scanf whitespace inside a function - c programming

Question

From previous threads, I know that you can achieve this by using the method shown here: How do you allow spaces to be entered using scanf?

It works in Main but when I put it inside a function, it doesn't work.

This is my code:

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

void dataInput();
struct record {
    char title[50];
    char author[50];
    char category[20];
    float price;
}; 

struct record info[500];

void main(){
    dataInput();
}

void dataInput(){
    int x;
    for(x = 0; x <= 10; x++ ){
        scanf("%s", &info[x].title);
        printf("%s\n", &info[x].title);
    }
}

Output:

Output1

If I use regex:

scanf("%50[0-9a-zA-Z ]s", &info[x].title);

Output:

Output2

Hmm whats wrong here

I've remove and added them, seems to be nothing wrong. & should only be in scanf though. — othniel_oum, Jun 02 '14 at 08:09
It shouldn't be in either in fact. scanf takes a char* for %s. — Quentin, Jun 02 '14 at 08:11
Probably fine in this case b/c the `&` address of a `char[]` is still the beginning of the array. — Felix Frank, Jun 02 '14 at 08:16
@Felix true. That will be a different story with a pointer, so better get it right now though :) — Quentin, Jun 02 '14 at 08:23

score 1 · Answer 1 · answered Jun 02 '14 at 08:18

1

I haven't got the most robust reference manual by hand, but from Wikipedia:

%s : Scan a character string. The scan terminates at whitespace. A null character is stored at the end of the string, which means that the buffer supplied must be at least one character longer than the specified input length.

So %s does very different things in scanf and printf, and will in fact only read one word in scanf, whereas in printf it will yield the null terminated content pointed to.

answered Jun 02 '14 at 08:18

Felix Frank

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"%s : Scan a character string", well _almost_. In C, a string is an array of `char` up to and including the `'\0'`. `"%s"` informs `scanf()` to 1) scan in and discard leading white-space, 2) scan in `char` until a white-space is encountered 3) unget that white-space for subsequent I/O 4) If any non--write-space was scanned append an `'\0'` to the destination. Curiously if a rare `'\0'` is scanned, it is treated as a non-white-space character and scanning continues. – chux - Reinstate Monica Jun 14 '14 at 14:05

score 1 · Answer 2 · answered Jun 02 '14 at 08:19

1

Your problem is pretty much explained here:

http://www.cplusplus.com/reference/cstdio/scanf/

For the %s format you have this:

Any number of non-whitespace characters, stopping at the first whitespace character found. A terminating null character is automatically added at the end of the stored sequence.

That is why when you using a regex the problem disappears.

answered Jun 02 '14 at 08:19

Cristina

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Yes it only works at main, but if it is inside a function, it just skipped the whole for loop. So am I stuck to have it inside main? – othniel_oum Jun 02 '14 at 08:25
So if you put the whole dataInput function in the main() your code works properly? – Cristina Jun 02 '14 at 08:30
Yes, all of the content inside the datainput function, it just works. outside, it doesn't. – othniel_oum Jun 02 '14 at 08:35

score 1 · Answer 3 · answered Jun 02 '14 at 08:55

1

Try

scanf("%[^\n]", &info[x].title);

answered Jun 02 '14 at 08:55

MegaMind

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**also u dont need "&" in printf** – MegaMind Jun 02 '14 at 08:56
Same thing, it just exited the for loop. – othniel_oum Jun 03 '14 at 00:46
2

Dude the thing is, that when you use %s it accepts only upto the space, this is a protocol. – MegaMind Jun 03 '14 at 07:33

score 0 · Answer 4 · edited Feb 21 '17 at 13:31

0

scanf (" %[^\n]%*c", &info[x].title);

Notice that there is a SPACE before %[, this will work correctly

edited Feb 21 '17 at 13:31

illright

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answered Feb 21 '17 at 12:29

scanf whitespace inside a function - c programming

4 Answers4