Exit status -1

@mwilki7 Sorry I dont know about repl putting of code, I just pasted my code down and that removed * before the c and e thus I am pasting the pic of it as well.
Sorry for that mistake so *c is a character pointer let me know why that didnt work.

@Yuvikiit I see, repl formatted your asterisks out.
Try one backwards tick mark ` for inline code, or three for code with multiple lines.

As for your code, when you declare a char pointer:
char *c;

You need space to store stuff in it.
Arrays do this automatically because you tell it from the start how much space it should have like so:
char c[20]; // space for 20 chars already made at compile

To do this with pointers:

char *c; // declare pointer
c = (char*)malloc(sizeof(char) * 20); // make space for 20 chars for this pointer

scanf("%s", c); // now I can use it, just make sure to limit your input to 20 or 
                // whatever you decide to change 20 to

OR you can steal the space made by an array if you don't want to use malloc

like so:

char *c;
char arr[20];

c = arr; // use arr's space for c

scanf("%s", c); // read to c

Here is a sample program that does this:

Program (allocate with array sharing)

#include <stdio.h>

int main()
{
    char *c;       // declare pointer
    char arr[20];  // declare array
    
    c = arr;       // array shares space with c
    
    scanf("%s", c);// write stuff to c
    
    printf("c: %s, arr: %s",c,arr); // print results
    
    return 0;    
}

Input

test

Output

c: test, arr: test

@mwilki7 Sankyu(Thanks) Sir for the reply, just let me know one more thing
Like why this works for a single string.
and how does it work:
I thought in the manner that being a character pointer
the pointer would store 1 character as I am entering and then will move to next byte for latter entry and as soon as I press enter it puts a '\0' at the end and completes the string,

And I also don't know like why this should be a problem even
is it because that the pointer has some garbage address stored in it and it is not a good programming practice to store our string at some garbage place though, the memory chunks might be continuous (as the char pointer
does that bit by default.)

//Please answer:)
#include<stdio.h>
#include<string.h>
int main()
{
  int i;
  char *c,d,*e;
  scanf("%s",c); //to take in the 1st string  
  //scanf("%c",&d);//after string the newline char needs to get in the buffer
  //scanf("%s",e); //to take in another string
  printf("%s\n",c); //to print both the strings
   return 0;
}

and
let me know is it some compiler dependant thing because I tried the same thing in Codeblocks and it didn't work.

Thanks once again for the help.

@mwilki7 Thank you for help I will try to be a good programmer with this practice.