fgets question

**Ben Bacarisse** · Jun 27 '08, 07:40 PM

Re: fgets question

"Bill Cunningham" <nospam@nspam.c omwrites:

I was talking with someone about fgets and he said that fgets puts the
\n in a string but not \0.

Where do you meet people to chat about C?

I decided to test this assumption because my
documentation didn't say if fgets put \0 after a string literal. Strlen was
what I used to decide the \0 was not added.

What made you think that? There must be a null byte or strlen won't
work.

How can it be added for a
string?

It is there already. If fgets runs out of room it will stop one short
just so it can always put the terminating null in the buffer. You
may not get a '\n' but you will always get a 0.

My code.
>
int main (void) {
char input[10];
fgets (input, sizeof(input), stdin);
printf ("%i\n", strlen(input));

Better is:
printf ("%i\n", (int)strlen(inp ut));

since %i needs an int not a size_t. Alternatively use "%zu" as the
format since you seem to be using somthing close to C99.

}
>
One more character than what was typed printf reported.

The '\n' is a character. If you type abc and hit return, fgets put
five things into the input array: 'a', 'b', 'c', '\n' and '\0'. This
is a four character string and strlen reports 4.

--
Ben.

**Jack Klein** · Jun 27 '08, 07:40 PM

Re: fgets question

On Mon, 09 Jun 2008 01:54:12 GMT, "Bill Cunningham" <nospam@nspam.c om>
wrote in comp.lang.c:

I was talking with someone about fgets and he said that fgets puts the
\n in a string but not \0. I decided to test this assumption because my
documentation didn't say if fgets put \0 after a string literal.

Who was this person, and what credentials did he have to make you
think he was right?

More importantly, what does a string literal have to do with anything?
fgets() has nothing at all to do with string literals. It reads input
from a stream. String literals are quoted strings in your source
code, not input read from a stream.

Strlen was
what I used to decide the \0 was not added. How can it be added for a
string? My code.

I surely don't understand what you were trying to do above. If you
had an array of characters that does not contain at least one '\0'
within the bounds of the array, then it is not a string. Calling
strlen() on an array of characters that is not a string, due to the
lack of a '\0', produces undefined behavior. strlen() is for strings,
not arbitrary arrays of garbage characters.

int main (void) {
char input[10];
fgets (input, sizeof(input), stdin);

In this call, fgets() will accept up to 9 characters from stdin. Fewer
if it receives a '\n' before the ninth character.

It will store these characters into consecutive elements of the array.
Regardless of whether it stores 9 or fewer characters in the array, it
will place a '\0' into the array after the last character it stores.

Here is specifically what the standard says about fgets():

"The fgets function reads at most one less than the number of
characters specified by n from the stream pointed to by stream into
the array pointed to by s. No additional characters are read after a
new-line character (which is retained) or after end-of-file. A
null character is written immediately after the last character read
into the array."

So your "someone" is wrong in their opinion.

printf ("%i\n", strlen(input));

The strlen() function returns a value of type size_t, and since
printf() is a variadic function, the function call does not perform
any conversion. "%i" is a conversion specifier for signed int, and
size_t can never be a signed int since it is always an unsigned type.

If you need to print a reasonably small value of size_t, you could do
this:

printf("%u", (unsigned)strle n(input));

}
>
One more character than what was typed printf reported.

I can't parse the sentence you have written above.

It would be much better to:

1. Show us why you typed into the program

2. Show us what the output of the program was.

Regardless of what you think you proved to yourself with your program,
the fgets() function most surely does produce a '\0' terminated proper
string, except under two exceptional cases:

"The fgets function returns s if successful. If end-of-file is
encountered and no characters have been read into the array, the
contents of the array remain unchanged and a null pointer is returned.
If a read error occurs during the operation, the array contents are
indeterminate and a null pointer is returned."

Either of these exceptional cases is easily recognized by the fact
that fgets() returns NULL, instead of returning the pointer that was
passed in to it.

--
Jack Klein
Home: http://JK-Technology.Com
FAQs for
comp.lang.c http://c-faq.com/
comp.lang.c++ http://www.parashift.com/c++-faq-lite/
alt.comp.lang.l earn.c-c++

http://www.club.cc.cmu.edu/~ajo/docs/FAQ-acllc.html

**Bill Cunningham** · Jun 27 '08, 07:40 PM

Re: fgets question

"Ben Bacarisse" <ben.usenet@bsb .me.ukwrote in message
news:87abhvmjlz .fsf@bsb.me.uk. ..

Where do you meet people to chat about C?

On a yahoogroup called c-prog. It's for C++ too.

>I decided to test this assumption because my
>documentatio n didn't say if fgets put \0 after a string literal. Strlen
>was
>what I used to decide the \0 was not added.

>
What made you think that? There must be a null byte or strlen won't
work.
>

>How can it be added for a
>string?

>
It is there already. If fgets runs out of room it will stop one short
just so it can always put the terminating null in the buffer. You
may not get a '\n' but you will always get a 0.
>

>My code.
>>
>int main (void) {
> char input[10];
> fgets (input, sizeof(input), stdin);
> printf ("%i\n", strlen(input));

>
Better is:
printf ("%i\n", (int)strlen(inp ut));
>

Oh that looks like a cast. What are you asking printf to do?

since %i needs an int not a size_t. Alternatively use "%zu" as the
format since you seem to be using somthing close to C99.
>

>}
>>
> One more character than what was typed printf reported.

>
The '\n' is a character. If you type abc and hit return, fgets put
five things into the input array: 'a', 'b', 'c', '\n' and '\0'. This
is a four character string and strlen reports 4.
>
--
Ben.

**Bill Cunningham** · Jun 27 '08, 07:40 PM

Re: fgets question

"Jack Klein" <jackklein@spam cop.netwrote in message
news:mb4p44po62 5s7hr9odpvms41r 017glm1ch@4ax.c om...

On Mon, 09 Jun 2008 01:54:12 GMT, "Bill Cunningham" <nospam@nspam.c om>
wrote in comp.lang.c:
>

> I was talking with someone about fgets and he said that fgets puts
>the
>\n in a string but not \0. I decided to test this assumption because my
>documentatio n didn't say if fgets put \0 after a string literal.

>
Who was this person, and what credentials did he have to make you
think he was right?
>
More importantly, what does a string literal have to do with anything?
fgets() has nothing at all to do with string literals. It reads input
from a stream. String literals are quoted strings in your source
code, not input read from a stream.

OK I see then.

>Strlen was
>what I used to decide the \0 was not added. How can it be added for a
>string? My code.

>
I surely don't understand what you were trying to do above. If you
had an array of characters that does not contain at least one '\0'
within the bounds of the array, then it is not a string. Calling
strlen() on an array of characters that is not a string, due to the
lack of a '\0', produces undefined behavior. strlen() is for strings,
not arbitrary arrays of garbage characters.
>

>int main (void) {
> char input[10];
> fgets (input, sizeof(input), stdin);

>
In this call, fgets() will accept up to 9 characters from stdin. Fewer
if it receives a '\n' before the ninth character.
>
It will store these characters into consecutive elements of the array.
Regardless of whether it stores 9 or fewer characters in the array, it
will place a '\0' into the array after the last character it stores.

[snip]

So is that extra element that isn't reported then where \0 goes ?

Bill

**Keith Thompson** · Jun 27 '08, 07:40 PM

Re: fgets question

"Bill Cunningham" <nospam@nspam.c omwrites:

I was talking with someone about fgets and he said that fgets puts the
\n in a string but not \0. I decided to test this assumption because my
documentation didn't say if fgets put \0 after a string literal. Strlen was
what I used to decide the \0 was not added. How can it be added for a
string? My code.
>
int main (void) {
char input[10];
fgets (input, sizeof(input), stdin);
printf ("%i\n", strlen(input));
}

You had "#include <stdio.h>", right?

One more character than what was typed printf reported.

I doubt that. For example, if you typed "abc" you almost certainly
typed 4 character: 'a', 'b', 'c', and the enter key, and strlen(input)
should have given you 4.

--
Keith Thompson (The_Other_Keit h) kst-u@mib.org <http://www.ghoti.net/~kst>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

**Keith Thompson** · Jun 27 '08, 07:40 PM

Re: fgets question

"Bill Cunningham" <nospam@nspam.c omwrites:

"Jack Klein" <jackklein@spam cop.netwrote in message
news:mb4p44po62 5s7hr9odpvms41r 017glm1ch@4ax.c om...

>On Mon, 09 Jun 2008 01:54:12 GMT, "Bill Cunningham" <nospam@nspam.c om>
>wrote in comp.lang.c:

[...]

>>int main (void) {
>> char input[10];
>> fgets (input, sizeof(input), stdin);

>>
>In this call, fgets() will accept up to 9 characters from stdin. Fewer
>if it receives a '\n' before the ninth character.
>>
>It will store these characters into consecutive elements of the array.
>Regardless of whether it stores 9 or fewer characters in the array, it
>will place a '\0' into the array after the last character it stores.

[snip]
>
So is that extra element that isn't reported then where \0 goes ?

The '\0' goes immediately after the last stored character from the
input, as Jack just told you.

If you typed 'a', 'b', 'c, and <return(4 character), then fgets()
will store a total of *five* characters in your array: 'a', 'b', 'c',
'\n', and '\0'.

strlen()'s result is the length of the string *excluding* the
terminating '\0' (in this case, 4).

--
Keith Thompson (The_Other_Keit h) kst-u@mib.org <http://www.ghoti.net/~kst>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

**pete** · Jun 27 '08, 07:40 PM

Re: fgets question

Ben Bacarisse wrote:

The '\n' is a character. If you type abc and hit return, fgets put
five things into the input array: 'a', 'b', 'c', '\n' and '\0'. This
is a four character string and strlen reports 4.

It's a five character string; and the length is four.

--
pete

**Richard** · Jun 27 '08, 07:40 PM

Re: fgets question

"Bill Cunningham" <nospam@nspam.c omwrites:

>>
>Better is:
> printf ("%i\n", (int)strlen(inp ut));
>>

>
Oh that looks like a cast. What are you asking printf to do?

Why not look up the documentation for printf and work it out yourself?

**Ben Bacarisse** · Jun 27 '08, 07:40 PM

Re: fgets question

"Bill Cunningham" <nospam@nspam.c omwrites:

"Ben Bacarisse" <ben.usenet@bsb .me.ukwrote in message
news:87abhvmjlz .fsf@bsb.me.uk. ..
>

>Where do you meet people to chat about C?

>
On a yahoogroup called c-prog. It's for C++ too.

Ah. I asked because if you could talk to someone face-to-face about C
I think you'd make better progress. Online chat is not at all the same.

--
Ben.

**Mark McIntyre** · Jun 27 '08, 07:40 PM

Re: fgets question

Bill Cunningham wrote:

> printf ("%i\n", (int)strlen(inp ut));
>>

>
Oh that looks like a cast. What are you asking printf to do?

%i requires an int.
strlen returns a size_t
the two are not necessarily the same width.

**CBFalconer** · Jun 27 '08, 07:40 PM

Re: fgets question

pete wrote:

Ben Bacarisse wrote:
>

>The '\n' is a character. If you type abc and hit return, fgets
>put five things into the input array: 'a', 'b', 'c', '\n' and
>'\0'. This is a four character string and strlen reports 4.

>
It's a five character string; and the length is four.

No, it is a five char array, which will fit in a five (or longer)
char array, and it holds a 4 char string.

--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home .att.net>
Try the download section.

** Posted from http://www.teranews.com **

**Bill Cunningham** · Jun 27 '08, 07:40 PM

Re: fgets question

"Richard" <rgrdev@gmail.c omwrote in message
news:g2iqq5$4em $5@registered.m otzarella.org.. .

Why not look up the documentation for printf and work it out yourself?

I'm pretty familiar with printf. There are things though I've seen I've
wondered about like [] in the first parameter of printf for example. I've
not seen those in the docs. And I have some pretty though stuff about
printf.

Bill

**Richard Tobin** · Jun 27 '08, 07:40 PM

Re: fgets question

In article <46h3k.7711$lE3 .5816@trnddc05> ,
Bill Cunningham <nospam@nspam.c omwrote:

I'm pretty familiar with printf. There are things though I've seen I've
>wondered about like [] in the first parameter of printf for example. I've
>not seen those in the docs.

What sort of "docs" are you reading? You can't expect to write C
without a proper reference, and any "doc" that doesn't explain all the
conversion specifiers is not a proper reference. Get K&R, or the
standard. Even the unix man pages tell you everything you need to
know about printf.

>And I have some pretty though stuff about printf.

No you don't.

-- Richard
--
In the selection of the two characters immediately succeeding the numeral 9,
consideration shall be given to their replacement by the graphics 10 and 11 to
facilitate the adoption of the code in the sterling monetary area. (X3.4-1963)

**Keith Thompson** · Jun 27 '08, 07:40 PM

Re: fgets question

CBFalconer <cbfalconer@yah oo.comwrites:

pete wrote:

>Ben Bacarisse wrote:
>>

>>The '\n' is a character. If you type abc and hit return, fgets
>>put five things into the input array: 'a', 'b', 'c', '\n' and
>>'\0'. This is a four character string and strlen reports 4.

>>
>It's a five character string; and the length is four.

>
No, it is a five char array, which will fit in a five (or longer)
char array, and it holds a 4 char string.

It depends on what you mean by "a 4 char string".

A "string" is, by definition,

a contiguous sequence of characters terminated by and including
the first null character.

The "length of a string" is, by definition:

the number of bytes preceding the null character

(Definitions are from C99 7.1.1p1.)

To emphasize the point, the terminating '\0' is part of the string.

So the string in question consists of exactly 5 characters, and has a
length of 4.

--
Keith Thompson (The_Other_Keit h) kst-u@mib.org <http://www.ghoti.net/~kst>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

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