incredibly simple problem (I think)

Unregistered · 12-04-2001

Code:

#include <stdio.h>
#include <ctype.h>

int main()
{

	char phrase[30];
	char *y;
	
	printf("What un-leet phrase shall I translate?\n");
	
	scanf("%c", phrase);


	y = phrase;

	while(*y != '\0')
	{
		
		switch(*y)
		{
		case 'e':
			putchar('3');
			y++;
			break;
		case 'E':
			putchar('3');
			y++;
			break;
		case ' ':
			putchar(' ');
			y++;
			break;
		default:
			putchar(*y);
			y++;
			break;
		}
						
	}
	putchar('\n');
}

the problem with this code is when it takes the string and tries to convert it to "leet speak" (Lame, I know, but I'm just practicing!) it will only do one word and stops after that. how do I get it to continue? (I thought that the spaces would be treated as characters but something goes awry in the program!)

Thanks for any help

Unregistered · 12-04-2001

At the end of any group of characters, a \0 is appended. Therefore, in the while loop, when testing *y with '\0', it will end the while loop when it reaches the end of a group of chars. I could be wrong that's my best guess.

**goran** · 12-04-2001

use gets(phrase) instead of scanf("%c", phrase).

hope that helps.

cheers,

Unregistered · 12-04-2001

Originally posted by goran
use gets(phrase) instead of scanf("%c", phrase).

hope that helps.

cheers,

Thanks! now, could someone explain why that works and scanf doesn't?

**dbaryl** · 12-04-2001

If I'm not mistaken, when you use scanf to input a string you use "%s"

like so:

Code:

scanf("%s", phrase)

**Prelude** · 12-04-2001

>use gets(phrase) instead of scanf("%c", phrase).
Use fgets instead of gets. gets is buggy and unsafe.
fgets(phrase, sizeof(phrase), stdin);

>could someone explain why that works and scanf doesn't?
Well, for starters the way you posted it scanf was only reading the first character of the phrase. But if you were to use %s it would still only read the first word as scanf uses a space character as a delimiter for when to stop reading.
gets and fgets use '\n' and EOF as the delimiter, so you read until the end of the line or until the end of the file if there are no newlines.

>If I'm not mistaken, when you use scanf to input a string you use "%s"
To input a single string yes, but if you want to input a phrase or sentence you have to either use a line input function or fiddle with scanf and loops.

To be honest, I'm surprised that you don't get warnings or errors when compiling. But it compiles fine and on my compiler causes some terrible garbage spewing when run.

-Prelude

**dbaryl** · 12-04-2001

it worked fine when I did it with %s...

**Prelude** · 12-04-2001

Yes, but when you ran the program did you enter in a single word and get a single word as output or enter a phrase with spaces and get a single word as output.

When I ran it with scanf and %s, here's what I got:
input - Speak to me my aged friend.
output - Sp3ak

But with fgets i got this:
input - Speak to me my aged friend;
output - Sp3ak to m3 my ag3d fri3nd.

Somewhat of a critical difference in the proper workings, no?

-Prelude

Unregistered · 12-04-2001

Originally posted by dbaryl
If I'm not mistaken, when you use scanf to input a string you use "%s"

like so:

Code:

scanf("%s", phrase)

yes I originally had that and since it didn't work I tried %c for the hell of it.

Thanks for the info Prelude, I am not yet familiar with fgets so I didn't know that at all.. and if I may ask, why is gets unsafe? (last night when messing around I used gets and in linux the compiler said it was unsafe.. made me wonder)

grr.. time to register or remember my old user/pass.

**Prelude** · 12-04-2001

Here's why:

Bugs:

Because it is impossible to tell without knowing the data
in advance how many characters gets() will read, and
because gets() will continue to store characters past the
end of the buffer, it is extremely dangerous to use. It
has been used to break computer security. Use fgets() instead.

And here's where you can find info on all standard functions in C.

And last but not least, a way to do it with scanf. I'm still debatting whether or not this is good style though. I still prefer fgets.

Code:

/*read everything until you get to a newline into phrase*/
scanf("%[^\n]", phrase);

-Prelude

**C_Coder** · 12-04-2001

I have to agree with prelude on this, scanf is best used by not using it, espicially with number inputs as newline character are left in the input buffer. So NEVER use scanf!!!

**Prelude** · 12-04-2001

Does that mean you don't usually agree with me?

Just kidding

-Prelude

**C_Coder** · 12-04-2001

Does that mean you don't usually agree with me?

Well I dont agree with that or do I.
I cant agree with myself most of the time

**dbaryl** · 12-04-2001

Can you guys tell me how this works:

Code:

scanf("%[^\n]", phrase);

how does the [^\n] part of it work... I'm kinda new to all this...

**quzah** · 12-05-2001

%[ ]

This is used to read until a character other than what is enclosed is encountered. Use a ^ before any characters to reverse this effect: to read until one of the enclosed characters is encountered. Thus:

%[abcdefg]

Would read those characters, in any order, into a string. For example, the following line of charaacters would all be read and stored:

abcedefefefefabcegabecabgac

However, the same would stop at the 'h' character in the following line:

abaegeabahabaeg ...the 'abaeg' at the end would not be read.

With the ^ symbol:

%[^abcdefg]

We would read until we encounter any of the enclosed characters:

t34903j3k3j4r2la

We'd stop when we hit the 'a'.

Additionally, you can use '*' to read something, but not put it into a variable. For example, if we had this line:

John 45 dog

And all we wanted was the second column, we could:

fscanf( fp, "%*s %d %*s", &age );

To read only the second column...

Quzah.

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