Why it's comes to error?

Question

#include<stdio.h>
int main(){
char ch[20];
ch="study To night";
printf("%s",ch);
}

Answer 1

String assignment does not exist in th C language,
You can use void strcpy(char str1[], char str2[]) from string.h.
This is equal to str1=str2 in C++;

So in your case :
strcpy( ch, "study to night");

Don't forget including <string.h>

Answer 2

String is not assignable to ch this is invalid . You need to use strcpy .
Or you can do this if you want to print your string

#include<stdio.h>
int main(){
char ch[20]="study To night";
printf("%s",ch);
}

Answer 3

It will give you an error because you are trying to assign a string "literally" to a char array, C doesn't support that. But you can assign it to a char* (well because it is a pointer, not an allocated memory of chars). So this is allowed,
char *s;
s = "study To night";
And it becomes a string literal which you can't modify directly.