Thread: question about these functions

I'm wondering if anyone could explain how a couple statements in this code works.

Code:

int main()
{
	char c, c1, c2;
	printf("enter your sentence\n");
	while( (c = getchar()) != EOF ){
		if( c == '%' ){
			c1 = getchar();
			c2 = getchar();
			if( c1 == EOF || c2 == EOF )  exit(0);
			c1 = tolower(c1);
			c2 = tolower(c2);
			if( ! isxdigit(c1) || ! isxdigit(c2) )  exit(0);
			if( c1 <= '9' )
				c1 = c1 - '0';
			else
				c1 = c1 - 'a' + 10;
			if( c2 <= '9' )
				c2 = c2 - '0';
			else
				c2 = c2 - 'a' + 10;
			putchar( 16 * c1 + c2 );
		} else if( c == '+' )
			putchar(' ');
		else
			putchar(c);
	}
	exit(0);
}

===============================================

int main()
{
	int c;
	char *h = "0123456789abcdef";

	while( (c = getchar()) != EOF ){
		if( 'a' <= c && c <= 'z'
		|| 'A' <= c && c <= 'Z'
		|| '0' <= c && c <= '9'
		|| c == '-' || c == '_' || c == '.' )
			putchar(c);
		else if( c == ' ' )
			putchar('+');
		else {
			putchar('%');
			putchar(h[c >> 4]);
			putchar(h[c & 0x0f]);
		}
	}
	exit(0);
}

I know what the code does I've worked several examples, but I'm a bit unsure how it works specifically.

Code:

if( ! isxdigit(c1) || ! isxdigit(c2) )  exit(0);
			if( c1 <= '9' )
				c1 = c1 - '0';
			else
				c1 = c1 - 'a' + 10;
			if( c2 <= '9' )
				c2 = c2 - '0';
			else
				c2 = c2 - 'a' + 10;
			putchar( 16 * c1 + c2 );

why is there a ! in the isxdigit(c1) statement?
and how does comparing '0' and 'a' and '9' in the if else statements change a Hex value to ASCII?

My other question is in the second part

Code:

putchar(h[c >> 4]);
putchar(h[c & 0x0f]);

I don't understand how these two statements function why is there >> and & in them? I've seen them used for nibble operations in microcontrollers but I'm confused what they are doing here. Thank you very much.

the '!' is negation. so its saying if either c1 or c2 are not hex numbers then exit. regarding how it changes the hex number to ascii i dont know, i havent really examined and thought about the code. but it just looks like a bunch of arithmetic im sure you could figure out if you checked out the integer values for the characters.

the second part involves bitwise operators, ive never used them myself but maybe this tutorial will help you: http://www.cprogramming.com/tutorial...operators.html

hope it helps.

Code:

if( 'a' <= c && c <= 'z'
|| 'A' <= c && c <= 'Z'
|| '0' <= c && c <= '9'
|| c == '-' || c == '_' || c == '.' )

Unfortunately the code is not portable since it assumes ASCII is the only character table out there.

Originally Posted by cpjust

Code:

if( 'a' <= c && c <= 'z'
|| 'A' <= c && c <= 'Z'
|| '0' <= c && c <= '9'
|| c == '-' || c == '_' || c == '.' )

Unfortunately the code is not portable since it assumes ASCII is the only character table out there.

It would work for any character set where a..z are consecutive and in order, A..Z are consecutive and in order, and 0..9 are consecutive and in order. Which is essentially 100% of the character sets now in modern use.

It would work for any character set where a..z are consecutive and in order, A..Z are consecutive and in order, and 0..9 are consecutive and in order. Which is essentially 100% of the character sets now in modern use.

Are you sure about that?
http://www.legacyj.com/cobol/ebcdic.html