-faq.com/aryptr/aryvsadr.html) when used in this manner, so it is correct to not include the `&`. — Timothy Jones, Aug 12 '14 at 05:33

Sathish · Accepted Answer · 2014-08-12T05:31:04.767

6

Use scanf("%[^\n]",buffer);. It will accept white space.

Sample program-

int main()
{
    char buffer[2048];
    printf("Enter the string\n");
    scanf("%[^\n]",buffer);

    printf("%s\n", buffer);
    return 0;
}

output-

root@sathish1:~/My Docs/Programs# ./a.out 
Enter the string
abc def ghi ijk
abc def ghi ijk
root@sathish1:~/My Docs/Programs#

edited Aug 12 '14 at 05:31

answered Aug 12 '14 at 05:25

Sathish

3,740
1
17
28

Timothy Jones · Answer 2 · 2014-08-12T05:36:55.850

Scanf isn't good for dealing with format that you're expecting to have a particular amount of whitespace. From the scanf man page:

White space (such as blanks, tabs, or newlines) in the format string match any amount of white space, including none, in the input.

And:

[

Matches a nonempty sequence of characters from the specified set of accepted characters; the next pointer must be a pointer to char, and there must be enough room for all the characters in the string, plus a terminating NUL character. The usual skip of leading white space is suppressed.

This means you can do something like

scanf("%[^\n]",buffer);

which says: "Read everything but the newline at the end of the string".

Or, if you're wanting to skip the first space, you can do:

scanf("%*[ ]%[^\n]",buffer);

This says "Read but ignore a space character, then read everything else into buffer".

score 1 · Answer 3 · answered Aug 12 '14 at 06:00

While you can use scanf, if you are reading lines of text, getline is preferred and provides the advantage of dynamic memory allocation (when line = NULL). getline does read/save the newline character, so if that isn't desired, it can be easily stripped. The following example illustrates the point:

#include <stdio.h>

int main (void) {

    char *line = NULL;
    ssize_t read = 0;
    size_t n = 0;

    printf ("\nEnter a line of text: ");
    read = getline (&line, &n, stdin);
    line [read - 1] = 0;  /* strip newline from string (optional) */
    read--;

    printf ("\n  read '%zd' characters: '%s'\n\n", read, line);

    return 0;

}

output:

./bin/getln

Enter a line of text: this is a line of text   with  white ..   ..  space.

  read '52' characters: 'this is a line of text   with  white ..   ..  space.'

score 0 · Answer 4 · edited Jul 10 '17 at 08:39

0

fgets(string, sizeof(string), stdin) will accept or scanf("%[\n]s",string); will accept

edited Jul 10 '17 at 08:39

Community

1
1

answered Aug 12 '14 at 08:16

Shreemay Panhalkar

194
1
11

scanf() doesn't accept whitespace

4 Answers4

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