Thread: Difference between two codes(very basic)

Code 1

Code:

#include<stdio.h>
typedef struct 
{
  int n;
  char *string;
}sct;

void fun(sct *);

int main(void)
{
  sct one;
  fun(&one);
  printf("%s\n",one.string);
  return 0;
}

void fun(sct *p)
{
  char two[]="funny";
  p->string=two;
  printf("%s\n",p->string);
}

Output 1

funny
��

Code 2

Code:

#include<stdio.h>
typedef struct 
{
  int n;
  char *string;
}sct;

void fun(sct *);

int main(void)
{
  sct one;
  fun(&one);
  printf("%s\n",one.string);
  return 0;
}

void fun(sct *p)
{
  //char two[]="funny";
  p->string="funny";
  printf("%s\n",p->string);
}

Output 2

funny
funny

So where my basics are wrong??
I need output 2 but it should in code 1's way.
What changes can i make in code 1?
thanks

In the first example you assign string to a local variable. This variable doesn't exist anymore after return from fun().
in the second example you assign a char literal that is propably stored somewhere in read only memory.
Kurt

ok the variable two doesn't exist but the address of the string "funny" is already copied in
p->string.

Originally Posted by subhasnagre

ok the variable two doesn't exist but the address of the string "funny" is already copied in
p->string.

Um, no. You are not copying the address of the string "funny", you are initializing the array with the contents of the string "funny". As Zuk pointed out, the array ceases to exist at the end of the function.

THIS would be copying the address of the string "funny":

Code:

void fun(sct *p)
{
  const char *two = "funny";
  p->string=two;
  printf("%s\n",p->string);
}

Ok I got it. Thanks for the replies.