Thread: Scanf

Originally Posted by meandmyipod

How can i re-code such that once 0 is detected, it will break the loop?

If "a" is an integer type, that's what your loop should do. Unless...

You might what to try checking exactly what is going on. Perhaps you have assumed something happens a certain way when it actually does not?

Code:

int c;
do {
	c = scanf("%d,%d" &a, &b);
	printf("%d %d %d\n", c, a, b);
} while (a != 0)

Well you can't just use scanf(), because you're trying to treat one of the inputs as optional.

Consider something like this.

Code:

char buff[BUFSIZ];
while ( fgets( buff, BUFSIZ, stdin ) != NULL ) {
  if ( sscanf(buff,"%d,%d", &a, &b ) == 2 ) {
    // success
  } else
  if ( sscanf( buff, "%d", &a ) == 1 && a == 0 ) {
    // just a 0 on a line by itself - exit
    break;
  } else {
    // something else
  }
}

Originally Posted by meandmyipod

Erm,

To terminate the loop now, i'll need to key in 0 0 instead of only one 0.

0 0 // This's what i need to do with my above code to terminate.

Erm, this might seem to be evidence of what I just mentioned:

Originally Posted by MK27

Perhaps you have assumed something happens a certain way when it actually does not?

Did you try adding reporting like I recommended? If that is too much effort for you to make, give up programming now.

I can tell you that this:

Code:

#include <stdio.h>

int main(void) {
	int a, b, c;
	do {
		c = scanf("%d, %d", &a, &b);
		printf("%d %d %d\n", c, a, b);
	} while (a != 0);
	return 0;
}

Works in the way you describe ("0" will end). Maybe the code you have posted is not the exact code you are using?

Originally Posted by Salem

Well you can't just use scanf(), because you're trying to treat one of the inputs as optional.

As long as the optional ones come after (as in this case) that should be okay, presuming the input buffer is flushed by a newline. Note the OP wants to use "0", not "0,0" but claims the later works while the former doesn't.

Of course, if the inputs have ended up "out of predicted order", then you will have a problem. Methinks the OP should find out why that has happened. The other possibility is that the OS does not flush the input buffer on a newline, but I doubt that or the scanf() would not have worked in the first place.

How so?

Granted, if the user knows what the implementation value of BUFSIZ is, then they could press the space bar say BUFSIZ-5 times and type in 12345,6789 and get some weird ........ to happen.

But that is why there is a final 'else' (which you nicely omitted in your quote before saying it was garbage), to catch all the nonsense input the user might try to type in.

If you want to be pedantic, then you should be using strtol() / strtoul() instead of sscanf, since the latter does NOT detect overflow.

Originally Posted by BillyTKid

You dont know what i mean.
If a user put more than BUFSIZ-1 characters in inputbuffer, fgets keeps the '\n' and a next inputread will ignored.

Hmm....this post and your last one. Are you really going after Salem? Have fun....

Originally Posted by BillyTKid

You dont know what i mean.
If a user put more than BUFSIZ-1 characters in inputbuffer, fgets keeps the '\n' and a next inputread will ignored.

Why would it ignore it?

Code:

#include<stdio.h>
#undef BUFSIZ
#define BUFSIZ 5
int main( void )
{
    char buff[BUFSIZ];
    int a, b;
    
    printf( "format is #,#\ngo:\n" );
    while ( fgets( buff, BUFSIZ, stdin ) != NULL ) {
        if ( sscanf(buff,"%d,%d", &a, &b ) == 2 ) {
            // success
            printf( "a: %d, b: %d\n", a, b );
        } else
        if ( sscanf( buff, "%d", &a ) == 1 && a == 0 ) {
            // just a 0 on a line by itself - exit
            break;
        } else {
            // something else
            printf( "something else:\'%s\'\n", buff );
        }
    }
    return 0;
}

Try it.


Quzah.

Originally Posted by BillyTKid

If a user put more than BUFSIZ-1 characters in inputbuffer, fgets keeps the '\n' and a next inputread will ignored.

Rather, the single line of input would appear to be multiple lines, leading to an incorrect interpretation of the data. This could well be caught by the error checking anyway, and even when it isn't, the program would not suffer from writing to the array out of bounds, although it would give incorrect output. Given that what the OP has been "taught is very limited up till this point", and that the only improvement comes from error handling for less likely cases, I would say this is an acceptable example.

Originally Posted by BillyTKid

You dont know what i mean.
If a user put more than BUFSIZ-1 characters in inputbuffer, fgets keeps the '\n' and a next inputread will ignored.

Originally Posted by quzah

Why would it ignore it?

Code:

#include<stdio.h>
#undef BUFSIZ
#define BUFSIZ 5
int main( void )
{
    char buff[BUFSIZ];
    int a, b;
    
    printf( "format is #,#\ngo:\n" );
    while ( fgets( buff, BUFSIZ, stdin ) != NULL ) {
        if ( sscanf(buff,"%d,%d", &a, &b ) == 2 ) {
            // success
            printf( "a: %d, b: %d\n", a, b );
        } else
        if ( sscanf( buff, "%d", &a ) == 1 && a == 0 ) {
            // just a 0 on a line by itself - exit
            break;
        } else {
            // something else
            printf( "something else:\'%s\'\n", buff );
        }
    }
    return 0;
}

Try it.


Quzah.

Originally Posted by laserlight

Rather, the single line of input would appear to be multiple lines, leading to an incorrect interpretation of the data. This could well be caught by the error checking anyway, and even when it isn't, the program would not suffer from writing to the array out of bounds, although it would give incorrect output. Given that what the OP has been "taught is very limited up till this point", and that the only improvement comes from error handling for less likely cases, I would say this is an acceptable example.

Hmm........................