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Hi experts,
I'm trying to write a program to solve the following exercise:
Accept three strings from the user. Find the first occurrence of the
first string in the third string. If it is present replace the second
string in its place.
Note: First and Second string might not be of the same length.
E.g.: First string: "cat"
Second string: "camel"
Third string: "concatenate"
Output: "concamelenate"
I've made a complete working program. The following is the main() part
of my program.
int main()
{
char s1[80], s2[80], s3[80];
int pos;
/*Code for accepting 3 strings from user*/
/*and storing in s1, s2 and s3*/
pos=find(s1, s3); /*find() returns the position at which */
/*s1 occurs in s3, -1 otherwise*/
delete(pos, s3, strlen(s1)); /*Deletes strlen(s1) characters from s3
starting at pos*/
/*Does not delete if pos<0*/
insert(pos, s3, s2);/*Inserts s2 in s3 at location specified by pos*/
/*Does nothing if pos<0*/
puts(s3);
}
So, the three operations I've used for finding and replacing are:
(1)find
(2)delete and
(3)insert
My question is: Is there a better way of doing this?
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On Sun, 15 Oct 2006, Registered User wrote:
>
Hi experts,
I'm trying to write a program to solve the following exercise:
Accept three strings from the user. Find the first occurrence of the
first string in the third string. If it is present replace the second
string in its place.
Note: First and Second string might not be of the same length.
E.g.: First string: "cat"
Second string: "camel"
Third string: "concatenate"
Output: "concamelenate"
I've made a complete working program. The following is the main() part
of my program.
int main()
{
char s1[80], s2[80], s3[80];
int pos;
/*Code for accepting 3 strings from user*/
/*and storing in s1, s2 and s3*/
pos=find(s1, s3); /*find() returns the position at which */
/*s1 occurs in s3, -1 otherwise*/
delete(pos, s3, strlen(s1)); /*Deletes strlen(s1) characters from s3
starting at pos*/
/*Does not delete if pos<0*/
insert(pos, s3, s2);/*Inserts s2 in s3 at location specified by pos*/
/*Does nothing if pos<0*/
What if strlen(s3)+strlen(s2) >= 80? Does 'insert' behave differently
then, or have you got a buffer overflow? http://en.wikipedia.org/wiki/Buffer_overflow
Also, you could speed up some cases by deleting or inserting only
as many characters as you need to, and writing over the others. For
example, to get from "concatenate" to "concamelenate", you'd delete
zero characters, insert two characters ("concatxxenate"), and then
overwrite all five characters ("concamelenate"). To get from
"concatenate" to "condogenate", you wouldn't insert or delete any
characters; just write "dog" at the right point in "concatenate"
and you're done.
HTH,
-Arthur | |
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On Sun, 15 Oct 2006, Arthur J. O'Dwyer wrote:
>
On Sun, 15 Oct 2006, Registered User wrote:
>>
Hi experts,
I'm trying to write a program to solve the following exercise:
Accept three strings from the user. Find the first occurrence of the
first string in the third string. If it is present replace the second
string in its place.
Note: First and Second string might not be of the same length.
E.g.: First string: "cat"
Second string: "camel"
Third string: "concatenate"
Output: "concamelenate"
I've made a complete working program. The following is the main() part
of my program.
int main()
{
char s1[80], s2[80], s3[80];
int pos;
/*Code for accepting 3 strings from user*/
/*and storing in s1, s2 and s3*/
pos=find(s1, s3); /*find() returns the position at which */
/*s1 occurs in s3, -1 otherwise*/
delete(pos, s3, strlen(s1)); /*Deletes strlen(s1) characters from s3
starting at pos*/
/*Does not delete if pos<0*/
insert(pos, s3, s2);/*Inserts s2 in s3 at location specified by pos*/
/*Does nothing if pos<0*/
What if strlen(s3)+strlen(s2) >= 80? Does 'insert' behave differently
then, or have you got a buffer overflow?
http://en.wikipedia.org/wiki/Buffer_overflow
Also, you could speed up some cases by deleting or inserting only
as many characters as you need to, and writing over the others. For
example, to get from "concatenate" to "concamelenate", you'd delete
zero characters, insert two characters ("concatxxenate"), and then
overwrite all five characters ("concamelenate"). To get from
"concatenate" to "condogenate", you wouldn't insert or delete any
characters; just write "dog" at the right point in "concatenate"
and you're done.
One could also merge the the three operations together by
using strstr(), memmove() and memcpy() in a single function:
#include <string.h>
char *found = strstr(s3, s1);
size_t len = strlen(s2);
if (found) {
/* Exercise: insert bounds checking code here */
memmove(found + len,
found,
strlen(s3) - strlen(s1) - (found - s3));
memcpy(found, s2, len);
}
Tak-Shing | |
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On Mon, 16 Oct 2006, Tak-Shing Chan wrote:
On Sun, 15 Oct 2006, Arthur J. O'Dwyer wrote:
>>
On Sun, 15 Oct 2006, Registered User wrote:
>>>
Hi experts,
I'm trying to write a program to solve the following exercise:
Accept three strings from the user. Find the first occurrence of the
first string in the third string. If it is present replace the second
string in its place.
Note: First and Second string might not be of the same length.
E.g.: First string: "cat"
Second string: "camel"
Third string: "concatenate"
Output: "concamelenate"
I've made a complete working program. The following is the main() part
of my program.
int main()
{
char s1[80], s2[80], s3[80];
int pos;
/*Code for accepting 3 strings from user*/
/*and storing in s1, s2 and s3*/
pos=find(s1, s3); /*find() returns the position at which */
/*s1 occurs in s3, -1 otherwise*/
delete(pos, s3, strlen(s1)); /*Deletes strlen(s1) characters from s3
starting at pos*/
/*Does not delete if pos<0*/
insert(pos, s3, s2);/*Inserts s2 in s3 at location specified by pos*/
/*Does nothing if pos<0*/
What if strlen(s3)+strlen(s2) >= 80? Does 'insert' behave differently
then, or have you got a buffer overflow?
http://en.wikipedia.org/wiki/Buffer_overflow
Also, you could speed up some cases by deleting or inserting only
as many characters as you need to, and writing over the others. For
example, to get from "concatenate" to "concamelenate", you'd delete
zero characters, insert two characters ("concatxxenate"), and then
overwrite all five characters ("concamelenate"). To get from
"concatenate" to "condogenate", you wouldn't insert or delete any
characters; just write "dog" at the right point in "concatenate"
and you're done.
One could also merge the the three operations together by
using strstr(), memmove() and memcpy() in a single function:
#include <string.h>
char *found = strstr(s3, s1);
size_t len = strlen(s2);
if (found) {
/* Exercise: insert bounds checking code here */
memmove(found + len,
found,
strlen(s3) - strlen(s1) - (found - s3));
memcpy(found, s2, len);
}
Sorry, the memmove() line above is wrong. It should be:
memmove(found + len,
found + strlen(s1),
strlen(s3) - strlen(s1) - (found - s3));
Tak-Shing | |
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P: n/a
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On Mon, 16 Oct 2006, Tak-Shing Chan wrote:
On Mon, 16 Oct 2006, Tak-Shing Chan wrote:
>On Sun, 15 Oct 2006, Arthur J. O'Dwyer wrote:
>>>
On Sun, 15 Oct 2006, Registered User wrote:
Hi experts,
I'm trying to write a program to solve the following exercise:
Accept three strings from the user. Find the first occurrence of the
first string in the third string. If it is present replace the second
string in its place.
Note: First and Second string might not be of the same length.
E.g.: First string: "cat"
Second string: "camel"
Third string: "concatenate"
Output: "concamelenate"
I've made a complete working program. The following is the main() part
of my program.
int main()
{
char s1[80], s2[80], s3[80];
int pos;
/*Code for accepting 3 strings from user*/
/*and storing in s1, s2 and s3*/
pos=find(s1, s3); /*find() returns the position at which */
/*s1 occurs in s3, -1 otherwise*/
delete(pos, s3, strlen(s1)); /*Deletes strlen(s1) characters from s3
starting at pos*/
/*Does not delete if pos<0*/
insert(pos, s3, s2);/*Inserts s2 in s3 at location specified by pos*/
/*Does nothing if pos<0*/
What if strlen(s3)+strlen(s2) >= 80? Does 'insert' behave differently
then, or have you got a buffer overflow?
http://en.wikipedia.org/wiki/Buffer_overflow
Also, you could speed up some cases by deleting or inserting only
as many characters as you need to, and writing over the others. For
example, to get from "concatenate" to "concamelenate", you'd delete
zero characters, insert two characters ("concatxxenate"), and then
overwrite all five characters ("concamelenate"). To get from
"concatenate" to "condogenate", you wouldn't insert or delete any
characters; just write "dog" at the right point in "concatenate"
and you're done.
One could also merge the the three operations together by
using strstr(), memmove() and memcpy() in a single function:
#include <string.h>
char *found = strstr(s3, s1);
size_t len = strlen(s2);
if (found) {
/* Exercise: insert bounds checking code here */
memmove(found + len,
found,
strlen(s3) - strlen(s1) - (found - s3));
memcpy(found, s2, len);
}
Sorry, the memmove() line above is wrong. It should be:
memmove(found + len,
found + strlen(s1),
strlen(s3) - strlen(s1) - (found - s3));
Oops, I forgot the terminating '\0':
memmove(found + len,
found + strlen(s1),
strlen(s3) - strlen(s1) - (found - s3) + 1);
Tak-Shing | |
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P: n/a
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In comp.lang.c Tak-Shing Chan <t.****@gold.ac.ukwrote:
Sorry, the memmove() line above is wrong. It should be:
Oops, I forgot the terminating '\0':
Perhaps posting on Mondays is a bad idea :-)
--
C. Benson Manica | I *should* know what I'm talking about - if I
cbmanica(at)gmail.com | don't, I need to know. Flames welcome. | |
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Is there a better way of doing this?
Your method looks fine. In the real world, time is money. An argument can
be made that whatever method works for you, especially if it's simple to
code, understand, and maintain, can be considered "better." Are there more
elegant solutions? Possibly. Are the "better?" Not necessarily.
I don't code in C but here's one way to do it PowerBASIC. (This probably
won't help you, since C almost surely doesn't have its own EXTRACTS$
function, but I couldn't resist.)
first$ = "cat" : second$ = "dog"
third$ = "Actually, I believe cat is man's best friend!"
a$ = EXTRACT$(third$, first$)
IF a$ <third$ THEN
a$ = a$ + second$ + RIGHT$(third$, LEN(third$) - LEN(a$) - LEN(first$))
END IF
In the above example, the variable a$ now is equal to "Actually, I believe
dog is man's best friend!"
Ed Collins
"Registered User" <in*******************@gmail.comwrote in message
news:11*********************@m73g2000cwd.googlegro ups.com...
Hi experts,
I'm trying to write a program to solve the following exercise:
rest snipped...
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