What do square brackets mean in array initialization in C?

Question

static uint8_t togglecode[256] = {     [0x3A] CAPSLOCK,     [0x45] NUMLOCK,     [0x46] SCROLLLOCK }; 

What's the meaning of [0x3A] here? I have only learned statements like int a[2] = {1, 2};

Answer 1

It means initialise the n-th element of the array. The example you've given will mean that:

togglecode[0x3A] == CAPSLOCK togglecode[0x45] == NUMLOCK togglecode[0x46] == SCROLLLOCK 

These are called "designated initializers", and are actually part of the C99 standard. However, the syntax without the = is not. From that page:

An alternative syntax for this which has been obsolete since GCC 2.5 but GCC still accepts is to write [index] before the element value, with no =.

Answer 2

According to the GCC docs this is ISO C99 compliant. They refer to it as "Designated Initializers":

To specify an array index, write `[index] =' before the element value. For example,

 int a[6] = { [4] = 29, [2] = 15 }; 

is equivalent to

 int a[6] = { 0, 0, 15, 0, 29, 0 }; 

I've never seen this syntax before, but I just compiled it with gcc 4.4.5, with -Wall. It compiled successfully and gave no warnings.

As you can see from that example, it allows you to initialize specific array elements, with the others being set to their default value (0).