Thread: storing dynamically allocated pointers

the well seems to getting deeper and deeper...

> arrays cannot be assigned.
so, this means the :

Code:

  
   array[i] = name ;

is itself wrong ?
but, we can assign one pointer to another and use the other pointer for referencing data for normal variables.. The array we have is a set of pointers in itself .. So, I got mislead into the notion that i can assign the "name " pointer into one of the array elements and then do what ever needed..

yeah, ok I shud have done that long time back

Code:

  void input ( char *names[]){
	 int i;
	char name[21];
	
	for(i=0; i<3; i++)
	{
		printf("enter the name \n");
		fgets(name,20,stdin);

		names[i] =  malloc (sizeof (char) * strlen(name) + 1);

		names[i] = name ;
		fputs(names[i],stdout);
		printf("i=%d\n",i);
		
	}
return ;
}

that is my function ..

In main I just print them out :

Code:

          for(count=0;count < 3; count++)
	   fputs(names[count],stdout);

I just hope ,i havent made any real blunder with my sleepy eyes here..

Code:

  void input ( char *names[]){
	 int i;
	char name[21];
	
	for(i=0; i<3; i++)
	{
		printf("enter the name \n");
		fgets(name,20,stdin);

		names[i] =  malloc (sizeof (char) * strlen(name) + 1);

		names[i] = name ;
		fputs(names[i],stdout);
		printf("i=%d\n",i);
		printf("%u\n",names[i]);
	}
return ;
}

Do you see the memory leak in bold, and the difference between similar names in different colors? And do you understand what is going on?

If it's still not clear, it seems a bit like wondering what happens to the 42:

Code:

#include <stdio.h>

int main (void)
{
   int x;
   x = 42;
   printf("x = %d\n", x);
   x = -1;
   printf("x = %d\n", x);
   return 0;
}

/* my output
x = 42
x = -1
*/