Thread: fgets

strchr

Although i don't know what exactly your program is supposed to do - are you
trying to signal end of input with EOF, and if a newline is pressed instead, quit?

array[0], or maybe all the element in the array to see if any one of them is a newline.

ugh.. that works but whenever i get rid of the \0 at the end of the string it doesn't work anymore. i.e. after i do:

Code:

*p;
if((p = strchr(string, '\n')))
	*p = '\0'

and i can't use strcmp on a specific array if (strcmp(string[0], "\n") == 0))) element because i just get ../test.c:38: warning: passing arg 1 of `strcmp' makes pointer from integer without a cast

ctrying to signal end of input with EOF, and if a newline is pressed instead, quit?

yes, basically i just want it to keep running till enter is pressed (it's to exit out of the program)

Read one character at a time, or do as suggested, and check your array for \n a character at a time. You know, if you're not going to read what people write, why are you even here?


Quzah.

Now i'm confused so I'm gonna try to clear things up:

You want to read in a string of data from the user - the user will press enter when they have finished, and it will prompt them to enter a string again, unless the string they enter contains only a newline - right?

In that case, what was wrong with your original solution?

Now, to explain what you did wrong with your last approaches:

1. If you read the link on strchr, the whole thing about changing the '\n' to '\0' isn't particularly relevant - what is relevant, is that strchr will return the location in a string of a character, or NULL if the character is not in the string. Now assuming that your quit condition is that the user enters a only a newline, what you want to do is check the address of strchr against the address of the first character in your string - i.e.

Code:

if (strchr (string, '\n') == string)
{
    /*do stuff to quit*/
}

2. strcmp(string[0], "\n") - that's wrong because strcmp expects a char * as it's first argument - string[0] is a char.

Maybe if you print the string after fgets you'll understand what is going on.

But to try to explain:

Code:

the loop is entered
"Enter something :" is printed
You type a string and hit enter
16 characters of what you typed is consumed by fgets (or less if fewer typed)
strcmp stanza evaluates to true (by luck)
the loop is entered
"Enter something :" is printed
input is still in the stdin buffer from your previous typing
16 characters of what you typed is consumed by fgets (or less if fewer left over)
....

Code:

while (strcmp(string, "\n") ==1);

This is wrong.
strcmp may never return 1.
You probably want != 0, or at least >= 1 which still doesn't make as much sense to me


BTW you typed more like 57 characters (in the posted example) which is why there were enough characters for 4 fgets calls

~/devel/c> bc -l
57/16
3.56250000000000000000

Like so

Code:

#include<stdio.h>
#include<string.h>

int main ()
{
    char buff[17];
    while ( fgets( buff, sizeof buff, stdin ) != NULL ) {
        if ( buff[0] == '\n' ) {
            printf( "Blank line entered - bye\n" );
            break;
        } else if ( strchr(buff,'\n') == NULL ) {
            int ch;
            printf( "Scrubbing extra long input\n" );
            while ( (ch=getchar()) != EOF && ch != '\n' );
        } else {
            printf("A good line - %s", buff );
        }
    }
    return 0;
}

$ ./a.exe
hello world
A good line - hello world
12345678901234567890
Scrubbing extra long input

Blank line entered - bye

>ok fflush worked
Then you got lucky. fflush is fundamentally not capable of clearing the input buffer. It's designed for flushing an output buffer, which is a completely different task.

>but is there any other way?
The best way is to design your code so that you don't need to. Otherwise, you can wimp out and call something like this:

Code:

void jsw_flush ( void ) 
{ 
  int ch;

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

  clearerr ( stdin );
}

Sadly, we see that kind of code all too often even from Cprog members who should know better.