Neither Array nor Pointer.

[Luke Victory]

Hey, I've got a problem with some code I wrote.
I keep getting the error: "subscripted value is neither array nor pointer", I wont post the original code, 'cause it's kinda long, but I replicated the error with this:

#include <stdio.h>

int main()
{	
	printf("\n");
	int i;
	char Guess;
	char Ans[1][4]={"A""B""C""D"};
	
	printf("Enter Guess: ");
	scanf("%c",&Guess);
	printf("\n");
	
	for (i=0;i<4;i++)
	{	if(Guess[i]==Ans[i])
		{	printf("Letter %d is correct!\n",(i+1));
		}
	}
	printf("\n");


return 0;
}

The error points to

	if(Guess[i]==Ans[i])

I'm kinda new to coding, and i suspect it is rather obvious, but... help!

[grumpy]

Guess is of type char, so using the array subscript on it (as in Guess[i]) is invalid if i is an integral type (which it is).

An array of X is not an X, so the two cannot be directly compared. You can loop and compare each element of an array with a value, but that's different.

[AndrewHunter]

I would like to begin by saying great job on isolating your down to the simplest example of what gave you the error. This is an important part of the troubleshooting process that many people find difficult to do.

Some things to note:

1. Read our FAQ-Define main so you understand what those empty parenthesis mean.

2. Based on your posted example code, you may find it of some value to read cprogramming.com's quick tutorial on arrays.

3. Also, you may also find some value in a more verbose treatment of the subject: Chapter 4.1 - Arrays

[Luke Victory]

Hey, I fixed the problem.

The links were also really useful Andrew, so thanks, that was great.

- Vic