Silly Pointer Problem

[vinothg]

Hi ,

I just wanna know why such a wierd output is seen for the following program.I know it is pretty simple but i can't make it.Just a clue could help me to sort out the issue.

#include <iostream.h>
#include <stdio.h>
 
void func(char *s)
{
   printf("%x \n",s);
   s="senthil";
   printf("%s %x\n",s,s);
}
 
int main()
{
   char* s ="manikandan";
   printf("%s %x\n",s,s);
   func(s);
   cout<<s;
}

zcarh0s5-135> !.
./a.out
manikandan 4eb0
4eb0 >>>>>>>>>> Same address is passed to func()
senthil 4ea0 >>>>>Here the address is diff !!!
manikandan


Can't we change the value of a string using char pointer.

[Banfa]

Can't we change the value of a string using char pointer.

No.

You initialised a pointer to an address in memory
You printed the contents of that address and the address itself
In the function
You print the address again
You change the value of the pointer passed to the function to a different address in memory
You printed the contents of the new address and the new address itself

Note that the function changes nothing in main, the original variable s still points to the original address.

[horace1]

you need to use strcpy() or strncpy() so

#include <iostream.h>
#include <stdio.h>
 
void func(char *s)
{
printf("%x \n",s);
strcpy(s,"senthil");
printf("%s %x\n",s,s);
 
}
int main()
{
char   s[] ="manikandan";
printf("%s %x\n",s,s);
func(s);
cout<<s;
cin.get();
}
 

note that you have to define an array in the calling program of sufficent size to receive the charcaters

[willakawill]

Can't we change the value of a string using char pointer.

Yes if you do it this way

void func(char *s)
{
    printf("%x \n",s);
    strcpy(s, "senthil\0");
    printf("%s %x\n",s,s);
 
}
int main()
{
    char s[50] ="manikandan\0";
    printf("%s %x\n",s,s);
    func(s);
    cout<<s;
    getchar();
    return 0;
}

[LSB]

I would suggest to run your code in debugger: you will see what your pointers are and what data they represent.
Very, very useful !!!