C char array v C char* initialization

Question

The following is accepted as valid c code by gcc version 6.3:

char white[] = { 'a', 'b', 'c' };
char blue[]  = "abc";
char *red    = "abc";

However the following fails:

 char *green = { 'a', 'b', 'c' };   // gcc error

I am sure there is a perfectly rational reason for this to be the case, but I am wondering what it is. This question is motivated by the case when having to initialize an array of bytes (so unsigned char rather than char), it is very tempting to write something like { '\x43', '\xde', '\xa0' } rather than "\x43\xde\xa0", and as soon as you forget to write my_array[] instead of *my_array, you get caught by the compiler.

Answer 1

The following will produce an error

char *green = { 'a', 'b', 'c' };

Because the initializer for green isn't an array of characters as you believe. It doesn't have a type, it's just a brace-enclosed initializer list. The thing that it initializes in the previous samples (i.e. white) determines how it's interpreted. The same initialzier can be used to initialize any aggregate that is capable of holding 3 characters.

But green is a pointer, and not an aggregate, so you can't use a brace-enclosed initializer list as it's initial value.¹

Now, the following two work but with very different semantics:

char blue[]  = "abc";
char *red    = "abc";

blue is an array. It will hold the same contents as the literal "abc". red is a pointer that points at the literal "abc".

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1. You can use a compound literal expression:

   char *green = (char[]){ 'a', 'b', 'c' };
   

   It tells the compiler to create an unnamed object (the life time of which depends on the scope of the declaration), that is of character array type and is initialized with those three characters. The pointer is then assigned the address of that object.