Thread: is there a way to replace EOF in this loop..

Originally Posted by Adak

Try replacing it with -1. iirc, that was the actual numerical value of EOF.

EOF is better to use, of course, but -1 may serve you well, if you can't use EOF.

EOF is only guaranteed to be negative, but that means that comparing for "less than 0" will do.

Originally Posted by laserlight

EOF is only guaranteed to be negative, but that means that comparing for "less than 0" will do.

But that is REALLY not the right solution. If the original poster is not supposed to use EOF, then comparing with negative values is just cheating - you are assuming that EOF is a negative number [I'm not sure the spec says so - it probably does], and that anything else ever coming out of getchar that is negative is a failure.

I'd like to tell the teacher that set these rules that he/she is wrong in forbidding correctly written code (presumably "because we haven't talked about it yet") - it's one thing if the code is "cheating" by for example using more advanced functions when the student is supposed to learn how to do it at a lower level.

--
Mats

Originally Posted by transgalactic2

did i substituted correctly the EOF condition

Code:

for (i = 0; i < 39 && (ch = getchar()) != '\n' && ch >=0; ++i)
            {
                input2[i] = ch;
            }

            input2[i] = '\0';

Yes, that should work - but be aware that it's BAD code. But I guess it's a case of "anything to get past the teacher".

--
Mats

Originally Posted by matsp

But that is REALLY not the right solution. If the original poster is not supposed to use EOF, then comparing with negative values is just cheating

Yes, it is cheating, but I am in Adak's good company

Originally Posted by matsp

you are assuming that EOF is a negative number [I'm not sure the spec says so - it probably does], and that anything else ever coming out of getchar that is negative is a failure.

Ah yes, that is true: it is guaranteed by the standard that EOF is negative, but it is not guaranteed that anything that is negative is EOF. Consequently, the "cheat" is not guaranteed to work.

EDIT:
Oops, that is not true: the C standard does guarantee that anything that is negative is EOF, since "the fgetc function obtains that character as an unsigned char converted to an int" and "the getc function is equivalent to fgetc" and "the getchar function is equivalent to getc with the argument stdin".

Originally Posted by transgalactic2

if i am going to enter a negative number into my string it will stop the input?

If you enter a negative number as text, no. It would merely be interpreted as the characters of a string, within the ASCII range.

Actually, read my edit. The "cheat" is guaranteed to work (but that does not make it any less of a cheat).

thanks
the final nail in a coffin

Code:

for (i = 0; i < 39 && (ch = getchar()) != '\n' && ch >=0; ++i)
            {
                input2[i] = ch;
            }

            input2[i] = '\0';

comparing with -1 will simply EXPECT EOF to be -1 (which it probably is on 98% or more of all compiler&C Library implementations). Since getchar() is just a wrapper around fgetc() and fgetc() returns only positive numbers for valid characters, and EOF is the only defined negative value, comparing with >=0 should be fine.

However, you are still doing an EOF comparison - just avoiding the name EOF. Which is why we are calling it a cheat. Yes, it will work - but it's harder to understand, and you are still "using EOF" - just not by name. (And NOT using EOF means that you are actually hiding what that part of the code does - making it harder to understand. For example, the reader must know that EOF is a negative number).

--
Mats

You could do this, without assuming anything about the range of char

Code:

for (i = 0; i < 39 && (ch = getchar()) != '\n' ; ++i)
            {
                if ( feof(stdin) ) break;
                input2[i] = ch;
            }

Throw in an ferror() for good measure.

Originally Posted by Salem

You could do this, without assuming anything about the range of char

Code:

for (i = 0; i < 39 && (ch = getchar()) != '\n' ; ++i)
            {
                if ( feof(stdin) ) break;
                input2[i] = ch;
            }

Throw in an ferror() for good measure.

But SURELY if you are not supposed to use EOF, then feof() is also bad, right? I mean the poor chap(ess) is not allowed to use fgets, hence the loop here [avoiding gets()].

--
Mats