scan arbitrary number of chars in c

Question

I want to read an unknown number of characters, which are not greater than 10.

char word[10];
for( i=0;i<10;i++){
    if( !scanf("%c",&word[i])){ //terminate with 0
        getchar();
        break;
    }   
}

The problem is that number is also an character, so the if statement won't be executed. Is there any other solution to terminate the input of characters for example with 0.

hint: `scanf()` does not return the scanned entry, it returns the number of items scanned and matched scussfully. — Sourav Ghosh, Jun 24 '20 at 10:42
Why not? but it does not return a 0, in case the scanning is successful. — Sourav Ghosh, Jun 24 '20 at 10:46
I suggest to use getchar() for reading single chars. This post is related https://stackoverflow.com/questions/14419954/reading-a-single-character-in-c — Jonny Schubert, Jun 24 '20 at 10:52
"*The problem is that number is also an character, so the if statement won't be executed.*" - Could you elaborate that? The `if` statement will *always* get executed. Not clear what you mean with "number is also an character" as well. Please add more details, actual input and output and expected output. — RobertS supports Monica Cellio, Jun 24 '20 at 10:59

Sourav Ghosh · Answer 1 · 2020-06-24T10:54:50.573

0

You can use a do..while loop. Something like (pseudo-code)

int keep_looping = 1;
int counter = 0;
do {
    ret = scanf(" %c",&word[counter]);
    if (!ret) continue;  //can use more cleanup and error check
    if (word[counter] == '0') keep_looping =0;
    counter++;
    }
while (keep_looping && counter < 10)

edited Jun 24 '20 at 10:54

answered Jun 24 '20 at 10:45

Sourav Ghosh

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@RobertSsupportsMonicaCellio maybe I don;t, care to enlight? – Sourav Ghosh Jun 24 '20 at 10:54
Not possible. I'm drawing blanks, too. ;-) – RobertS supports Monica Cellio Jun 24 '20 at 10:55
Thank you for your solution. Why does the while loop terminate if keep_looping is 0? – Boris Jun 24 '20 at 11:02
because the controlling expression has to evaluate to a TRUE value (1) to continue looping. – Sourav Ghosh Jun 24 '20 at 11:04
@Boris If A subcondition of `&&` evaluates to `0`, the main condition is not `true` and thus the loop doesn't iterate anymore. – RobertS supports Monica Cellio Jun 24 '20 at 11:04
But the value of 1 outside the loop will be substituted with 0? – Boris Jun 24 '20 at 11:06
Thank you. I got it. – Boris Jun 24 '20 at 11:22

score 0 · Answer 2 · answered Jun 24 '20 at 10:50

0

you can check the character that you just read (at word[i]) and if not valid (e.g not alpabetical) than break.

answered Jun 24 '20 at 10:50

LiozN

1

score 0 · Accepted Answer · answered Jun 25 '20 at 08:01

Suggest:

char word[10];

if( scanf("%9s",word) != 1 )
{
    fprintf( stderr, "scanf for (max 9 char word) failed\n" );
    exit( EXIT_FAILURE );
}

using %9s because the %s input format conversion specifier always appends a NUL byte to the input.

If the input is less than 9 characters, that is properly handled.

If the input is greater than 9 characters, the 9 modifier will stop the input, so the input buffer is not overflowed. Such an overflow would result in undefined behavior.

So in order to input 10 characters, I have to change %9 to %10 and char word[11], so that in position word[10] is the NUL byte? — Boris, Jun 28 '20 at 09:49

scan arbitrary number of chars in c

3 Answers3