pointer to string

[jerico]

Hi.We have the following program..

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
void fun(char*);
int main()
{
char *t="hello";
fun(t);
printf("%s",t);
return 0;
}
void fun(char *r)
{
r=(char*)malloc(15);
strcpy(r,"world");
printf("%s",r);
}

Here though we have passed the address of t and modifying its value in the function fun(),then why the change is not reflected?Please explain how different types of memory are used here?

Jerico

[Banfa]

Here though we have passed the address of t

You have not passed the address of t. The address of t is

&t

you are passing

t

So you are passing t as am absolute value.

You are passing the address of the string literal "Hello" because this is what t contains.

[jerico]

ok.But what about the memory allocation for this program?

Jerico

[horace1]

when you malloc() inside function fun() you don't change the value of the pointer t in the main() program - to do so you need to pass the address of t (i.e. &t) and the fun() function header then becomes
void fun(char **r)
where the parameter r is a pointer to char *

other changes are then required in fun() to dereferrence parameter r correctly - the resultant program is

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
void fun(char**);
int main()
{
char *t="hello";
fun(&t);
printf("%s",t);
return 0;
}
// function parameter is a pointer to char *
void fun(char **r)
{   
*r=(char*)malloc(15);
strcpy(*r,"world");
printf("%s",*r);
}
 

when run it now gives
worldworld

[jerico]

Thanks.but I have two question-

1.Isn't the string in the main() a constant string?Why are we not getting any eror here even sfter modifying it?
2.Can you give me the sequence in which various addresses are pushed and popped out from memory?

Jerico

[horace1]

Thanks.but I have two question-

1.Isn't the string in the main() a constant string?Why are we not getting any eror here even sfter modifying it?
2.Can you give me the sequence in which various addresses are pushed and popped out from memory?

Jerico

in the statement
char *t="hello";
the string "hello" is constant, t is a char * (pointer to char)

to make t constant you would have
const char *t="hello";

and you would get a compiler message on the call
fun(&t);
because fun() could change the value of t

program with some explaination and which prints addresses as well as the strings

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
void fun(char**);
int main()
{
// define t as char* (pointer to char) and point it at "hello" 
char *t="hello";     
// print the address in t and what points at, i.e. "hello"
printf("t contains address %x which contains %s\n", t, t);
// call fun() passing the address of t   
fun(&t);
// print the address in t and what points at, i.e. "hello"
printf("t contains address %x which  contains  %s\n", t, t);
return 0;
}
// function parameter is a pointer to char *
void fun(char **r)
{   
// malloc storage and assign address to char* r points to 
*r=(char*)malloc(15);
printf("malloc storage is at address %x \n", *r);
// copy "world" into storage pointed at by r
strcpy(*r,"world");
printf("*r contains address %x which  contains  %s\n",*r, *r);
}
 
 

a run gave
t contains address 404000 which contains hello
malloc storage is at address 3e5e50
*r contains address 3e5e50 which contains world
t contains address 3e5e50 which contains world

[jerico]

Thanks.

Jerico