fgets and overrun buffer

[zio_mangrovia]

I want to input string by fgets(buffer, 5, stdin) e.g. "stop", "lame", "rear"

char buffer[5];

If I type string which has a length of 20 characters, the next calls to fgets() keep partial content of previous input, what can I avoid it into Ansi C please?

[salem_c]

Your three basic steps are
- input
- validation
- conversion.

That you only want 4 characters falls into the latter stages.

So normally, I would do

char buff[BUFSIZ];
if ( fgets(buff, BUFSIZ, stdin) ) {  // input
    if ( strlen(buff) > 4 ) {  // validation 
        strmcpy(buffer, buff, 4);  // conversion
        buffer[4] = '\0';
    }
}

So if they did type 20 chars, the initial call to fgets would have consumed the excess for you from the outset.


Only people trying to DoS your program would ever get near trying to fill BUFSIZ characters without pressing return.

[zio_mangrovia]

Your three basic steps are
Only people trying to DoS your program would ever get near trying to fill BUFSIZ characters without pressing return.

I'm trying to limit input only for X number of characters, so that, while user is typing, He can show only at maximum X characters.
The excessive input should be hidden

[salem_c]

Well there's no way in ANSI C to just input 4 characters and NOT display anything else on the line until return is pressed.

[2kaud]

If supported, the closest you're going to get is to use getch() (or _getch() if using MS VS) from conio.h - but this isn't 'standard' c. This function will accept 1 char without echo. So you'd need a loop etc to obtain 4 chars
https://learn.microsoft.com/en-us/cp...?view=msvc-170