What is wrong in this "C" code?

Try this example from TutorialsPoint:

#include <stdio.h>

int main () {
   char str1[20], str2[30];

   printf("Enter name: ");
   scanf("%19s", str1);

   printf("Enter your website name: ");
   scanf("%29s", str2);

   printf("Entered Name: %s\n", str1);
   printf("Entered Website:%s", str2);
   
   return(0);
}

Although it does not seem to work in the TutorialsPoint live editor: Online C Compiler[^]

[EDIT] it seems you have to type the input first in the STDIN tab.

To add to what Greg has said, with an array, you don't need to use the "address of" operator at all as the C specification says that the name of an array is a pointer to the first element. So this:

scanf("%s", &name[0]);

is the same as this:

scanf("%s", name);

And this:

scanf("%s", &name);

is too as the address of the array variable is the same as the address of the first character in it. Looks odd? Yes - but C is a very old language!

Try this and you'll see what I mean:

char name[20] = "OriginalGriff";
printf("%s\n", name);
printf("%s\n", &name[0]);
printf("%s\n", &(name[0]));
printf("%s\n", &name);

They all print the same thing, so use the clearest one - the first! (And the last will rightly give you a warning on some compilers!)

In fact you can print the pointer values, and you'll see they are all the same:

printf("%p\n", name);
printf("%p\n", &name[0]);
printf("%p\n", &(name[0]));
printf("%p\n", &name);

#include <stdio.h>
int main()
{
  char name[20];  
  printf("Enter Your Name : "); 
  scanf("%s",&name[20]);  
  printf("Your name is : %s", name);  
return 0;

 }</stdio.h>

#include <stdio.h>
int main()
{
  char name[20];  // Here, 20 is the size of array name
  printf("Enter Your Name : "); 
  scanf("%s",&name char[20]);  // here 20 is a position in array name
                               // Position 20 is 1 after the end of array
  printf("Your name is : %s", name);  
 }