Thread: a do...while loop is the right answer??

Look:

Code:

		for (i = 0; i <10; ++i) {		// for (initial; condition; increment)
                [...............]
	} while (astring[i] != 'a');

What will be the value of i at the end of the do..while(), when the condition is executed?

This loop will repeat infinitely unless astring[9] happens to be an a. Also, you should be using strlen() to set the condition for the for() loop rather than using the array size.

Sorry, that was a rhetorical question. The answer is "9", which I thought would be obvious but don't worry, I'll try and explain why:

Code:

		for (i = 0; i <10; ++i) {		// for (initial; condition; increment)
                [...............]
	} while (astring[i] != 'a');

What will "i" equal at the end of the for loop? What part of astring do you think "astring[i]" refers to in the while condition? Remember, the for loop is nested inside the while loop, so the entire for loop is executed (according to it's conditions) before the while loop finishes the first time.

Yeah, sorry, I always think ++i will end differently than i++ here but nope.

So it's 10, not 9. That's even worse, because this is now out of bounds.

Thanks,
I wasn't too sure where to put the new while statement

Code:

while(astring[i] != 'a' && astring[i] != '\0') i++;

and the if else loop.

edit:Now it just enters the string twice and the second time prints"you have an a!" regaurdless.
I need some input as to how to arrange the statements.

is this the correct order:
1. enter the string
2. perform string check with for statement
3. while != a and != /0
4. loop again
5. else print "your word contains an a"

I know this code below isn't in this order, I posted this before the edit when I thought it was working (before checking a few times)

Code:

int main(){
	
	char astring[10];	//declare the array
	int i = 0;			//initialise the variable
	
	// while (astring[i] != 'a' && astring[i] != '\0') i++;
	{
		printf("\nEnter a word:\n");
		scanf("%s", astring);			//put the string in a memory location, no &
		
		while (astring[i] != 'a' && astring[i] != '\0') i++;
		{
		//for (i = 0; i <10; ++i) {		// for (initial; condition; increment)
			if (astring[i] != 'a') {
				printf("\nEnter a word:\n");
				scanf("%s", astring);	
			}
			
			else (astring[i] == 'a'); {	// if onr of the elemnts contain 'a'
				printf("Your word contains a!\n");
				
			}
		}
		
	
}
}

Code:

while (astring[i] != 'a' && astring[i] != '\0') i++;

That is the body of that loop.

Code:

{
	printf(" enter a word\n");
scanf("%s", astring);
	if (astring[i] == 'a')
	{
		printf ("you entered an a!\n");
	//break;
	}
}

This is its own code block, which has nothing to do with any loop.


Quzah.

Is there a reason you don't just use strchr?

You aren't wrapping the part that reads a string up in your loop:

Code:

int main( void )
{
    ...declare variables...
    do
    {
        printf( "enter a word of 9 or less characters\n" );
        scanf( "%s", astring );
        if( findana( astring ) == FOUNDANA )
            break;
    } while( ...some condition... );
}

That's the general idea.

Quzah.

so I have this code,
but am looking to break out of it if the string does contain 'a'

EDIT: got it, finally!!

Code:

#include <stdio.h>

int main(){
	
	char astring[10];	//declare the array
	int i = 0;			//initialise the variable
	
	do {
		printf("\nEnter a word:\n");
		scanf("%s", astring);			//put the string in a memory location, no &
		
		for (i = 0; i <10; ++i) {		// for (initial; condition; increment)
			if (astring[i] == 'a') {	// if onr of the elemnts contain 'a'
				printf("Your word contains a!\n");
break;
			}
		}
		
	} //while (astring[i] != 'a');
		while (astring[i] != 'a' && astring[i] != '\0'); i++;
	}