Unable to find error in C code

Question

When I execute the following code program exits before I get to input any string for word 2. i.e scanf-word2 is not executed. Why? I am using mac and I can't make the 2nd scanf execute in both codelite and terminal(using make)

#include <stdio.h>

int main()
{
    char word1[20], word2[20];

    printf("enter word 1 : ");

    scanf("%[^\n]s",&word1);

    printf("enter word 2 : ");

    scanf("%[^\n]s",&word2);

}

Possible duplicate of [scanf: "%\[^\n\]" skips the 2nd input but " %\[^\n\]" does not. why?](http://stackoverflow.com/questions/6083045/scanf-n-skips-the-2nd-input-but-n-does-not-why), [Simple C scanf does not work?](https://stackoverflow.com/questions/3744776/simple-c-scanf-does-not-work), [C: Multiple scanf's, when I enter in a value for one scanf it skips the second scanf](https://stackoverflow.com/questions/9562218/c-multiple-scanfs-when-i-enter-in-a-value-for-one-scanf-it-skips-the-second-s)... etc. etc. etc. — Jeff Mercado, Jul 16 '16 at 18:45

Cherubim · Accepted Answer · 2016-07-16T19:33:55.577

4

You are sending wrong argument type into scanf() for scanning strings and not checking for whitespaces... Instead use scanf this way :

scanf("%19[^\n]",word1); 
//and also
scanf(" %19[^\n]",word2); 
// a space in front to consume white spaces and don't put & before string variable

While scanning a string using scanf() the argument to be sent must be of the type char* but not char
here, word1 and word2 are of the type char* whereas word1[i] and word2[i] are of the type char
%19[^\n] would allow scanning only 19 characters including whitespaces (i.e, '\0', '\n' ' ') and will leave the last index for terminating null character '\0'. Here, 19 iss known as width field (click to know more)

Further reading :

Why does everyone say not to use scanf? What should I use instead?

edited Jul 16 '16 at 19:33

answered Jul 16 '16 at 18:42

Cherubim

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19 is a random value you chose? – Uttaran Jul 16 '16 at 18:58
Nope... Since 20 is the max size of string... %19[^\n] allow scanning only 19 characters including whitespaces and will leave the last block for terminating null character – Cherubim Jul 16 '16 at 19:00

Ed Heal · Answer 2 · 2016-07-16T19:06:25.967

2

If you read the manual page scanf

Then this bit does not make sense

scanf("%[^\n]s",&word1);

It should be

if (scanf("%19[^\n]",word1) == 1) ....

And ditto for word2

EDIT

For word 2

if (scanf("\n%19[^\n]",word2) == 1) ....

edited Jul 16 '16 at 19:06

answered Jul 16 '16 at 18:41

Ed Heal

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OP would need to take care of newline character in `stdin` , adding check won't do that . – ameyCU Jul 16 '16 at 19:03
The check ensures that `word` is filled in – Ed Heal Jul 16 '16 at 19:06
Yes, specifically I mentioned about the second `scanf` . Ahh you have updated the answer :). – ameyCU Jul 16 '16 at 19:12

chqrlie · Answer 3 · 2016-07-16T19:50:47.760

Prevent buffer overflow by passing the maximum number of characters to scanf and skip the '\n' with a dummy format:

#include <stdio.h>

int main(void) {
    char word1[20], word2[20];

    printf("Enter word 1: ");
    fflush(stdout);  // make sure prompt is output
    if (scanf("%19[^\n]%*c", word1) < 1)
        return 1;

    printf("Enter word 2: ");
    fflush(stdout);  // make sure prompt is output
    if (scanf("%19[^\n]%*c", word2) < 1)
        return 1;

    return 0;
}

Note that using scanf to read lines has some side effects: you cannot read an empty line this way and you have to hard code the buffer size in the format string in an awkward way.

I would instead suggest using fgets() and stripping the trailing '\n' this way:

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

int main(void) {
    char word1[20], word2[20];

    printf("enter word 1 : ");
    fflush(stdout);  // make sure prompt is output
    if (fgets(word1, sizeof word1, stdin) == NULL)
        return 1;

    word1[strcspn(word1, "\n")] = '\0'; // strip the \n if any

    printf("enter word 2 : ");
    fflush(stdout);  // make sure prompt is output
    if (fgets(word2, sizeof word2, stdin) == NULL)
        return 1;

    word2[strcspn(word2, "\n")] = '\0'; // strip the \n if any

    printf("word1: |%s|, word2: |%s|\n", word1, word2);
    return 0;
}

Perhaps checking the return values `scanf` you be a good idea — Ed Heal, Jul 16 '16 at 18:47
@EdHeal: indeed it would be a good idea, I'll update the answer, but it would be good to use `word1` and `word2` too;-) — chqrlie, Jul 16 '16 at 19:44
Perhaps adding `fflush(stdout);` after the `printf` may be required — Ed Heal, Jul 16 '16 at 19:46
@EdHeal: Flushing `stdout` is usually not required if both `stdin` and `stdout` are bound to the terminal, but I guess it may be useful on some systems. — chqrlie, Jul 16 '16 at 19:52

score -1 · Answer 4 · answered Jul 16 '16 at 18:36

-1

You need to add getchar() before second scanf(); There is already a '\n' char in the buffer. In the second scanf() operation it reads that character and terminates the program.

answered Jul 16 '16 at 18:36

Denis

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Unable to find error in C code

4 Answers4