strcpy error with arrays

I am currently using the Visual C++ compiler on a WIN 2000 system. I am trying to scan in user input and then assigning what was entered into one of the dimensioanl arrays.

In the Visual C++ compiler, it gave me this error:
error C2664: 'strcpy' : cannot convert parameter 2 from 'char [3][20]' to 'const char *'

So, I tried to copy and paste the same exact code into Miracle C (another C compiler) and it worked.

Is this a Visual Studio bug? How can I work around it? HELP PLEASE!!!

Code:

---

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

char charname[3][20];
int i;

int main()
{
  for (i=0; i<3; i++)
  {
    printf("Enter in something: ");
    scanf("%s", charname);

    strcpy(charname[i],charname);
 
    printf("%s\n",charname[i]);
  }
 
  return(0);
}

---

Look where you are storing the user input. It's the 2 dimensional array. Just make a one dimensional temporary array and then copy that in to your two dimensional one as indexed by [i]. That should be fine.

Well, I'm trying to accept 3 different string values and then have them displayed on the screen.

That's why they are in 2 dimensional arrays.

Can you show me an example of what you are asking me to do?

Edit to add:

I figured it out!!!!! Thank you so much!!!!!!!!

Code:

---

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


char charname[3][20];
char charname2[20];
int i, x;

int main()
{
  for (i=0; i<3; i++)
  {
    printf("Enter in something: ");
    scanf("%s", charname2);

    strcpy(charname[i],charname2);
 
    printf("%s\n",charname[i]);
  }
 
  return(0);
}

---

Something like this..

Code:

---

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

char charname[3][20];
int i;

int main()
{
  char buffer[20];

  for (i=0; i<3; i++)
  {
    printf("Enter in something: ");
    scanf("%s", buffer);

    strcpy(charname[i],buffer);
 
    printf("%s\n",charname[i]);
  }
 
  return(0);
}

---

THANK YOU SO MUCH!!!! IT WORKED!!