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I am not able to understand the difference. I use %[^\n]s, for taking phrases input from the user. But this was not working when I needed to add two phrases. But the above one did. Please help me understanding me the difference.
The %[\n] directive tells scanf() to match newline characters, and the * flag signals that no assignment should be made, so %*[\n] skips over any leading newline characters (assuming there is at least one leading \n character: more on this in a moment). There is a space following this first directive, so zero or more whitespace characters are skipped before the final %[^\n] directive, which matches characters until a newline is encountered. These are stored in input_string[], and the newline character is left behind in the input stream. Subsequent calls using this format string will skip over this remaining newline character.
But, there is probably no need for the %*[\n] directive here, since \n is a whitespace character; almost the same thing could be accomplished with a leading space in the format string: " %[^\n]".
One difference between the two: "%*[\n] %[^\n]" expects there to be a newline at the beginning of the input, and without this the match fails and scanf() returns without making any assignments, while " %[^\n]" does not expect a leading newline, or even a leading whitespace character (but skips them if present).
If you used "%[^\n]" instead, as suggested in the body of the question (note that the trailing s is not a part of the scanset directive), the first call to scanf() would match characters until a newline is encountered. The matching characters would be stored in input_string[], and the newline would remain in the input stream. Then, if scanf() is called again with this format string, no characters would be matched before encountering the newline, so the match would fail without assignment.
Please note that you should always specify a maximum width when using %s or %[] in a scanf() format string to avoid buffer overflow. With either of %s or %[], scanf() automatically adds the \0 terminator, so you must be sure to allow room for this. For an array of size 100, the maximum width should be 99, so that at most 99 characters are matched and stored in the array before the null terminator is added. For example: " %99[^\n]".
In scanf function, '*' tells the function to ignore a character from input.
%*[\n]
This tells the function to ignore the first '\n' character and then accept any string
Run the code and first give "ENTER" as input and then give "I am feeling great!!!"
Now print the buffer. You will get I am feeling great!!! as output
Try this code snippet
int main()
{
char buffer[100];
printf("Enter a string:"),scanf("%*[\n] %[^\n]', buffer),printf("buffer:%s\n", buffer);
return 0;
}
%[^\n] is an edit conversion code for scanf() as an alternative of gets(str).
Unlike gets(str), scanf() with %s cannot read more than one word.
Using %[^\n], scanf() can read even the string with whitespace.
It will terminate receiving string input from the user when it encounters a newline character.
https://i.imgur.com/FLxF9sP.png
As shown in the link above I have to input '<' twice instead of once, why is that? Also it seems that the first input is ignored but the second '<' is the one the program recognizes.
The same thing occurs even without a loop too.
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
int main(){
int randomGen, upper, lower, end, newRandomGen;
char answer;
upper = 100;
lower = 1;
end = 1;
do {
srand(time(0));
randomGen = rand()%(upper + lower);
printf("%d\n", randomGen);
scanf("%s\n", &answer);
}while(answer != '=');
}
Whitespace in scanf format strings, like the \n in "%c\n", tries to match any amount of whitespace, and scanf doesn’t know that there’s no whitespace left to skip until it encounters something that isn’t whitespace (like the second character you type) or the end of input. You provide it with =\n, which fills in the %c and waits until the whitespace is over. Then you provide it with another = and scanf returns. The second time around, the character could be anything and it’d still work.
Skip leading whitespace instead (and use the correct specifier for one character, %c, as has been mentioned):
scanf(" %c", &answer);
Also, it’s good practice to make sure you actually succeeded in reading something, especially when failing to read something means leaving it uninitialized and trying to read it later (another example of undefined behaviour). So check scanf’s return value, which should match the number of conversion specifiers you provided:
if (scanf(" %c", &answer) != 1) {
return EXIT_FAILURE;
}
As has been commented, you should not use the scanf format %s if you want to read a single character. Indeed, you should never use the scanf format %s for any purpose, because it will read an arbitrary number of characters into the buffer you supply, so you have no way to ensure that your buffer is large enough. So you should always supply a maximum character count. For example, %1s will read only one character. But note: that will still not work with a char variable, since it reads a string and in C, strings are arrays of char terminated with a NUL. (NUL is the character whose value is 0, also sometimes spelled \0. You could just write it as 0, but don't confuse that with the character '0' (whose value is 48, in most modern systems).
So a string containing a single character actually occupies two bytes: the character itself, and a NUL.
If you just want to read a single character, you could use the format %c. %c has a few differences from %s, and you need to be aware of all of them:
The default maximum length read by %s is "unlimited". The default for %c is 1, so %c is identical to %1c.
%s will put a NUL at the end of the characters read (which you need to leave space for), so the result is a C string. %c does not add the NUL, so you only need to leave enough space for the characters themselves.
%s skips whitespace before storing any characters. %c does not ignore whitespace. Note: a newline character (at the end of each line) is considered whitespace.
So, based on the first two rules, you could use either of the following:
char theShortString[2];
scanf("%1s", theShortString);
char theChar = theShortString[0];
or
char theChar;
scanf("%c", &theChar);
Now, when you used
scanf("%s", &theChar);
you will cause scanf to write a NUL (that is, a zero) in the byte following theChar, which quite possibly is part of a different variable. That's really bad. Don't do that. Ever. Even if you get away with it today, it will get you into serious trouble some time soon.
But that's not the problem here. The problem here is with what comes after the %s format code.
Let's take a minute (ok, maybe half an hour) to read the documentation of scanf, by typing man scanf. What we'll see, quite near the beginning, is: (emphasis added)
A directive is one of the following:
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.
So when you use "%s\n", scanf will do the following:
skip over any white-space characters in the input buffer.
read the following word up to but not including the next white-space character, and store it in the corresponding argument, followed by a NUL.
skip over any white-space following the word which it just read.
It does the last step because \n — a newline — is itself white-space, as noted in the quote from the manpage.
Now, what you actually typed was < followed by a newline, so the word read at step 2 will be just he character <. The newline you typed afterwards is white-space, so it will be ignored by step 3. But that doesn't satisfy step 3, because scanf (as documented) will ignore "any amount of white space". It doesn't know that there isn't more white space coming. You might, for example, be intending to type a blank line (that is, just a newline), in which case scanf must skip over that newline as well. So scanf keeps on reading.
Since the input buffer is now empty, the I/O library must now read the next line, which it does. And now you type another < followed by a newline. Clearly, the < is not white-space, so scanf leaves it in the input buffer and returns, knowing that it has done its duty.
Your program then checks the word read by scanf and realises that it is not an =. So it loops again, and the scanf executes again. Now there is already data in the input buffer (the second < which you typed), so scanf can immediately store that word. But it will again try to skip "any amount of white space" afterwards, which by the same logic as above will cause it to read a third line of input, which it leaves in the input buffer.
The end result is that you always need to type the next line before the previous line is passed back to your program. Obviously that's not what you want.
So what's the solution? Simple. Don't put a \n at the end of your format string.
Of course, you do want to skip that newline character. But you don't need to skip it until the next call to scanf. If you used a %1s format code, scanf would automatically skip white-space before returning input, but as we've seen above, %c is far simpler if you only want to read a single character. Since %c does not skip white-space before returning input, you need to insert an explicit directive to do so: a white-space character. It's usual to use an actual space rather than a newline for this purpose, so we would normally write this loop as:
char answer;
srand(time(0)); /* Only call srand once, at the beginning of the program */
do {
randomGen = rand()%(upper + lower); /* This is not right */
printf("%d\n", randomGen);
scanf(" %c", &answer);
} while (answer != '=');
scanf("%s\n", &answer);
Here you used the %s flag in the format string, which tells scanf to read as many characters as possible into a pre-allocated array of chars, then a null terminator to make it a C-string.
However, answer is a single char. Just writing the terminator is enough to go out of bounds, causing undefined behaviour and strange mishaps.
Instead, you should have used %c. This reads a single character into a char.
I am not able to understand the difference. I use %[^\n]s, for taking phrases input from the user. But this was not working when I needed to add two phrases. But the above one did. Please help me understanding me the difference.
The %[\n] directive tells scanf() to match newline characters, and the * flag signals that no assignment should be made, so %*[\n] skips over any leading newline characters (assuming there is at least one leading \n character: more on this in a moment). There is a space following this first directive, so zero or more whitespace characters are skipped before the final %[^\n] directive, which matches characters until a newline is encountered. These are stored in input_string[], and the newline character is left behind in the input stream. Subsequent calls using this format string will skip over this remaining newline character.
But, there is probably no need for the %*[\n] directive here, since \n is a whitespace character; almost the same thing could be accomplished with a leading space in the format string: " %[^\n]".
One difference between the two: "%*[\n] %[^\n]" expects there to be a newline at the beginning of the input, and without this the match fails and scanf() returns without making any assignments, while " %[^\n]" does not expect a leading newline, or even a leading whitespace character (but skips them if present).
If you used "%[^\n]" instead, as suggested in the body of the question (note that the trailing s is not a part of the scanset directive), the first call to scanf() would match characters until a newline is encountered. The matching characters would be stored in input_string[], and the newline would remain in the input stream. Then, if scanf() is called again with this format string, no characters would be matched before encountering the newline, so the match would fail without assignment.
Please note that you should always specify a maximum width when using %s or %[] in a scanf() format string to avoid buffer overflow. With either of %s or %[], scanf() automatically adds the \0 terminator, so you must be sure to allow room for this. For an array of size 100, the maximum width should be 99, so that at most 99 characters are matched and stored in the array before the null terminator is added. For example: " %99[^\n]".
In scanf function, '*' tells the function to ignore a character from input.
%*[\n]
This tells the function to ignore the first '\n' character and then accept any string
Run the code and first give "ENTER" as input and then give "I am feeling great!!!"
Now print the buffer. You will get I am feeling great!!! as output
Try this code snippet
int main()
{
char buffer[100];
printf("Enter a string:"),scanf("%*[\n] %[^\n]', buffer),printf("buffer:%s\n", buffer);
return 0;
}
%[^\n] is an edit conversion code for scanf() as an alternative of gets(str).
Unlike gets(str), scanf() with %s cannot read more than one word.
Using %[^\n], scanf() can read even the string with whitespace.
It will terminate receiving string input from the user when it encounters a newline character.
I am having some doubt. The doubt is
What is the difference between the following two scanf statements.
scanf("%s",buf);
scanf("%[^\n]", buf);
If I am giving the second scanf in the while loop, it is going infinitely. Because the \n is in the stdin.
But in the first statement, reads up to before the \n. It also will not read the \n.
But The first statement does not go in infinitely. Why?
Regarding the properties of the %s format specifier, quoting C11 standrad, chapter §7.21.6.2, fscanf()
s Matches a sequence of non-white-space characters.
The newline is a whitespace character, so only a newlinew won't be a match for %s.
So, in case the newline is left in the buffer, it does not scan the newline alone, and wait for the next non-whitespace input to appear on stdin.
The %s format specifier specifies that scanf() should read all characters in the standard input buffer stdin until it encounters the first whitespace character, and then stop there. The whitespace ('\n') remains in the stdin buffer until consumed by another function, like getchar().
In the second case there is no mention of stopping.
You can think of scanf as extracting words separated by whitespace from a stream of characters. Imagine reading a file which contains a table of numbers, for example, without worrying about the exact number count per line or the exact space count and nature between numbers.
Whitespace, for the record, is horizontal and vertical (these exist) tabs, carriage returns, newlines, form feeds and last not least, actual spaces.
In order to free the user from details, scanf treats all whitespace the same: It normally skips it until it hits a non-whitespace and then tries to convert the character sequence starting there according to the specified input conversion. E.g. with "%d" it expects a sequence of digits, perhaps preceded by a minus sign.
The input conversion "%s" also starts with skipping whitespace (and that's clearer documented in the opengroup's man page than in the Linux one).
After skipping leading whitespace, "%s" accepts everything until another whitespace is read (and put back in the input, because it isn't made part of the "word" being read). That sequence of non-whitespace chars -- basically a "word" -- is stored in the buffer provided. For example, scanning a string from " a bc " results in skipping 3 spaces and storing "a" in the buffer. (The next scanf would skip the intervening space and put "bc" in the buffer. The next scanf after that would skip the remaining whitespace, encounter the end of file and return EOF.) So if a user is asked to enter three words they could give three words on one line or on three lines or on any number of lines preceded or separated by any number of empty lines, i.e. any number of subsequent newlines. Scanf couldn't care less.
There are a few exceptions to the "skip leading whitespace" strategy. Both concern conversions which usually indicate that the user wants to have more control about the input conversion. One of them is "%c" which just reads the next character. The other one is the "%[" spec which details exactly which characters are considered part of the next "word" to read. The conversion specification you use, "%[^\n]", reads everything except newline. Input from the keyboard is normally passed to a program line by line, and each line is by definition terminated by a newline. The newline of the first line passed to your program will be the first character from the input stream which does not match the conversion specification. Scanf will read it, inspect it and then put it back in the input stream (with ungetc()) for somebody else to consume. Unfortunately, it will itself be the next consumer, in another loop iteration (as I assume). Now the very first character it encounters (the newline) does not match the input conversion (which demands anything but the newline). Scanf therefore gives up immediately, puts the offending character dutifully back in the input for somebody else to consume and returns 0 indicating the failure to even perfom the very first conversion in the format string. But alas, it itself will be the next consumer. Yes, machines are stupid.
First scanf("%s",buf); scan only word or string, but second scanf("%[^\n]", buf); reads a string until a user inputs is new line character.
Let's take a look at these two code snippets :
#include <stdio.h>
int main(void){
char sentence[20] = {'\0'};
scanf("%s", sentence);
printf("\n%s\n", sentence);
return 0;
}
Input : Hello, my name is Claudio.
Output : Hello
#include <stdio.h>
int main(void){
char sentence[20] = {'\0'};
scanf("%[^\n]", sentence);
printf("\n%s\n", sentence);
return 0;
}
Input : Hello, my name is Claudio.
Output : Hello, my name is Claudio.
%[^\n] is an inverted group scan and this is how I personally use it, as it allows me to input a sentece with blank spaces in it.
Common
Both expect buf to be a pointer to a character array. Both append a null character to that array if at least 1 character was saved. Both return 1 if something was saved. Both return EOF if end-of-file detected before saving anything. Both return EOF in input error is detected. Both may save buf with embedded '\0' characters in it.
scanf("%s",buf);
scanf("%[^\n]", buf);
Differences
"%s" 1) consumes and discards leading white-space including '\n', space, tab, etc. 2) then saves non-white-space to buf until 3) a white-space is detected (which is then put back into stdin). buf will not contain any white-space.
"%[^\n]" 1) does not consume and discards leading white-space. 2) it saves non-'\n' characters to buf until 3) a '\n' is detected (which is then put back into stdin). If the first character read is a '\n', then nothing is saved in buf and 0 is returned. The '\n' remains in stdin and explains OP's infinite loop.
Failure to test the return value of scanf() is a common code oversight. Better code checks the return value of scanf().
IMO: code should never use either:
Both fail to limit the number of characters read. Use fgets().
You can think of %s as %[^\n \t\f\r\v], that is, after skipping any leading whitespace, a group a non-whitespace characters.
I heve a following code
while ( a != 5)
scanf("%s", buffer);
This works well but takes no space in between the mentioned words or in other words, scanf terminates if we use spaces to scan
If I use this
while( a != 5)
scanf("%[^\n]", buffer);
It works only for once which is bad
I never use gets() because I know how much nasty it is..
My last option is this
while( a != 5)
fgets(buffer, sizeof(buffer), stdin);
So my questions are
Why the second command is not working inside the loop?
What are the other options I have to scan a string with spaces?
"%[^\n]" will attempt to scan everything until a newline. The next character in the input would be the \n so you should skip over it to get to the next line.
Try: "%[^\n]%*c", the %*c will discard the next character, which is the newline char.
Why the second command is not working inside the loop
Becuase, for the first time what you scan until \n, the \n is remaining in the input buffer. You need to eat up (or, in other word, discard) the stored newline from the buffer. You can make use of while (getchar()!=\n); to get that job done.
What are the other options I have to scan a string with spaces?
Well, you're almost there. You need to use fgets(). Using this, you can
Be safe from buffer overrun (Overcome limitation of gets())
Input strings with spaces (Overcome limitation of %s)
However, please keep in mind, fgets() reads and stores the trailing newline, so you may want to get rid of it and you have to do that yourself, manually.