How to stimulate EOF without preceding newline in C

问题

Lets say I have the below C code:

int getLine (char line[])
{
    int c, i=0;
    while( (c=getchar()) != EOF )
           line[i++]=c;
    line[i++] = c;
    return i;
}

>> Enter: 007
>> ^Z
>> Output:

If we closely observe the way I give output above, I am pressing Enter before stimulating EOF. This means, the length of string is 4 not 3 (excluding EOF).

When I am doing my exercises, I am really facing some trouble with that extra \n.

How do I stimulate EOF without newline? Is it possible at all?

>> Enter: 007^Z
>> ^Z
>> Output: length=6

回答1:

Converting my comment into an answer:

On which platform? On Unix and derivatives, you would type the EOF 'character' twice — usually control-D rather than control-Z, though. That may also work on Windows; I don't know, but it is worth a try.

(A response comment affirms that the platform is Windows.)

On Unix, control-D makes the data on the line available to the program. The first control-D gives it what you've typed already; the second gives it zero bytes to read, which is the indication of EOF.

回答2:

Then avoid storing the newline, in the loop. It's not as if you're being forced to store all characters regardless of value. :)

Also, you're not terminating the string correctly. This:

line[i++] = c;

should be:

line[i] = '\0';

And of course, it's sensitive to buffer overflow.

In general, you'd be better of using fgets().

EDIT: I might be missing the point, but it seems to be that the entire focus on EOF is ... misguided, if all you want to do is read a line. Lines are not generally terminated with EOF, but with \n. So the function should probably just store characters until either EOF or \n is encountered.

来源：https://stackoverflow.com/questions/14199485/how-to-stimulate-eof-without-preceding-newline-in-c

标签

eof