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Full Discussion: EOF & precedence of !=
Top Forums Programming EOF & precedence of != Post 50658 by alan on Wednesday 28th of April 2004 05:58:33 PM
Old 04-28-2004
Question EOF & precedence of !=

Gurus,

I am teaching myself C and have a question.

I wrote a small prog that reads characters as entered at the prompt and checks the value for EOF.

Unless I am 100% wrong, the value will be '1' until getchar() has anything to read in my stream.

/* PROG 1 */
#include <stdio.h>

main()
{
int c;

while (c = getchar() != EOF)
printf("%d\n", c);
}

I compile this code, I run it at the prompt and type a string "1234<enter>" and get this output
1
1
1
1
1

So far, so good, right? For grins, I changed the number of brackets in my while() clause like below:

/* PROG 2 */
...
while ((c = getchar()) != EOF)
...

I re-compile this code and run it: at the prompt, I type the same string "1234<enter>" and get this output
49
50
51
52
10

Now, I obtain the decimal values for the stream I entered (line feed included).

Can someone please explain what is going on?

Thanks in advance! Al.
 

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