Beginner Question: Problem with scanf and strings. Help!

First off how does scanf know that this: %[^\n] is supposed to be formatted for a String? Shouldn't %s need to be in there? According to the syntax it's not supposed to be.

Anyway, my real question is when I run this program (below) it lets me type in my first name, but then when I press Return it skips over letting me enter my last name and shows this:

Code:

What is your FIRST name: Bryan
What is your LAST name: Your FULL name is Bryan  and it is 6 long.

Code:

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

int main (int argc, const char * argv[]) {

	char		firstname[31];
	char		lastname[31];
	char		fullname[63];
	
	printf( "What is your FIRST name: " );
	scanf( "%[^\n]", firstname );      // accepts everything until it sees a \n
	
	printf( "What is your LAST name: " );
	scanf( "%[^\n]", lastname );      // accepts everything until it sees a \n
	
	strcpy( fullname, firstname );
	strcat( fullname, " " );
	strcat( fullname, lastname );
	
	printf( "Your FULL name is %s and it is %d long.", fullname, strlen(fullname) );
		  
    return 0;
}

See this link for information about scanf(): scanf [C++ Reference]

In short, the characters between the brackets, ^\n, tells scanf() to suck in everything EXCEPT a newline character. The NEWLINE character is left in the input buffer.

When the next scanf() is issued, it reads everything up to the NEWLINE, which is nothing, and returns immediately.

If you use the bracket notation, you'll next an extra line of code to remove the NEWLINE from the input buffer.

scanf( "%[^\n]", firstname ); // accepts everything until it sees a \n

Yep. Now check your docs for this angle:

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

Or, you could just use %s, and then, as Dino is trying to say, remove the '\n' from your string if desired.

I figured using a fflush(stdin); would clear the buffer but it doesn't seem to make a difference... I'm clueless how to get this working as expected

Code:

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

int main (int argc, const char * argv[]) {

	char		firstname[31];
	char		lastname[31];
	char		fullname[63];
	
	printf( "\nWhat is your FIRST name: " );
	scanf( "%[^\n]", firstname );
	fflush(stdin);                       // this doesn't seem to work
	
	printf( "\nWhat is your LAST name: " );
	scanf( "%[^\n]", lastname );
	
	strcpy( fullname, firstname );
	strcat( fullname, " " );
	strcat( fullname, lastname );
	
	printf( "\nYour FULL name is %s and is %d characters long.", fullname, strlen(fullname) );
		  
    return 0;
}

Originally Posted by lucidrave

I figured using a fflush(stdin); would clear the buffer

Never use fflush(stdin) -- it is not defined by the C standard:

[#2] If stream points to an output stream or an update
stream in which the most recent operation was not input, the
fflush function causes any unwritten data for that stream to
be delivered to the host environment to be written to the
file; otherwise, the behavior is undefined.

Read my previous post; the "*" flag means ignore. In the context of your scanf line, that %c will be the newline.

ohh ok... so it means ignore any characters after it sees the new line?

(in this code anyway)

Originally Posted by lucidrave

ohh ok... so it means ignore any characters after it sees the new line?
(in this code anyway)

No, it means ignore the newline, which is a single character ('\n'). So if you typed:

mk27\n

(the newline being because of the enter key), and you capture everything up to but not including the newline with %[^\n], '\n' is left in the buffer. %*c captures one more character and throws it away; because of the context, in this case that character would have to be the newline (get it?).

I guess what's confusing me is I thought %[^\n] was to ignore the new line, but then %*c also ignores the new line. isn't that redundant?

if %[^\n] doesn't work by itself should I not use it?

Also is the * part of %*c acting as a "wildcard"? It's a bit confusing know what purpose it serves since it could be multiply (which wouldn't make any sense in this case) or referencing a pointer.

Originally Posted by MK27

No, it means ignore the newline, which is a single character ('\n'). So if you typed:

mk27\n

(the newline being because of the enter key), and you capture everything up to but not including the newline with %[^\n], '\n' is left in the buffer. %*c captures one more character and throws it away; because of the context, in this case that character would have to be the newline (get it?).

Originally Posted by lucidrave

I guess what's confusing me is I thought %[^\n] was to ignore the new line, but then %*c also ignores the new line. isn't that redundant?

As mentioned, %[^\n] means "get everything that isn't a newline". If you do that, well, the newline is still there since you got everything that wasn't a newline. You need to now do something with that newline, instead of what you seem to be doing which is "ignore it and hope that it goes away".