scanf reads previous line instead of getting new one

Question

I have the following code:

#include <stdio.h>
int opt;
scanf("%d",&opt);
while(opt!=0)
{
    switch(opt)
    {
    case 1:func1();
        break;
    case 2:func2();
        break;
    case 3:return 0;
    }
    printf("\n");
    scanf("%d",&opt);
}

I want to computer to get from user everytime different digit (1 or 2 or 3) and in every time the computer will run on seperate function. but when I press 2, func2 gets 3 ints and prints chars on the screen in a new line. the computer reads them instead of reading next input (1 or 2 or 3). how can I fix it?

EDIT: The problem rises only after specific inputs (I build a program converts bases).

Input:

OUTPUT:

ERROR

INPUT:

UNWANTED OUTPUT:

I'm not sure from where the computer scans the second num it uses to print the last unwanted output.

Providing function definitions of `func1()` and `func2()` would be helpful to us to answer your question. — haccks, Nov 20 '13 at 12:26
It's hard to tell without actually knowing what the functions do, if you replace them with dummy function, the code works, obviously. — pAndrei, Nov 20 '13 at 12:26
your second scanf is supposed to block but it isn't because opt is not empty — H_squared, Nov 20 '13 at 12:33
@hhachem what do you mean? to remove the ampersand? in the function func2 it works great except for the case I edited in original post (and obviously func2 includes some scanfs). — Danis Fischer, Nov 20 '13 at 12:38
You need to check the return value of `scanf` and take it into consideration. Read about it here http://en.cppreference.com/w/cpp/io/c/fscanf — StoryTeller - Unslander Monica, Nov 20 '13 at 12:43

score 1 · Answer 1 · answered Nov 20 '13 at 12:35

1

your second scanf is supposed to block but it isn't because the input buffer is not empty after the first call! This is why it is always returning the previous value and looping infinitely. Use a combination of fgets and sscanf instead.

Also check this : http://c-faq.com/stdio/scanfprobs.html

answered Nov 20 '13 at 12:35

H_squared

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I don't i'm allowed to do so. however what is the difference between scanf and fgets except security issues? – Danis Fischer Nov 20 '13 at 12:39
1

fgets reads input from stdin and stores it in an array. Then you can parse it using sscanf. scanf is really bad, because once you hit `enter` it won't read new inputs because it will always detect `enter`pressed in the buffer. Check the link in my answer it can explain why in more detail. – H_squared Nov 20 '13 at 12:42

score 1 · Answer 2 · answered Nov 20 '13 at 12:40

1

If your functions use scanf() inside them to ask for characters or you are using fgets() to read a while line, you might find that scanf() / fgets() does run without user input. This is because the newline character you entered at scanf("%d",&opt); is still there and is used as new input for the next scanf()/fgets(). You have to erase it by issuing

fflush (stdin);

Just prior to call scanf() or fgets()

answered Nov 20 '13 at 12:40

mcleod_ideafix

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`fflush (stdin);` is undefined behavior http://stackoverflow.com/questions/2187474/i-am-not-able-to-flush-stdin – StoryTeller - Unslander Monica Nov 20 '13 at 12:50
@StoryTeller Correct. Although he is right about clearing stdin, I would rather use `fgets` and insert an `\0` at the beginning of the buffer, so the next `fgets` call blocks. `sscanf` can then be used to parse the integer – H_squared Nov 20 '13 at 12:55
`fflush(stdin)` has a documented behavior for both input and output streams, at least for Windows, according to the MSDN: http://msdn.microsoft.com/en-us/library/9yky46tz.aspx – mcleod_ideafix Nov 20 '13 at 13:44
1

@mcleod_ideafix It is extended by microsoft as explained in the example code. It is non-standard in C and therefore not portable. `fflush` can only be used on output streams as explained here: http://c-faq.com/stdio/stdinflush.html – H_squared Nov 20 '13 at 14:43
@Coargu Aliquis `char buffer[512]=""; fgets(buffer,sizeof(buffer),stdin);` this reads the input and stores it in buffer as a series of characters (aka string). You will have to use `sscanf` afterwards to get the integer since `scanf` does not allow you to specify from where you want to read. Also once you are done with the input you will have to delete it by `buffer[0]='\0';` Haccks's solution should work by the way. – H_squared Nov 20 '13 at 14:50

score 1 · Answer 3 · answered Nov 20 '13 at 12:48

you may need something like this..

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

int main(int argc, const char *argv[])
{
    int opt;

    do{
        printf("Enter the option\n");
        scanf("%d",&opt);
        switch(opt){
            case 1: func1();
                    break;
            case 2: func2();
                    break;
            case 3: 
                    return 0;
        }

    }while(opt != 3);

    return 0;
}

score 0 · Answer 4 · edited May 23 '17 at 10:25

0

One problem that is pointed out by hhachem in comment that you are not exiting from your while loop.
You can solve this problem by doing (as you want to exit only after pressing 3):

 while(1)
{
    switch(1)
    {
        case 1:func1();
            break;
        case 2:func2();
            break;
        case 3: exit(0);
    }
    printf("\n");
    scanf("%d",&opt);
}

Test program:

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

int main()
{
    int opt;
    scanf("%d",&opt);
    while(1)
    {
        switch(opt)
        {
            case 1:printf("1");
                   break;
            case 2:printf("2");
                   break;
            case 3:exit(0);
         }

         printf("\n");
         scanf("%d",&opt);
    }

    return 0;
}

See the output.

edited May 23 '17 at 10:25

Community

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1

answered Nov 20 '13 at 12:32

haccks

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I want to exit it only after I press 3. I can exit from it, can't I? – Danis Fischer Nov 20 '13 at 12:33
returning `0` from `main` (which is curiously absent from the OP, but I assume this is the code inside it) will exit just fine. – StoryTeller - Unslander Monica Nov 20 '13 at 12:35
@CoarguAliquis; Yes, you can. See the updated answer. – haccks Nov 20 '13 at 12:40
This doesn't address the problem. – StoryTeller - Unslander Monica Nov 20 '13 at 12:41
@haccks, I don't know what question you are answering, but your revised code sample still suffers the same problem as the OP and intoduces a new one a-la `switch(1)`. It doesn't matter how the OP exits, as long as it can't parse it's input correctly. – StoryTeller - Unslander Monica Nov 20 '13 at 12:45
No i misread the code earlier. Check my answer below. The issue is scanf and the input buffer being populated by the first scanf call. So all following scanf calls do not read anymore because the input buffer has not been cleared. – H_squared Nov 20 '13 at 12:45
@haccks I see. it runs on my system too. Nonetheless, I'm not a fan of `scanf`, there are far better options out there. – H_squared Nov 20 '13 at 13:12
@hhachem; I am also not a big fan of it but OP wants to use this function in his program (as he stated in the comment in responce to your answer. – haccks Nov 20 '13 at 13:19
1

@haccks Oh yes I missed that. – H_squared Nov 20 '13 at 13:23

scanf reads previous line instead of getting new one

4 Answers4