String Reverse Question

[Joe Wright]

Dread, as I already said in another post: Sunday silliness.... :-/




Sorry, no, you invoke undefined behaviour by *computing* the invalid
value (Cf. ISO/IEC 9899:1999 6.5.6p8).




It's the only one of the three version, that does not invoke undefined
behaviour when fed with an empty string.

I started some of this, sorry. How about this last one?

char *revstr(char *s) {
char t, *b = s, *e;
if (s && *s) {
e = s + strlen(s);
while (b < e)
if (e - b > 1)
t = *b, *b++ = *--e, *e = t;
else
--e;
}
return s;
}

It is a NOP in the case of s == NULL or *s == '\0'. Also, unlike its
younger bretheren, it avoids reversing a one-character string or the
center character of an odd-length string.

 

[Irrwahn Grausewitz]

I started some of this, sorry. How about this last one?

char *revstr(char *s) {
char t, *b = s, *e;
if (s && *s) {
e = s + strlen(s);
while (b < e)
if (e - b > 1)
t = *b, *b++ = *--e, *e = t;
else
--e;
}
return s;
}

It is a NOP in the case of s == NULL or *s == '\0'. Also, unlike its
younger bretheren, it avoids reversing a one-character string or the
center character of an odd-length string.

That looks OK. However, now that you've added the extra check
for *s != '\0', you can't invoke undefined behaviour anyway, and
thus can again simplify, without loss of efficiency, to:

char *revstr(char *s) {
char t, *b = s, *e;
if (s && *s) {
e = s + strlen(s);
while (b < --e)
t = *b, *b++ = *e, *e = t;
}
return s;
}

Funny, isn't it?

Regards

 

[Richard Delorme]

Irrwahn Grausewitz a écrit :

Sorry, no, you invoke undefined behaviour by *computing* the invalid
value (Cf. ISO/IEC 9899:1999 6.5.6p8).

Right, I am another victim of Sunday silliness ;-)

 

[Christopher Benson-Manica]

p = malloc( (len + 1) * sizeof *p);

Of course, if sizeof *p == 1 as it does here, then these expressions
are the same, but the point of "sizeof *p" is to be able to avoid problems
related to what the type of "p" is.

Yep, I blew it, and was saved from disaster only by luck :|

 

[Christopher Benson-Manica]

James McIninch <[email protected]> spoke thus:

Please don't top-post. Thanks.

#include <stdlib.h>
#include <string.h>
char * strrev (const char *s)
{
char * p;
int i, j;
j = strlen(s)
if (p = calloc(1, j+1))

ITYM

if( p=calloc(j+1,1) );

The results, of course, are identical, but it matches the semantics of
calloc()'s prototype.

for (i=0; i<j; ++i)
p = s[j-i];

Per Barry's post, ITYM

p=s[j-i-1];

although of course changing the value of j appropriately before
entering the loop would be viable as well.

 

[Christopher Benson-Manica]

Just a bit curious here - Is it ok to allocate locally here, and
return it. Would that not lead to memory leaks.

Certainly not, as long as you're careful to free it where you returned
it.