Member Avatar for manishanibhwani 
int n;
scanf("%d\n", &n);
printf("you typed %d\n", n);

it seems to hang until I type one extra line of input....Can anyone explain this behaviour

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  • Latest Post Latest Post by WaltP

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The program “hangs” because the trailing \n in the format string is a whitespace directive for the scanf family — it tells the scanner to keep reading and discarding whitespace until it finds a non‑whitespace character (or EOF). In an interactive session a user pressing Enter only supplies a newline (which is whitespace), so scanf keeps waiting for the next non‑whitespace token. This is exactly what , and were pointing out. C11 (N1570), scanf man page. (port70.net)

More precisely: whitespace in a scanf format matches a maximal run of input white‑space (as classified by isspace) and stops only when a non‑whitespace character is seen (that character is left unread). Also note that most conversion specifiers (e.g. %d, %s, %f) already skip leading whitespace; the exceptions are %c, %[ and %n, which do not. The newline character '\n' counts as whitespace per isspace. (port70.net)

Practical fixes: remove the trailing \n from the format string (use scanf("%d", &n) and check the return value), or—more robust—read a whole line and parse it. Example pattern (safer for interactive and file input):

char buf[128];
if (fgets(buf, sizeof buf, stdin)) {
    char *end;
    long v = strtol(buf, &end, 10);
    if (end != buf) {
        n = (int)v; /* successful conversion */
    }
}

The scanf man page itself recommends reading full lines and parsing with sscanf/strtol for reliable input handling. (man7.org)

Troubleshooting tips: with a format like "%d\n" the call will return only when a non‑whitespace is typed or the stream is closed; sending EOF (Ctrl+D on Unix, Ctrl+Z then Enter on Windows) ends the waiting. For interactive programs prefer line‑based input to avoid these surprising blocks. (port70.net)

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All 13 Replies

Member Avatar for Member #957352

I don't think so there should be any problem. But try to give the code in which you are facing this problem, then that will be easier to judge where is the problem. thanks.

Member Avatar for sepp2k

A whitespace character (including \n) in the format string will cause scanf to discard any whitespace in the input stream upto the next token. If a whitespace character is the last character in the format string that means that the user must already enter the next token, so that scanf knows upto where it needs to discard whitespace.

Genereally you simply don't want to have whitespace characters at the end of your format string.

Member Avatar for WaltP

IOW, get rid of the \n in the scanf() call.

Member Avatar for Member #957352

sir, will you please explain it more that how '\n' is making problem here. please explain it . thanks.

Member Avatar for WaltP

sepp2k explained it.

Member Avatar for Member #957352

but, i didn't get it na... plzz it's a request to u .. please

Member Avatar for WaltP

Because it won't work the way you have it. You cannot end the format string with \n.

Member Avatar for sepp2k

You cannot end the format string with \n

Well, you can. It just means that scanf won't return until the user has entered another token. (Not that I could imagine a scenario where that's ever useful).

</pedantry>

Member Avatar for *sharath*

Quoted Text Here

It just means that scanf won't return until the user has entered another token

does this mean that the user needs to enter another input (may be a string or an integer or so) so that the scanf returns?

Member Avatar for deceptikon

does this mean that the user needs to enter another input (may be a string or an integer or so) so that the scanf returns?

It means there needs to be at least one non-whitespace character after all of the whitespace characters. When you put whitespace in the format string, you can think of it as a specifier saying "read and discard all contiguous whitespace". If a newline is the last character then scanf() doesn't have any way of knowing it's the last and will wait for you to give it more.

Member Avatar for Member #957352

@jame sir, but why is it so ? i mean why scanf is waiting for input ? why is it expecting a character ?

and secondly , new line is a white space or non-white space character ? i am confused!

thanks to you.

Member Avatar for sepp2k

i mean why scanf is waiting for input ? why is it expecting a character ?

Because, as has been explained repeatedly now, a white space in scanf's format string means "consume any whitespace characters until you find the next non-whitespace character". Why does it mean that? Because the C standard says that that's what it means.

new line is a white space or non-white space character ?

The newline character is a whitespace character. For more information on what is and isn't a whitespace character see the documentation of the isspace function.

Member Avatar for WaltP

For more information on what is and isn't a whitespace character see the documentation of the isspace function.

Or better yet, write a quick program to loop from 0 to 255 and call all the isXXX() functions and print the results in a chart:

char isspace isdigit isalpha ispunct ....
0x00    F       F       F      F
...
0x20    T       F       F      F
...

Then you can have a list at hand if you ever have doubts.

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