Trying to compare a hard defined string with a user defined string - c

Question

Problem Statement
Today is Newton School's first class of this year. Nutan, a student at
Newton School, has received his first assignment. He will be given a
string as input. His task is to print "Gravity'' if the input is
"Apple''; otherwise, he will have to print "Space''.
Can you help Nutan in solving his first assignment? Note that the
quotation marks are just for clarity. They are not part of the input
string, and should not be a part of your output string. Input The
input consists of a single line that contains a string S (1 ≤ length
of S ≤ 10). The string only consists of lowercase and uppercase
letters. Output Print "Gravity'' or "Space'' according to the input.
What I am trying to do:
Basically, I am taking a user-defined string and trying to compare it with the hard input string i.e "Apple". If both the string matches then it will print "Gravity" or else it will print "Space"
#include <stdio.h> // header file for Standard Input Output
#include <stdlib.h> // header file for Standard Library
#include <string.h> // for strcmp() function
int main() {
char str1[10]="Apple";
char str2[20];
int value;
printf("Enter the input ");
scanf("%s", &str2[20]);
value = strcmp(str1, str2);
if(value==0)
printf("Gravity");
else
printf("Space");
return 0;
}

Answer 1

scanf("%s", &str2[20]);
may invoke undefined behavior by out-of-range access. You should:
Pass the pointer to the first element of the array, not one to the next element of the last element. (most) arrays in expressions are automatically converted to pointes to their first elements.
Specify the maximum length to read to avoid buffer overrun.
Check if reading succeeded.
The line should be:
if (scanf("%19s", str2) != 1) {
puts("read error");
return 1;
}

Answer 2

Some improvements:
Don't use "%s", use "%<WIDTH>s", to avoid buffer-overflow
Instead of using bare return 0;, use return EXIT_SUCCESS;, which is defined in the header file stdlib.h.
always check whether scanf() input was successful or not
Use const char * instead of char str1[10]
There's no need for int value;
SYNTAX ERROR: &str2[20]
There's no need for passing the address of str2 READ MORE
Initialize str2 with zeroes
Add 1 more space in your str2 for NULL ('\0') terminating character
Final Code:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void) {
const char *str1 = "Apple";
char str2[21] = {};
printf("Enter the input ");
if(scanf("%20s", str2) != 1)
{
fprintf(stderr, "bad input\n");
return EXIT_FAILURE;
}
if(strcmp(str1, str2) == 0)
printf("Gravity");
else
printf("Space");
return EXIT_SUCCESS;
}

Answer 3

You are entering a string starting from the memory address after the last element of the array str2
scanf("%s", &str2[20]);
and then trying to compare the character array str1 with the non-initialized array str2
value = strcmp(str1, str2);
Change the call of scanf like
scanf("%19s", str2);
And the program will be safer if at least the array str2 will be initially initialized
char str2[20] = "";
Also as the array str1 is not changed then instead of the array you could declare a pointer to the string literal like
const char *str1 = "Apple";
And instead of the calls of printf
printf("Gravity");
//..
printf("Space");
it is better to use calls of puts
puts("Gravity");
//..
puts("Space");
Pay attention to that neither declaration from the header <stdlib.h> is used in your program. So you may remove this include directive
#include <stdlib.h> // header file for Standard Library