Assigning strings to arrays of characters

When initializing an array, C allows you to fill it with values. So

char s[100] = "abcd";

is basically the same as

int s[3] = { 1, 2, 3 };

but it doesn't allow you to do the assignment since s is an array and not a free pointer. The meaning of

s = "abcd" 

is to assign the pointer value of abcd to s but you can't change s since then nothing will be pointing to the array.
This can and does work if s is a char* - a pointer that can point to anything.

If you want to copy the string simple use strcpy.

There is no such thing as a "string" in C. In C, strings are one-dimensional array of char, terminated by a null character \0. Since you can't assign arrays in C, you can't assign strings either. The literal "hello" is syntactic sugar for const char x[] = {'h','e','l','l','o','\0'};

The correct way would be:

char s[100];
strncpy(s, "hello", 100);

or better yet:

#define STRMAX 100
char    s[STRMAX];
size_t  len;
len = strncpy(s, "hello", STRMAX);

Initialization and assignment are two distinct operations that happen to use the same operator ("=") here.

1    char s[100];
2    s = "hello";

In the example you provided, s is actually initialized at line 1, not line 2. Even though you didn't assign it a value explicitly at this point, the compiler did.

At line 2, you're performing an assignment operation, and you cannot assign one array of characters to another array of characters like this. You'll have to use strcpy() or some kind of loop to assign each element of the array.