modify pointer to a string/character constant.

in c programming language.
I have 1 question
1)

Code:

---

#include <stdio.h>
#include <string.h>
int main()
{

char *x;
char t[10]= "telephone";
x=(char *)malloc( sizeof(char) *6  );
x=t;
printf("&#37;s\n",x);
free(x);
/* x[0]='t' will probably give segementation fault.. */
printf("%s\n",x);
return 0;

}

---

the output is:
telephone
telephone

even though I had free(x);

How do I properly free this, supposing I want the char *x to point to something else.
For example say I have the string "telephone" and want it to be "zelephone"

Code:

---

char *x;
char t[10]= "telephone";
x=(char *)malloc( sizeof(char) *6  );  // why 6 ?
x=t;        // this assignes x to the char array t, not an error but not a good idea
          // you have lost the pointer that malloc returned
free(x);  // trying to free an array

---

Use strcpy instead of assignement and allocate enough to hold the data.
Kurt

Accessing free()'d memory doesn't HAVE to cause a segmentation fault. It's simply undefined behavior, meaning anything can happen. BTW, you don't need to cast malloc(), and you should always #include <stdlib.h> when using malloc().

FAQ > Explanations of... > Casting malloc

If I stick to using strcpy to copy ... notice how here I can't and even strncpy can't work
instead I would need concatenation of this new character to the rest of the string.
I also tried char backup[strlen(s)+1]; it fails too.


to try again..

Code:

---

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void replaceFirstCharacter(char *s,char replaceWith);
int main()
{

char *x;
char t[10]= "telephone";
x=(char *)malloc( sizeof(char) *10  ); /*9 characters plus null byte.*/
strcpy(x,t);
/*so that it works with any string... changing the first character only*/
replaceFirstCharacter(x,'z');


printf("&#37;s\n",x);
return 0;

}

void replaceFirstCharacter(char *s,char replaceWith)
{
char *backup;
if (strlen(s) >=1)
{
backup=(char *)malloc( sizeof(char) * (strlen(s)+1));
strcpy(backup,s);
backup[0]=replaceWith; /*I guess I should somehow avoid this?*/
strcpy(s,backup);

}
}

---

You should include <stdlib.h> for malloc(), and avoid casting malloc() unless you prefer to do so for some reason.

sizeof(char) is always 1, so you can leave it out of your expressions.

Code:

---

void replaceFirstCharacter(char *s,char replaceWith)
{
char *backup;
if (strlen(s) >=1)
{
backup=(char *)malloc( sizeof(char) * (strlen(s)+1));
strcpy(backup,s);
backup[0]=replaceWith; /*I guess I should somehow avoid this?*/
strcpy(s,backup);

}
}

---

It seems to me that that function could be replaced with

Code:

---

void replaceFirstCharacter(char *s, char replaceWith) {
    if(strlen(s)) s[0] = replaceWith;
}

---

But anyway, if you really wanted to make s point to a different string, you'd need to pass in s as a char** pointer or return the modified s; because, as written, any changes to what s points to inside that function will not be reflected in the calling function.

Well let's be sure that we understand something here. If you just want to change the first character of whatever the user types than all you have to do is set up an array really, and then reassign the first character to 'z', or whatever.

If you want to treat user input as a constant, than you would need to explicitly copy anything the user types into a new string before you make changes, with functions like strdup() or sprintf(). And then when you are done with the copy you can throw it away.

But turning "telephone" into "zelephone" should be trivial work.

Perhaps this will resolve my question once and for all!...
I need to pass x to foo(char *s) and change it to "House".

If possible I would like to keep the main program intact, and change only the foo function to accomplish
this.

oddly void foo(char *s) { strcpy(s,"House"); }
somehow works even though it doesnt seem memory is allocated to it!


the current output is(Notice how the pass by reference fails to show in the main program):

before call:Fun
In foo:House
after call:Fun

Code:

---

#include <stdio.h>
void foo(char *s);
int main()
{
char x[4]="Fun";
printf("before call:&#37;s\n",x);
foo(x);
printf("after call:%s\n",x);
}

/*sets a given string to "House"*/
void foo(char *s)
{
free(s);
s=(char *)malloc( 6 * sizeof(char));
strcpy(s,"House");
printf("In foo:%s\n",s);

}

---

Well it can't be that different. Just create a new string and copy over what you need.

Code:

---

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>

char *foo( char str[], const char change_to[] )
{
  void *temp;
  assert( str != NULL && change_to != NULL );
  temp = realloc( str, strlen( change_to ) + 1 );
  if ( !temp )
      return NULL;
  str = temp;
  strcpy( str, change_to );
 
  return str;
}

int main( void )
{
  char *p = malloc( 4 );
  if( p ) {
      strcpy( p, "Fun" );
      if( foo( p, "House" ) )
        puts( p );

      free( p );
  }
  return 0;
}

---

okay, well I am not familiar with assert, so from a Google search, it looks like it gives errors if allocation fails.. anyways in
foo()
1)what exactly does str=temp; do?
2)does it allocate str to be able to hold as much as temp does?
or does it make str point to temp?

3)which I guess char str[] could be rewritten as char * str ?

Read about realloc. It can resize the allocated memory that a pointer points to, and in this case, I wanted to ensure that we had enough room to store the change_to string. But it's dubius just to reassign the pointer you gave realloc like here.

p = realloc( p, SOME_SIZE );

If realloc returns NULL and p pointed to some memory before, we would leak the old memory. So that's why I needed to use the temp pointer. I simply reassigned the str pointer after I was sure that realloc worked, and later returned it.

Absolutely. In terms of function parameters, char *foo is char foo[], more or less.

Code:

---

void foo(char *s)
{
free(s);
s=(char *)malloc( 6 * sizeof(char));
strcpy(s,"House");
printf("In foo:%s\n",s);

}

---

For this function to do what you want it to do, you need to pass a char** not a char*.
The address of variable s in foo() is not the same as the address of variable x in main(). If you change the value of s, it will not change the value of x. You need to pass the address of x like this: foo( &x ); then dereference s in foo()...
Something like this:

Code:

---

void foo( char** s )
{
    char* temp = malloc( 6 );

    if ( temp != NULL )
    {
        strcpy( temp, "House" );
        free( *s );
        *s = temp;
    }
}

---

well I'm given the function prototype like I said... any ideas on how to solve this ...
to have a
void foo(char *s) that changes a string in the main program...
It works when I use strcpy only... but when I use malloc and free it does not overwrite whats in the main... whats also bad
is when I try to allocate a large string it fails.

Code:

---

#include <stdio.h>
#include <stdlib.h>
void * foo(char *s);
int main()
{
char x[3]="Hi";
printf("before call:&#37;s\n",x);
foo(x);
printf("after call: %s",x);
printf("\n");
return 0;

}

  void * foo (char *s)
{
/* How do I dynamically allocate memory for s here, so that it will change it in the main WITHOUT changing the function prototype or call*/
free(s);
s=(char *)malloc(31*sizeof(char));
/*the above 2 statements fail. */
strcpy(s,"hello how are you doing today.");  /*this works with a small string*/
}

---

You can only free() what you got back from malloc()
So initially, you need to start with a malloc if your function is going to call free

Eg.

Code:

---

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

void *foo(char *s); /* returning char* would be better */

int main()
{
    char *x = malloc(3);
    strcpy(x,"Hi");
    printf("before call:&#37;s\n", x);
    x = foo(x);
    printf("after call: %s", x);
    free(x);
    printf("\n");
    return 0;

}

void *foo(char *s)
{
    free(s);
    s = malloc(31 * sizeof(char));
    strcpy(s, "hello how are you doing today.");
    return s;
}

---