CODEWITHSUNDEEP
cstring
Novaea
Novaea

Reputation: 45

Issues printing strings

We did this exercise in class today, but unfortunately, my code wasn't running properly. It won't print string1. My teacher also could not figure out why this was happening.

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

void main()
{
char string1[20];
char string2[] = "It is almost end of semester!";
size_t idx;
int size;

printf("Enter string1:\n");
scanf_s("%s", string1);
size = sizeof(string2) / sizeof(char);
printf("\nString 1 is : %s\n\n", string1);
for (idx = 0; idx < size; idx++)
{
    printf("%c ", string2[idx]);
}
puts("");
system("pause");;
}

Upvotes: 0

Views: 61

Answers (2)

David C. Rankin
David C. Rankin

Reputation: 84642

You have used scanf_s instead of scanf which, as Mr. Lynch correctly pointed out requires an additional parameter. You can accomplish the same thing with scanf itself. One approach is as follows:

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

int main () {
    char string1[20];
    char string2[] = "It is almost end of semester!";
    size_t idx;
    int size;

    printf ("Enter string1:\n");
    scanf ("%[^\n]%*c", string1);
    string[19] = 0;                            /* force null-termination of string1 */
    size = sizeof (string2) / sizeof (char);
    printf ("\nString 1 is : %s\n\n", string1);
    for (idx = 0; idx < size; idx++) {
        printf ("%c ", string2[idx]);
    }
    puts ("");
    return 0;
}

output:

$ ./bin/strprn
Enter string1:
scanf_s is not scanf

String 1 is : scanf_s is not scanf

I t   i s   a l m o s t   e n d   o f   s e m e s t e r !

Note: main() is type int not void no matter what ms lets you get away with. It returns a value as well.

Upvotes: 0

Bill Lynch
Bill Lynch

Reputation: 81996

scanf_s requires an extra parameter.

scanf_s("%s", string1, _countof(string1));

Upvotes: 1

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