entering more characters

[kk]

Hi all,

in the following code, i enter more characters into a character array.
array size declared as 80. i enter more characters then i get
segmentation fault. if anybody knows give me reply.
thanks in advance.

Regards
kk


main()
{
char s[80];
scanf("%[^\n]s",s);
printf("%s",s);
}

[Suman]

Re: entering more characters



kk wrote:[color=blue]
> Hi all,
>
> in the following code, i enter more characters into a character array.
> array size declared as 80. i enter more characters then i get
> segmentation fault. if anybody knows give me reply.[/color]
Are you entering more than 80 characters?
Writing past array bounds is your problem.[color=blue]
> thanks in advance.
>
> Regards
> kk
>
>
> main()
> {
> char s[80];
> scanf("%[^\n]s",s);
> printf("%s",s);
> }[/color]

[Barry]

Re: entering more characters

Why don't u do a bounce check? never get past an array limit


kk wrote:[color=blue]
> Hi all,
>
> in the following code, i enter more characters into a character array.
> array size declared as 80. i enter more characters then i get
> segmentation fault. if anybody knows give me reply.
> thanks in advance.
>
> Regards
> kk
>
>
> main()
> {
> char s[80];
> scanf("%[^\n]s",s);
> printf("%s",s);
> }[/color]

[Suman]

Re: entering more characters



Suman wrote:[color=blue]
> kk wrote:[color=green]
> > Hi all,
> >
> > in the following code, i enter more characters into a character array.
> > array size declared as 80. i enter more characters then i get
> > segmentation fault. if anybody knows give me reply.[/color]
> Are you entering more than 80 characters?[/color]
Ah! Off by one! You can't have more than 79 characters in your
`s' thingy.[color=blue]
> Writing past array bounds is your problem.[color=green]
> > thanks in advance.
> >
> > Regards
> > kk
> >
> >
> > main()
> > {
> > char s[80];
> > scanf("%[^\n]s",s);
> > printf("%s",s);
> > }[/color][/color]
The following might be of interest to you:

http://groups.google.co.in/group/comp.lang.c/browse_frm/

thread/4612b24419b17c7 8/27bbc52f5be7f3b e?
q=pete+%2B+scan f&rnum=6&hl=en# 27bbc52f5be7f3b e

[Kenny McCormack]

Encouraging illegal behavior is O/T in clc (Was: Re: entering more characters)

In article <1122466531.889 573.144440@o13g 2000cwo.googleg roups.com>,
Barry <prbharath@gmai l.com> wrote:[color=blue]
>Why don't u do a bounce check? never get past an array limit[/color]

I think bouncing checks is illegal.

[Lew Pitcher]

Re: entering more characters

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kk wrote:[color=blue]
> Hi all,
>
> in the following code, i enter more characters into a character array.
> array size declared as 80. i enter more characters then i get
> segmentation fault. if anybody knows give me reply.
> thanks in advance.[/color]

A truism that my mother taught me applies here...

"You cant stuff a size 10 foot into size 7 shoes. Either the shoes split
or your foot does."

In this case, you can't stuff more than 80 chars into an 80 char array.
If you do, you write over something (what, we cannot tell you) outside
of the array, and cause your program to fall over.

In other words (to quote Mel Brooks) "Don't do that."


- --

Lew Pitcher, IT Specialist, Enterprise Data Systems
Enterprise Technology Solutions, TD Bank Financial Group

(Opinions expressed here are my own, not my employer's)
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[Walter Roberson]

Re: entering more characters

In article <1122465724.574 698.280280@g44g 2000cwa.googleg roups.com>,
kk <kirancareers@g mail.com> wrote:[color=blue]
>in the following code, i enter more characters into a character array.
>array size declared as 80. i enter more characters then i get
>segmentation fault. if anybody knows give me reply.
>thanks in advance.[/color]
[color=blue]
>main()
>{
> char s[80];
> scanf("%[^\n]s",s);
> printf("%s",s);
>}[/color]


I -suspect- that the answer you are looking for is:

#include <stdio.h>
int main(void) {
char s[80];
scanf("%79[^\n]", s);
printf("%s",s);
}


Note that if you use this and the input stream has more than 79
characters before the newline, then the remaining characters will
remain in the input buffer. Also, watch out because even if a newline
was supplied, it is going to remain in the input buffer.

Also note that the format specifier ends at the ] and that the
s that you had after that would be treated as a literal character
to be matched -- the format specifier is not %[something]s
just %[something]
--
Ceci, ce n'est pas une idée.

[Jack Klein]

Re: entering more characters

On 27 Jul 2005 05:15:31 -0700, "Barry" <prbharath@gmai l.com> wrote in
comp.lang.c:

1. Stop top posting.
[color=blue]
> Why don't u do a bounce check? never get past an array limit[/color]

2. Don't post nonsense.

-- 'u' doesn't post here anymore, we kicked him out.

-- C doesn't have anything called a "bounce check".

-- There is no way to stop scanf "%s" without a length value from
overflowing a small buffer. On some implementations , there is no way
to prevent it from overflowing any size buffer.
[color=blue]
> kk wrote:[color=green]
> > Hi all,
> >
> > in the following code, i enter more characters into a character array.
> > array size declared as 80. i enter more characters then i get
> > segmentation fault. if anybody knows give me reply.
> > thanks in advance.
> >
> > Regards
> > kk
> >
> >
> > main()
> > {
> > char s[80];
> > scanf("%[^\n]s",s);
> > printf("%s",s);
> > }[/color][/color]

--
Jack Klein
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