fgetchar() doesnt work

[roaan]

I have this program to see all the char input and output functions. It works fine except at the fgetchar() line. The problem is that at that point it should wait for me to enter the character before it moves ahead but it simply seems to iterate through that without me even entering anything to the next line.

[insert]

#include <stdio.h>
#include <stdlib.h>

int main(void)
{
	char ch;

	printf("\n Press any key to continue");
	getch(); // will not echo the char
	//printf("\n %c", ch);

	printf("\n Type any character");
	ch = getche(); // will echo the char
	//printf("\n %c", ch);

	printf("\n Type any character");
	getchar(); // macro, will echo the char, requires enter key
	//printf("\n %c", ch);

	printf("\n Press Y or N");
	fgetchar(); // function, will echo the char, requires enter key
	//printf("\n %c", ch);

	printf("\n I am here");

	ch = 'A';
	putch(ch);
	putchar(ch);
	fputchar(ch);
	putch('z');
	putchar('z');
	fputchar('z');

	return 0;
}

[Spidey]

Cprogramming.com FAQ > Flush the input buffer

[roaan]

Yes the addition of these two lines just before fgetchar solves the problem

[insert]

	puts("Flushing input");

	while((ch = getchar()) != '\n' && ch != EOF);

So the while loop makes sure that the input stream is cleared before i start taking any further input from the user. Is this necessary to do this everytime or only when i start facing such issues shoudl i have this mechanism for my program.

[MK27]

Is this necessary to do this everytime or only when i start facing such issues shoudl i have this mechanism for my program.

Well, in theory you might predict them -- this is always a result of taking a single character from the input stream, so:

scanf("%c",&byte);

You type:

x\n

"\n" is left in the buffer. %d does not do this, for one reason or another.

So another way to deal with this:

scanf("%c%*c",&byte);

%* discards the input, in this case the \n. You could also just use two fgetchar()s in a row, one unassigned. The while loop has some obvious advantages tho.

[roaan]

Well, in theory you might predict them -- this is always a result of taking a single character from the input stream, so:

scanf("%c",&byte);

You type:

x\n

"\n" is left in the buffer. %d does not do this, for one reason or another.

So another way to deal with this:

scanf("%c%*c",&byte);

%* discards the input, in this case the \n. You could also just use two fgetchar()s in a row, one unassigned. The while loop has some obvious advantages tho.

In %* c , is * the same meaning as thought it denotes a pointer ?

[MK27]

In %* c , is * the same meaning as thought it denotes a pointer ?

Nope.

The conversion specifications in a scanf template string have the general form:

% flags width type conversion

In more detail, an input conversion specification consists of an initial `%' character followed in sequence by:

- An optional flag character `*', which says to ignore the text read for this specification. When scanf finds a conversion specification that uses this flag, it reads input as directed by the rest of the conversion specification, but it discards this input, does not use a pointer argument, and does not increment the count of successful assignments.

[roaan]

Thanks :-)

That clears up a lot of things !!!!!

[tabstop]

Well, in theory you might predict them -- this is always a result of taking a single character from the input stream, so:

scanf("%c",&byte);

You type:

x\n

"\n" is left in the buffer. %d does not do this, for one reason or another.

Of course %d does this too. The difference between %d and %c (which is even listed at that man page you posted) is that %d will skip any and all whitespace at the beginning of the input, so that the next time, when the input buffer now has \n175\n, the first \n will get discarded, 175 will get read in, and the \n is still there in the input buffer.

[MK27]

Okay, I'm wrong twice today