Print Page - warning #2238: Array index for 'char [7][7]' is out-of-bounds.

Title: warning #2238: Array index for 'char [7][7]' is out-of-bounds.
Post by: sannemander on May 02, 2012, 06:55:01 PM

Hi,

I don´t get why I get this warning, since the array cannot be bigger then 7x7!

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#include<stdio.h>

#define N 7

int main(void)
{

	char CheckerBoard [N][N];
	int row, col;

	
    for(row = 0; row < N; row++);
	  for(col = 0; col < N; col++);
		if ((row+col)/2 == row)
			CheckerBoard[row][col] = 'B';
		else
			CheckerBoard[row][col] = 'R';




return 0;

}

I also tried it with a while loop but kept the same error!

grtz!

Title: Re: warning #2238: Array index for 'char [7][7]' is out-of-bounds.
Post by: Bitbeisser on May 02, 2012, 07:24:11 PM

Very simple error!

Take a very close look at the (end of the) for() loop statements... ;)

Ralf

Title: Re: warning #2238: Array index for 'char [7][7]' is out-of-bounds.
Post by: sannemander on May 02, 2012, 07:25:58 PM

Ah thx I see it now! :-[

Title: Re: warning #2238: Array index for 'char [7][7]' is out-of-bounds.
Post by: Bitbeisser on May 02, 2012, 07:31:30 PM

Quote from: sannemander on May 02, 2012, 07:25:58 PM
Ah thx I see it now! :-[

You're welcome! 8)

Just as a tip, which helps IMHO to avoid such quirks, is to getting used to start writing such a statement as a template like

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for(;;){};and then fill out the "inner" parts.
The braces help to determine the scope of the for() loop and also in placing the terminating semicolon in the right place...

Ralf

Pelles C forum

C language => Beginner questions => Topic started by: sannemander on May 02, 2012, 06:55:01 PM