This is my program on swap of strings.i just used pointer for swapping
the string.
can anyone help me why it is not working?

#include<iostream.h>
#include<string.h>

void swap(char *char1, char *char2)
{

char *temp;
temp=char1;
char1=char2;
char2=temp;
}

void main()
{

char*char1="sidd";
char* char2="pras";
cout<<"string 1=" <<char1<<endl;
cout<<"string 2=" <<char2<<endl;

swap( char1, char2);

cout<<"string 1=" <<char1<<endl;
cout<<"string 2=" <<char2<<endl;

}

thanks in advance.

On 25 Dec 2005 22:27:20 -0800 in comp.lang.c++, "meendar"
<askjavaprogrammers@gmail.com> wrote,[color=blue]
>This is my program on swap of strings.i just used pointer for swapping
>the string.
>can anyone help me why it is not working?
>
>#include<iostream.h>
>#include<string.h>
>
> void swap(char *char1, char *char2)
> {
>
> char *temp;
> temp=char1;
> char1=char2;
> char2=temp;
> }[/color]

char1 and char2 are local variables in the function swap().
Altering their values does not change anything outside that
function.

thanks for your help.
But we are using pointer to denote these two variables.why doesn't it
work.

meendar wrote:[color=blue]
> This is my program on swap of strings.i just used pointer for swapping
> the string.
> can anyone help me why it is not working?
>
> #include<iostream.h>
> #include<string.h>
>
> void swap(char *char1, char *char2)
> {
>
> char *temp;
> temp=char1;
> char1=char2;
> char2=temp;
> }
>
> void main()
> {
>
> char*char1="sidd";
> char* char2="pras";
> cout<<"string 1=" <<char1<<endl;
> cout<<"string 2=" <<char2<<endl;
>
> swap( char1, char2);
>
> cout<<"string 1=" <<char1<<endl;
> cout<<"string 2=" <<char2<<endl;
>
> }
>
> thanks in advance.
>[/color]

C++ has strings as part of the standard library. Use those instead of
icky character arrays. Your code looks more like C than C++. Also, main
should return an integer, not void.

#include <iostream>
#include <string>
#include <cstdlib>

int main()
{
// Creating strings
std::string first = "sidd";
std::string second = "pras";

// Displaying strings
std::cout << first << std::endl;
std::cout << second << std::endl;

// Swapping strings
first.swap(second);

// Displaying strings
std::cout << first << std::endl;
std::cout << second << std::endl;

return EXIT_SUCCESS;
}

On 25 Dec 2005 22:33:04 -0800 in comp.lang.c++, "meendar"
<askjavaprogrammers@gmail.com> wrote,[color=blue]
>thanks for your help.
>But we are using pointer to denote these two variables.why doesn't it
>work.[/color]

Reason stated. Your pointer variables are local to the function.

On Mon, 26 Dec 2005 06:35:59 +0000, W Marsh
<wayneDOTmarshATgmailDOTcom@decipher> wrote:
[color=blue]
>meendar wrote:[color=green]
>> This is my program on swap of strings.i just used pointer for swapping
>> the string.
>> can anyone help me why it is not working?
>>[/color][/color]
(...)[color=blue][color=green]
>>[/color]
>
>C++ has strings as part of the standard library. Use those instead of
>icky character arrays. Your code looks more like C than C++. Also, main
>should return an integer, not void.
>[/color]
(...)

And, if you persist on using character pointers, do write the program
as folllows:


#include<iostream>


void swap(const char *& char1, const char *& char2)
{

const char *temp;
temp=char1;
char1=char2;
char2=temp;
}

void main()
{
using std::cout;

const char*char1="sidd";
const char* char2="pras";
cout<<"string 1=" <<char1<<endl;
cout<<"string 2=" <<char2<<endl;

swap( char1, char2);

cout<<"string 1=" <<char1<<endl;
cout<<"string 2=" <<char2<<endl;

}

But, of course, it is much better to use std::string, as stated W
Marsh

Regards,

-- Zara

David Harmon wrote:[color=blue]
> On 25 Dec 2005 22:27:20 -0800 in comp.lang.c++, "meendar"
> <askjavaprogrammers@gmail.com> wrote,
>[color=green]
>>This is my program on swap of strings.i just used pointer for swapping
>>the string.
>>can anyone help me why it is not working?
>>
>>#include<iostream.h>
>>#include<string.h>
>>
>> void swap(char *char1, char *char2)
>> {
>>
>> char *temp;
>> temp=char1;
>> char1=char2;
>> char2=temp;
>> }[/color]
>
>
> char1 and char2 are local variables in the function swap().
> Altering their values does not change anything outside that
> function.
>[/color]

Since the pointers are local in scope to the swap function, upon
returning from swap your variables are unchanged. Try this:

void swap(char ** char1, char ** char2)
{
char * temp = *char1;
*char1 = *char2;
*char2 = temp;
}

"Zara" <yozara@terra.es> wrote in message
news:n48vq1tot8cnc7facia8sj631pqerj3c84@4ax.com...[color=blue]
> On Mon, 26 Dec 2005 06:35:59 +0000, W Marsh
> <wayneDOTmarshATgmailDOTcom@decipher> wrote:
>[color=green]
>>meendar wrote:[color=darkred]
>>> This is my program on swap of strings.i just used pointer for swapping
>>> the string.
>>> can anyone help me why it is not working?
>>>[/color][/color]
> (...)[color=green][color=darkred]
>>>[/color]
>>
>>C++ has strings as part of the standard library. Use those instead of
>>icky character arrays. Your code looks more like C than C++. Also, main
>>should return an integer, not void.
>>[/color]
> (...)
>
> And, if you persist on using character pointers, do write the program
> as folllows:
>
>
> #include<iostream>
>
>
> void swap(const char *& char1, const char *& char2)[/color]

Will this const work, since the variables are actually being changed?
Doesn't he want them non constant?
[color=blue]
> {
>
> const char *temp;
> temp=char1;
> char1=char2;
> char2=temp;
> }
>
> void main()
> {
> using std::cout;
>
> const char*char1="sidd";
> const char* char2="pras";
> cout<<"string 1=" <<char1<<endl;
> cout<<"string 2=" <<char2<<endl;
>
> swap( char1, char2);
>
> cout<<"string 1=" <<char1<<endl;
> cout<<"string 2=" <<char2<<endl;
>
> }
>
> But, of course, it is much better to use std::string, as stated W
> Marsh
>
> Regards,
>
> -- Zara[/color]

On Mon, 26 Dec 2005 00:41:10 -0800, "Jim Langston"
<tazmaster@rocketmail.com> wrote:
[color=blue]
>
>"Zara" <yozara@terra.es> wrote in message
>news:n48vq1tot8cnc7facia8sj631pqerj3c84@4ax.com.. .[color=green]
>> On Mon, 26 Dec 2005 06:35:59 +0000, W Marsh
>> <wayneDOTmarshATgmailDOTcom@decipher> wrote:
>>[color=darkred]
>>>meendar wrote:[/color][/color][/color]
(...)[color=blue][color=green]
>>
>>
>> void swap(const char *& char1, const char *& char2)[/color]
>
>Will this const work, since the variables are actually being changed?
>Doesn't he want them non constant?
>[color=green]
>> {
>>
>> const char *temp;
>> temp=char1;
>> char1=char2;
>> char2=temp;
>> }
>>[/color][/color]
(...)


Yes, this function will work, because the chars are constant, not the
pointers. The chars are really constant, as they are assigned as
char * p="whatever";

I have added the const qualifying to be strict.

But. although chars pointed are constant, char pointers are variable
and may be swapped.


Best regards,

-- Zara