Why I get this error on Hackerrank in C language (~ no response on stdout ~)

Question

Input

12  
4.0   
is the best place to learn and practice coding!

Output

 ~ no response on stdout ~

Expected output

16, 4,  HackerRank is the best place to learn and practice coding!

I am getting this error when I tried scanf or fgets and I could not find a solution for it.

  *** buffer overflow detected ***: ./Solution terminated
Reading symbols from Solution...done.
[New LWP 1597965]
[Thread debugging using libthread_db enabled]
Using host libthread_db library "/lib/x86_64-linux-gnu/libthread_db.so.1".
Core was generated by `./Solution'.
Program terminated with signal SIGABRT, Aborted.
#0  __GI_raise (sig=sig@entry=6) at ../sysdeps/unix/sysv/linux/raise.c:50

#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {
    int i = 4;
    double d = 4.0;
    char s[] = "HackerRank ";
    
    int j;
    double k;
    char g[100];

 
    scanf(" %d %lf\n%[^\n]", &j, &k);
    
    fgets(g,100,stdin);
   

    printf("%d\n",(i+j));
    
    printf("%.1lf\n",(d+k));
    
    printf("%s\n",strcat(s, g));
    
    
    
    // Declare second integer, double, and String variables.
    
    // Read and save an integer, double, and String to your variables.
    
    // Print the sum of both integer variables on a new line.
    
    // Print the sum of the double variables on a new line.
    
    // Concatenate and print the String variables on a new line
    // The 's' variable above should be printed first.

    return 0;
}

Answer 1

You have multiple severe issues in your code.

If I feed your code into GCC I get valuable hints about some of them. Besides warnings about imcompatible implicit function declarations (because you do not include stdio.h) there is:

test.c: In function ‘main’:
test.c:14:24: warning: format ‘%[^
   ’ expects a matching ‘char *’ argument [-Wformat=]
   14 |     scanf(" %d %lf\n%[^\n]", &j, &k);
      |                     ~~~^~
      |                        |
      |                        char *

And I even did not add any flag to increase warning level to GCC.

You provide 3 format specifiers but only 2 addresses to store the result. This will result in taking the (more or less) random content of stack or registers and interpret it as address to store a string. This is very likey the reason for your crash.

You use fgets in the next line. Therefore I assume this is just a remains of your first attempts to read input.

As a fix, just remove that format specifier.

Then you want to concatenate your input string to some predefined string:

strcat(s, g)

This causes undefined behaviour because s cannot hold any more characters. You must provide an array that is large enough to hold the result.

You should also check the return value of scanf and other I/O functions.

A fixed version of your code looks like this:

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

int main() {
    int i = 4;
    double d = 4.0;
    char s[] = "HackerRank ";

    int j;
    double k;
    char g[100];

    int params = scanf(" %d %lf\n", &j, &k);
    if (params != 2)
    {
      fprintf(stderr, "Wrong input\n");
      return 1;
    }
    fgets(g,100,stdin);

    printf("%d\n",(i+j));
    printf("%.1lf\n",(d+k));

    size_t len = strlen(s) + strlen(g) + 1;
    char result[len];
    strcpy(result, s);
    strcat(result, g);
    printf("%s\n", result);

    return 0;
}

This fixes the crash.

Then you have wrong expected output.

16, 4,  HackerRank is the best place to learn and practice coding!

You take 4.0 as input and add a fixed value which is also 4.0. That is 8.0. Not 4.

Besides these issues I would suggest not to mix scanf and fgets but to use fgets for all the input and then use sscanf to do the parsing of numbers etc.