fgets doesn't work after scanf

Question

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

void delspace(char *str);

int main() {
    int i, loops;
    char s1[101], s2[101];

    scanf("%d", &loops);

    while (loops--) {
        fgets(s1, 101, stdin);
        fgets(s2, 101, stdin);
        s1[strlen(s1)] = '\0';
        s2[strlen(s2)] = '\0';

        if (s1[0] == '\n' && s2[0] == '\n') {
            printf("YES\n");
            continue;
        }

        delspace(s1);
        delspace(s2);

        for (i = 0; s1[i] != '\0'; i++)
            s1[i] = tolower(s1[i]);

        for (i = 0; s2[i] != '\0'; i++)
            s2[i] = tolower(s2[i]);

        if (strcmp(s1, s2) == 0) {
            printf("YES\n");
        }
        else {
            printf("NO\n");
        }
    }

    return 0;
}

void delspace(char* str) {
    int i = 0;
    int j = 0;
    char sTmp[strlen(str)];

    while (str[i++] != '\0') {
        if (str[i] != ' ') {
            sTmp[j++] = str[i];
        }
    }
    sTmp[j] = '\0';
    strcpy(str, sTmp);
}

After I entered "loops", "s1" was assigned a blank line automatically. How does it happen? I'm sure my keyboard works fine.

Answer 1

The following works if fgets() is being "skipped" after using scanf()

After saying:

scanf("%d", &loops);

Say:

char garbage[100];

fgets(garbage,100,stdin);

This will store anything left on the input buffer into the garbage variable.

This will effectively clear the input buffer and allow you to use fgets() afterwards.

EDIT: I have recently learned that there is an easier solution than the one above. If you say getchar() after scanf(), it will allow you to use fgets() without issues. getchar() will get the next character in the input buffer, which in this case will be '\n'. Once you remove '\n' from the input buffer, fgets should work just fine.

Answer 2

Vayn,

Geekoaur has answered your question well, I'm just pointing out another "problem" with your code.

The line s1[strlen(s1)] = '\0'; is a no-op if s1 is allready properly null terminated BEFORE it executes.

But if s1 is NOT allready properly null terminated BEFORE this line executes (and you're unlucky) it will cause:

• a SIGSEGV on a POSIX (*nix) system.
• a GPF on Windows.

This is because strlen basicaly finds the index of the existing null-terminator and returns it! Here's a valid, unoptimized implementation of strlen:

int strlen(char *string) {
    int i = 0;
    while(string[i] != '\0') {
        ++i;
    }
    return i;
}

So... If you're REALLY worried about strings NOT being null-terminated then you'd do something like:

• string[sizeof(string)]='\0'; on local automatic strings (where the compiler "knows" the size of the string);
• or string[SIZE_OF_STRING] for all other strings, where SIZE_OF_STRING is (most commonly) a #define'd constant, or a variable which you maintain specifically to store the current-SIZE (not length) of a dynamically allocated string.

And if you're REALLY, REALLY, REALLY worried about strings not being null-terminated (like you're dealing with "dirty" libary methods (like Tuxedo's ATMI for example) you ALSO "clear" your "return strings" before passing them to the suspect library methods with:

• before:memset(string, NULL, SIZE_OF_STRING);
• invoke: DirtyFunction(/*out*/string);
• after: string[SIZE_OF_STRING]='\0'

SIG11's are a complete biatch to find because (unless you "hook" them with a signal-processor and say otherwise, they cause unix to hard-terminate your program, so you can't log anything (after the fact) to help figure out where-in-the-hell-did-that-come-from... especially considering that in many cases the line of code which throws the SIG11 is no-where-near the actual cause of the string loosing it's null-terminator.

Does that make sense to you?

Cheers mate. Keith.

PS: WARNING: strncpy does NOT allways nullterminate... you probably meant strlcpy instead. I learned this the hard way... when a 60 million dollar billing run crashed.

---

EDIT:

FYI: Here's a "safe" (unoptimized) version of strlen which I'll call strnlen (I reckon this should be in stdlib. Sigh.).

// retuns the length of the string (capped at size-1)
int strnlen(char *string, int size) {
    int i = 0;
    while( i<size && string[i]!='\0' ) {
        ++i;
    }
    return i;
}

Answer 3

scanf() reads exactly what you asked it to, leaving the following \n from the end of that line in the buffer where fgets() will read it. Either do something to consume the newline, or (my preferred solution) fgets() and then sscanf() from that string.

Answer 4

I know this is very old. I'm new to c and wanted to check my method, which uses getchar:

#include <stdio.h>

int main()
{

    printf("Please enter your name\n");
    char string[10];

    scanf("%s", string);
    printf("Hello %s\n", string);

    //getchar();  # un commenting this line, fgets perfectly works!!
    printf("Please enter your name again\n");

    fgets ( string, 10, stdin );     

    printf("Hello again %s", string);

    getchar();
}

Answer 5

scanf leaves whitespace in the input buffer, including new-line characters. To use fgets to read the next line you need to manually remove the rest of the current line:

int c;
do{
    c = getchar();
}while(c != EOF && c != '\n');

Answer 6

I have solve your Problem. Now your program is working fine. If you have any doubt you can ask me.

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

    int main()
    {
        int i, age;
        char phone[10];
        char name[100];
        printf("ENTER YOUR NAME:");
        fgets(name , 100 , stdin);

        printf("ENTER YOUR AGE:");
        scanf("%d",&age);

        printf("ENTER YOUR PHONE NUMBER:");
        scanf(" ");
        fgets(phone,10,stdin);


        printf("\nStudent detail\n");
        printf("Name: ");
        fputs(name , stdout );
        printf("Age: %d\n",age);
        printf("Phone Number: ");
        puts(phone);
        printf("----\n");
        return 0;

    }