Assigning strings to arrays of characters

Question

I am a little surprised by the following.

Example 1:

char s[100] = "abcd"; // declare and initialize - WORKS 

Example 2:

char s[100]; // declare s = "hello"; // initalize - DOESN'T WORK ('lvalue required' error) 

I'm wondering why the second approach doesn't work. It seems natural that it should (it works with other data types)? Could someone explain me the logic behind this?

Answer 1

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.

Answer 2

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);