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I just realize this 'bug' of scanf now after 8 years with C.

Below scanf code will skip the leading whitespace characters from the second line of input.

int x;
char in[100];

scanf("%d\n",&x);
gets(in);

Input:

1
     s

x will contain 1, but in will be just "s" not " s"

Is this standard C or just gcc behaviour?

dragon135
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3 Answers3

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A whitespace character in your scanf format string will cause scanf to consume any (and all) white space till a non-whitespace char occurs.

This seems to be standard scanf behaviour and is not limited to gcc.

Klas Lindbäck
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  • *`skip any (and all) white space`* I think better say something like will consume all whitespaces till a non-whitespace char occurs. because scanf don't skip but remove from input buffer. – Grijesh Chauhan Feb 17 '14 at 12:42
  • + , Because both below answers are wrong .. you may also like to include that scanf only returns to calling place when it read `s` (a non-white space) but don't consume remove `s` from input buffer that is read by `gets()` in `in[ ]` array. So just output is `"s"`. – Grijesh Chauhan Feb 17 '14 at 12:49
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    @ajay The OP is doing `scanf("%d\n", &x);` not `scanf("%d", &x);`. The `\n` in the format string consumes whitespace after the number. – Klas Lindbäck Feb 17 '14 at 13:41
  • @KlasLindbäck thanks Klas for a moment I confused from link and faq – Grijesh Chauhan Feb 17 '14 at 13:45
1

Its not a Bug in scanf, the manual of scanf says,

A sequence of white-space characters (space, tab, newline, etc.; see isspace(3)). This directive matches any amount of white space, including none, in the input.

Which means any white space characters with directive as %d\n will read a number followed by consuming a sequence of white space characters in the input and only returns until you type a non white space character. That how you are able to see only "s" without a space before it.

Grijesh Chauhan
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Sunil Bojanapally
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  • *`but the newline character is still in the input buffer.`* - **no** it is consumed by scanf due to `\n` in format string – Grijesh Chauhan Feb 17 '14 at 12:36
  • *`gets() will immediately read this newline character giving you a space when you read a string late`* again **not** because gets just reads `s` that is the reason output is just `"s"` – Grijesh Chauhan Feb 17 '14 at 12:37
  • I like to share a tutorial over [behavior of scanf and printf](http://wpollock.com/CPlus/PrintfRef.htm) – Grijesh Chauhan Feb 17 '14 at 12:39
  • @GrijeshChauhan I was hurry in answering and made a mistake. Appreciate your link. Answer edited. – Sunil Bojanapally Feb 17 '14 at 13:43
  • @SunEric so it means that scanf("%d\t",&x) and scanf("%d ",&x) and scanf("%d\n",&x) behave exactly the same? I thought it will consume only until character '\t' or ' ' or '\n' respectively. – dragon135 Feb 18 '14 at 02:40
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The '\n' (and this is true for any whitespace character in the format string) in

scanf("%d\n", &x);

matches any number of whitespace characters in the input (characters for which isspace function returns 1, i.e, true, such as newline, space, tab etc.) and not just the newline character '\n'. This means that scanf will read all whitespace characters in the input and discard them till it encounters a non-whitespace character. This explains the behaviour you observed.

This is a part of the standard definition of the scanf function and not a gcc feature. Also, gets function is deprecated and unsafe. It does not check for buffer overrun and can lead to bugs and even program crash. In fact, gcc emits a warning against the use of gets on my machine. Use of fgets instead is recommended.

To do what you want, you can do the following:

int x;
char in[100];
scanf("%d", &x);

After scanf returns successfully, the input stream can contain any sequence of characters terminated by a newline depending on the input given by the user. Get rid of those extraneous characters before reading a string from the stdin.

char ch;
while((ch = getchar()) != '\n' || ch != EOF);  // null statement
fgets(in, 100, stdin);

The above fgets call means that it will read at most 100-1 = 99 (it saves one character space for the terminating null byte which it adds to the buffer being read into before exiting) characters from the stream pointed to by stdin and store them in the buffer pointed to by in. fgets will exit if it encounters EOF, '\n' or it has already read 100-1 characters - whichever of the three condition occurs first. If it reads a newline, it will store it into the buffer.

Is the user enters 100 characters or more in this case, then the extraneous characters would be lying around in the input buffer which can mess up with the subsequent input operation of characters or strings by scanf, fgets, getchar etc. calls. You can check for this checking the length of the string in.

if(strlen(in) > 99) {
    // extraneous chars lying around in the input buffer
    // read and discard them
    char ch;
    while((ch = getchar()) != '\n' || ch != EOF);  // null statement
}
ajay
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  • *`That newline character is read by the gets call`* No \n not read by gets – Grijesh Chauhan Feb 17 '14 at 12:44
  • Ajay please don't mind but still you need to correct `The newline character is left in the input buffer after you input an int and hit` because Scanf consumes all white spaces and remove from stdin (but don't store anywhere). – Grijesh Chauhan Feb 17 '14 at 12:52
  • @GrijeshChauhan `'\n'` **is** read by `gets` but is not saved into the buffer passed to it and discarded. – ajay Feb 17 '14 at 17:37
  • `gets()` read a `\n` **after** `s` and discard that. Read [`char * gets (char *s)`](http://www.gnu.org/software/libc/manual/html_node/Line-Input.html) But Question is regarding `\n` and spaces before `"s"` but not those encountered after `s` – Grijesh Chauhan Feb 18 '14 at 10:02