fgets doesn't work after scanf [duplicate]

Question

This question already has answers here:

scanf() leaves the new line char in the buffer (5 answers)

Closed last year.

#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

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 2

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 3

This is a more Simpler solution

scanf("%d",&loops);
while ((getchar()) != '\n'); //This will consume the '\n' char
//now you're free to use fgets
fgets(string,sizeof(string),stdin);

Answer 4

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 5

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 6

just put scanf("%d\n",&loops);

instead of scanf("%d",&loops);

Answer 7

Another way to ignore following newline character (due to hitting ENTER) after scanning the integer in variable label loops is:

scanf ("%d%*c", &loops);

Where, as per the man pages:

* Suppresses assignment. The conversion that follows occurs as usual, but no pointer is used; the result of the conversion is simply discarded.

This is very unlikely, but good habit to check for error during scanf:

errno = 0
scanf ("%d%*c", &loops);
if (errno != 0) perror ("scanf");
// act accordingly to avoid un-necessary bug in the code

For example in your code, loops is a local un-initialized variable, containing garbage. If scanf fails to fill the desired value, following while loop may run arbitrarily.

Answer 8

scanf() leave the following \n from the end of that line in the buffer where fgets() will read it. To slove this we have to do something to consume the newline. Let's see the problem.

The problem

#include<stdio.h>
void main()
{
    char input[10];
    printf("Enter in fgets: ");
    scanf("%s", input);
    getchar();
    printf("Enter in scanf: ");
    fgets(input, 10, stdin);
}

Enter in scanf: Hello
Enter in fgets:

As you can see that it dons't show the fgets part. Let's see the solution.

Put getchar() between scanf and fgets()

#include<stdio.h>
void main()
{
    char input[10];
    printf("Enter in fgets: ");
    scanf("%s", input);
    getchar();
    printf("Enter in scanf: ");
    fgets(input, 10, stdin);
}

Enter in scanf: Hello
Enter in fgets: Hello

If it's possible use scanf() after the fgets()

#include<stdio.h>
void main()
{
    char input[10];
    printf("Enter in fgets: ");
    fgets(input, 10, stdin);
    getchar();
    printf("Enter in scanf: ");
    scanf("%s", input);
}

Enter in fgets: Hello
Enter in scanf: Hello

Put fget() 2 times(This method don't work some time)

#include<stdio.h>
void main()
{
    char input[10];
    printf("Enter in fgets: ");
    scanf("%s", input);
    getchar();
    printf("Enter in scanf: ");
    fgets(input, 10, stdin);
    fgets(input, 10, stdin)
}

Enter in scanf: Hello
Enter in fgets: Hello

Put sscanf() inseted of scanf()

#include<stdio.h>
void main()
{
    char input[10];
    printf("Enter in fgets: ");
    sscanf(hello, "%s", input);
    getchar();
    printf("Enter in scanf: ");
    fgets(input, 10, stdin);
}

Enter in sscanf: Hello
Enter in fgets: Hello