String reversing problem

[Albert]

Why doesn't:

#include <stdio.h>

void reverse(char[], int);

main()
{
char s[5];

s[0] = 'h';
s[1] = 'e';
s[2] = 'l';
s[3] = 'l';
s[4] = 'o';
reverse(s, 5);

for (int i=0; i<=4; i++)
putchar(s[i]);
return 0;
}

void reverse(char s[], int num_elements)
{
int i, j;

for (i=0,j=num_elem ents-1; (i<=num_element s-1) && (j>=0); i++,j--)
s[i] = s[j];
}

output:

olleh

?

[tmp123]

Chuck F. wrote:

pai wrote:

I think an extra variable is needed to store, bcoz u cant
interchange 2 variable as such. or is there any way to do it

Include context, without which your message is meaningless. For
means on the broken google interface, see my sig below.

Try this, after #include <string.h>:

/* reverse string in place. Return length */
static size_t revstring(char *stg)
{
char *last, temp;
size_t lgh;

if ((lgh = strlen(stg)) > 1) {
last = stg + lgh; /* points to '\0' */
while (last-- > stg) {
temp = *stg; *stg++ = *last; *last = temp;
}
}
return lgh;
} /* revstring */

--
"If you want to post a followup via groups.google.c om, don't use
the broken "Reply" link at the bottom of the article. Click on
"show options" at the top of the article, then click on the
"Reply" at the bottom of the article headers." - Keith Thompson
More details at: <http://cfaj.freeshell. org/google/>

Hi,

I agree on the previous, and, of course, if some day I need to code
something similar I will write more or less the same (specially in an
answer to a beginner).

But, just for fun, and taken into account is the third time this
question has been posted, another version:

int revstring ( char *s )
{
char *e;
int r;
for( e=s+(r=strlen(s ))-1; s<e; *s^=*e^=*s^=*e, s++, e--);
return r;
}
Kind regards.

(hope google doesn't heats indentation).

[William J. Leary Jr.]

"Albert" <al************ *****@gmail.com > wrote in message
news:11******** *************@g 43g2000cwa.goog legroups.com...

What do you mean by 'desk checking'?

I'd have called it "paper check," or, when I'm feeling whimsical "let's play
computer."

Whatever way it's expressed, it means get out some paper and a pen(cil) and
perform, yourself, the steps the computer will take to execute your program.
Sometimes it's better (or quicker) than a debugger. Sometimes it's the only
way to debug (for very limited platforms, for example). Well worth practicing.

- Bill

[Chuck F.]

tmp123 wrote:

Chuck F. wrote:
.... snip ...

/* reverse string in place. Return length */
static size_t revstring(char *stg)
{
char *last, temp;
size_t lgh;

if ((lgh = strlen(stg)) > 1) {
last = stg + lgh; /* points to '\0' */
while (last-- > stg) {
temp = *stg; *stg++ = *last; *last = temp;
}
}
return lgh;
} /* revstring */

.... snip ...
But, just for fun, and taken into account is the third time this
question has been posted, another version:

int revstring ( char *s )
{
char *e;
int r;
for( e=s+(r=strlen(s ))-1; s<e; *s^=*e^=*s^=*e, s++, e--);
return r;
}

FYI your version invokes undefined behaviour. Try it with a string
of zero chars. My length test is not there for fun. I also have
evil suspicions about the xor operations.

--
"If you want to post a followup via groups.google.c om, don't use
the broken "Reply" link at the bottom of the article. Click on
"show options" at the top of the article, then click on the
"Reply" at the bottom of the article headers." - Keith Thompson
More details at: <http://cfaj.freeshell. org/google/>

[tmp123]

Chuck F. wrote:

tmp123 wrote:

int revstring ( char *s )
{
char *e;
int r;
for( e=s+(r=strlen(s ))-1; s<e; *s^=*e^=*s^=*e, s++, e--);
return r;
}

FYI your version invokes undefined behaviour. Try it with a string
of zero chars. My length test is not there for fun. I also have
evil suspicions about the xor operations.

Hi,

If length is 0, then e=s-1, thus s<e is false exiting loop.

However, if the usage of pointer comparation and pointer substraction
is not welcome, another version:

void revstring ( char *s )
{
char *e;
for( e=s+strlen(s); s!=e-- && s!=e; *s^=*e^=*s++^=* e);
}

Kind regards.

PS: some days ago, someone posted about if "C is easy". I do not known,
but it is tricky.

[Keith Thompson]

"tmp123" <tm****@menta.n et> writes:
[...]

void revstring ( char *s )
{
char *e;
for( e=s+strlen(s); s!=e-- && s!=e; *s^=*e^=*s++^=* e);
}
Undefined behavior.
PS: some days ago, someone posted about if "C is easy". I do not known,
but it is tricky.

It can be if you go out of your way to make it tricky.

--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.

[tmp123]

Keith Thompson wrote:

"tmp123" <tm****@menta.n et> writes:
[...]

void revstring ( char *s )
{
char *e;
for( e=s+strlen(s); s!=e-- && s!=e; *s^=*e^=*s++^=* e);
}

Undefined behavior.

Could be... but, could you prove your statement?

[Tim Rentsch]

"tmp123" <tm****@menta.n et> writes:

Keith Thompson wrote:

"tmp123" <tm****@menta.n et> writes:
[...]

void revstring ( char *s )
{
char *e;
for( e=s+strlen(s); s!=e-- && s!=e; *s^=*e^=*s++^=* e);
}

Undefined behavior.

Could be... but, could you prove your statement?

Even the first half of the control expression (namely, 's!=e--')
can yield undefined behavior if strlen(s) == 0.

[Keith Thompson]

"tmp123" <tm****@menta.n et> writes:

Keith Thompson wrote:

"tmp123" <tm****@menta.n et> writes:
[...]

> void revstring ( char *s )
> {
> char *e;
> for( e=s+strlen(s); s!=e-- && s!=e; *s^=*e^=*s++^=* e);
> }

Undefined behavior.

Could be... but, could you prove your statement?

The expression
*s^=*e^=*s++^=* e
modifies both *s and *e twice between sequence points.

--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.

[Christian Bau]

In article <11************ **********@g47g 2000cwa.googleg roups.com>,
"tmp123" <tm****@menta.n et> wrote:

Keith Thompson wrote:

"tmp123" <tm****@menta.n et> writes:
[...]

void revstring ( char *s )
{
char *e;
for( e=s+strlen(s); s!=e-- && s!=e; *s^=*e^=*s++^=* e);
}

Undefined behavior.

Could be... but, could you prove your statement?

It's kind of obvious. Even if it wasn't, whoever wrote that kind of code
should be slapped silly. Immediate removal from any programming team
that I am involved with.

[Chuck F.]

tmp123 wrote:

Chuck F. wrote:

tmp123 wrote:

> int revstring ( char *s )
> {
> char *e;
> int r;
> for( e=s+(r=strlen(s ))-1; s<e; *s^=*e^=*s^=*e, s++, e--);
> return r;
> }
FYI your version invokes undefined behaviour. Try it with a
string of zero chars. My length test is not there for fun. I
also have evil suspicions about the xor operations.

If length is 0, then e=s-1, thus s<e is false exiting loop.

However, if the usage of pointer comparation and pointer
substraction is not welcome, another version:

void revstring ( char *s )
{
char *e;
for( e=s+strlen(s); s!=e-- && s!=e; *s^=*e^=*s++^=* e);
}

Same problem and undefined behaviour. You are not allowed to
generate a pointer that points before s (although you are allowed
to generate one that points just after s).

--
"If you want to post a followup via groups.google.c om, don't use
the broken "Reply" link at the bottom of the article. Click on
"show options" at the top of the article, then click on the
"Reply" at the bottom of the article headers." - Keith Thompson
More details at: <http://cfaj.freeshell. org/google/>