scanf() leaves the newline character in the buffer

Question

I have the following program:

int main(int argc, char *argv[])
{
    int a, b;
    char c1, c2;
    printf("Enter something: ");
    scanf("%d", &a); // line 1
    printf("Enter other something: ");
    scanf("%d", &b); // line 2

    printf("Enter a char: ");
    scanf("%c", &c1); // line 3
    printf("Enter another char: ");
    scanf("%c", &c2); // line 4

    printf("Done"); // line 5

    system("PAUSE");

    return 0;
}

As I read in the C book, the author says that scanf() left a newline character in the buffer, therefore, the program does not stop at line 4 for user to enter the data, rather it stores the new line character in c2 and moves to line 5.

Is that right?

However, does this only happen with char data types? Because I did not see this problem with int data types as in line 1, 2, 3. Is it right?

Fernand

It is sometimes suggested that fflush(stdin) can be used before the call to scanf() for a single character. Please read Using fflush(stdin) for a discussion of the pros and cons and alternatives to that method (which works, more or less, on Windows, and does not work most other places). – Advertising

Could you please let us know which book you are referring to .? – Diptych

@JonathanLeffler Using fflush(stdin) is plain undefined behaviour. – Arlettaarlette

Answer 1

The scanf() function skips leading whitespace automatically before trying to parse conversions other than characters. The character formats (primarily %c; also scan sets %[…] — and %n) are the exception; they don't skip whitespace.

Use " %c" with a leading blank to skip optional white space. Do not use a trailing blank in a scanf() format string.

Note that this still doesn't consume any trailing whitespace left in the input stream, not even to the end of a line, so beware of that if also using getchar() or fgets() on the same input stream. We're just getting scanf to skip over whitespace before conversions, like it does for %d and other non-character conversions.

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Note that non-whitespace "directives" (to use POSIX scanf terminology) other than conversions, like the literal text in scanf("order = %d", &order); doesn't skip whitespace either. The literal order has to match the next character to be read.

So you probably want " order = %d" there if you want to skip a newline from the previous line but still require a literal match on a fixed string, like this question.

Latent

%c, %n, %[] are the 3 specified expectations that do not consume leading whitespace. – Potman

@chux So Does in Other Cases, The scanf clears all the whitespaces before in the buffer or ignores them for they input but they are still there? – Committeewoman

@SurajJain Yes, – Potman

See Trailing blank in scanf() format string and scanf() asking twice for input while I expect it to ask only once for a discussion of trailing blanks in format strings. They're a bad idea — astoundingly bad if you expect human interaction and bad for program interaction. – Advertising

This answer does a very good job of explaining that scanf() reads and then formats, and not just read according to the format. – Slipslop

Answer 2

Use scanf(" %c", &c2);. This will solve your problem.

Homerhomere

Answer 3

Another option (that I got from here) is to read and discard the newline by using the assignment-supression option. To do that, we just put a format to read a character with an asterisk between % and c:

scanf("%d%*c",&a); // line 1
scanf("%c%*c",&c1); // line 3

scanf will then read the next char (that is, the newline) but not assign it to any pointer.

In the end, however, I would second the FAQ's last option:

Or, depending on your requirements, you could also forget about scanf()/getchar(), use fgets() to get a line of text from the user and parse it yourself.

Birr

The trouble with this technique is that if the user types a then space then newline, the suppressed character conversion reads the space and still leaves the newline. If the user types supercalifragilisticexpialidocious when you expect a, you've got a lot of extra characters to deal with. You can never tell whether a trailing suppressed conversion succeeds, either — they're not counted in the return from scanf(). – Advertising

This only discards one character. Placing a space before % will discard any amount of leading whitespace (including none). – Battled

@WeatherVane: For a line like 123\n, "%d %*c" will have the space in the format string consume whitespace and then %*c consume the first non-whitespace byte of the next line. (Just trailing whitespace in the format string will wait until there's a non-whitespace character, which might be ok if reading from a file, but not for interactive input. Trailing " %*c" does that and then discards the first non-whitespace character, so if waiting for the next line now is acceptable at all, just use trailing whitespace.) – Tongs

Answer 4

To echo what I have posted in another answer about C++: I suggest to toss scanf() away, to never use it, and to instead use fgets() and sscanf().

The reason for this is, that at least in Unix-like systems by default, the terminal your CLI program runs on does some processing of the user input before your program sees it. It buffers input until a newline is entered, and allows for some rudimentary line editing, like making backspace work.

So, you can never get a single character at a time, or a few single characters, just a full line. But that's not what e.g. scanf("%d") processes, instead it processes just the digits, and stops there, leaving the rest buffered in the C library, for a future stdio function to use. If your program has e.g.

printf("Enter a number: ");
scanf("%d", &a);

printf("Enter a word: ");
scanf("%s", word);

and you enter the line 123 abcd, it completes both scanf()s at once, but only after a newline is given. The first scanf() doesn't return when a user has hit space, even though that's where the number ends (because at that point the line is still in the terminal's line buffer); and the second scanf() doesn't wait for you to enter another line (because the input buffer already contains enough to fill the %s conversion).

This isn't what users usually expect!

Instead, they expect that hitting enter completes the input, and if you hit enter, you either get a default value, or an error, with possibly a suggestion to please really just give the answer.

You can't really do that with scanf("%d"). If the user just hits enter, nothing happens. Because scanf() is still waiting for the number. The terminal sends the line onward, but your program doesn't see it, because scanf() eats it. You don't get a chance to react to the user's mistake.

That's also not very useful.

Hence, I suggest using fgets() or getline() to read a full line of input at a time. This exactly matches what the terminal gives, and always gives your program control after the user has entered a line. What you do with the input line is up to you, if you want a number, you can use atoi(), strtol(), or even sscanf(buf, "%d", &a) to parse the number. sscanf() doesn't have the same mismatch as scanf(), because the buffer it reads from is limited in size, and when it ends, it ends -- the function can't wait for more.

^{(fscanf() on a regular file can also be fine if the file format is one that supports how it skims over newlines like any whitespace. For line-oriented data, I'd still use fgets() and sscanf().)}

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So, instead of what I had above, use something like this:

printf("Enter a number: ");

fgets(buf, bufsize, stdin);
sscanf(buf, "%d", &a);

or, actually, check the return value of sscanf() too, so you can detect empty lines and otherwise invalid data:

#include <stdio.h>

int main(void)
{
    const int bufsize = 100;
    char buf[bufsize];
    int a;
    int ret;
    char word[bufsize];

    printf("Enter a number: ");
    fgets(buf, bufsize, stdin);

    ret = sscanf(buf, "%d", &a);

    if (ret != 1) {
        fprintf(stderr, "Ok, you don't have to.\n");
        return 1;
    }

    printf("Enter a word: ");
    fgets(buf, bufsize, stdin);

    ret = sscanf(buf, "%s", word);
    if (ret != 1) {
        fprintf(stderr, "You make me sad.\n");
        return 1;
    }

    printf("You entered %d and %s\n", a, word);
}

Of course, if you want the program to insist, you can create a simple function to loop over the fgets() and sscanf() until the user deigns to do what they're told; or to just exit with an error immediately. Depends on what you think your program should do if the user doesn't want to play ball.

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You could do something similar e.g. by looping over getchar() to read characters until a newline after scanf("%d") returned, thus clearing up any garbage left in the buffer, but that doesn't do anything about the case where the user just hits enter on an empty line. Anyway, fgets() would read until a newline, so you don't have to do it yourself.

Sami

fgets()/getline() and then sscanf() or strtol(), etc., also has another huge advantage over using scanf() directly. When the program encounters unexpected input, the input stream isn't left in an unknown state where recovery without losing data can be impossible. – Grochow

For further reading about alternatives to scanf: What can I use for input conversion instead of scanf? Also, this guide may be helpful: A beginners' guide away from scanf() – Unreliable

Answer 5

Use getchar() before calling second scanf().

scanf("%c", &c1);
getchar();  // <== remove newline
scanf("%c", &c2);

Dessau

This works provided that the user didn't type anything else — trailing blanks, for example. But it isn't as good as a loop that scans to the next newline: int c; while ((c = getchar()) != EOF && c != '\n') ; (written over three lines when not in a comment). It is often sufficient; it is not foolproof (and you have to remember that fools are very clever about crashing things). – Advertising

@JonathanLeffler do you mean written over two lines? int c; while ((c = getchar()) != EOF && c != '\n') ; – Minnesota

@john: It can be one, two, three lines or more — the compiler doesn't care. Left to my own devices, it would be three lines: the declaration, the loop control and the loop body. The semicolon denoting the empty loop body would be on a line on its own to emphasize that it is intentional (as recommended by K&R). – Advertising

Answer 6

Two workarounds. One is catch the extra whitespace trickly.

getchar(); // (1) add `getchar()`
scanf("%c", &c1);

scanf(" %c", &c1); // (2) add whitespace before %c

The other is using fgets() to instead scanf(). Note: gets() is unsafe, see why gets is dangerous

By contrast, I would prefer to recommend the second way. Because it's more readable and maintainable. For example, in first way, you must think a while before adding/moving some scanf().

Encephalon

Answer 7

/*Take char input using scanf after int input using scanf just use fflush(stdin) function  after int input */
#include<stdio.h>
#include<conio.h>
void main()
{
  int x;
  char y;
  clrscr();
  printf(" enter an int ");
  scanf("%d",&x);
  fflush(stdin);
  printf("\n Now enter a char");
  scanf("%c",&y);
  printf("\n X=%d and Y=%c",x,y);
  getch();
}

Hazelhazelnut

Note the caveats about Using fflush(stdin). – Advertising

conio.h is not part of ISO C, but is a platform-specific header. The functions clrscr and getch are also platform-specific. Note that the question is not tagged with any particular platform, so that the question does not apply to any particular platform. Therefore, if you decide to provide a platform-specific answer, you should at least clearly mark it as such and specify to which platform your answer applies. – Unreliable