Thread: Counting how many chars in a string

Could someone give me a little nugget of code to count how many charactors there are in a string?

Here's a function I was trying to get to count how many chars were in a string:

Code:

int current_char;
static char somestring[5] = { "Hello" };


int getstrlength(somestring)
{
	for (current_char = 0;current_char < 80;current_char++) {
		if (somestring[current_char] == '\0') {
			return (current_char);
			break;
		}
	}
}

My compiler, Dev-C++ 5 beta, gives the error: subscripted value is neither array nor pointer (and it doesn't compile). I put the line it reported this on in bold.

On a sidenote the ASCII tables on www.asciitable.com are not really correct, at least for my PC... some of the box drawing chars at the bottom are really accented A charactors, which turned out to be quite annoying for what I was doing. I'm looking for box drawing chars which join double lines to single lines on corners or T shapes etc.

Thanks for any help!

Ah, I've been dumb again.

Code:

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

...
printf("%d",strlen("Hello"));

Problem solved.

Anyway, any ideas on the missing box drawing chars?

Thanks.

Well I tried this in CodeWarrior. It failed in a QuickWin, but as a console app it worked as expected.

Code:

#include <stdio.h> 

int main(void) 
{ 
	printf("\xC9\xCD\xCD\xBB\n");
	printf("\xCC\xCD\xCD\xB9\n");
	printf("\xC8\xCD\xCD\xBC\n");
    
    return 0; 
}

An unsigned char should also work if you want a variable.

I find a good way of seeing an ASCII char from its value is

printf("%c",200);

where 200 is the ASCII value.

I tried some hex values and still couldn't find any double-line-to-single-line box chars. Thanks for your help though.