In C89 A.1.2.2 Declarations
direct-declarator [ constant-expression ]
In C99 A.2.2 Declarations
direct-declarator [ type-qualifier-list_opt assignment-expression_opt ]
I understand assignment-expression_opt as var = 1;. But arrays are not declared like int arr[i = 0];. Why does C99 use the term "assignment-expression" instead of "constant-expression"and what does it mean?
In C89 there was no support for variable length arrays. This means that array sizes must be fixed at compile time.
Starting with C99, declarations and statements may be mixed within a block, so full expressions that must be executed are now allowed as an initializer. This also is what allows for variable length arrays to be created.
So a declaration like int arr[i=0]; is valid syntax, although it's invalid because it created an array of size 0. int arr[i=2]; is valid and will create arr as an array of int of size 2, and it sets i to 2.
This is because C99 supports VLA and whatever goes between [] may then be pretty much any expression.
As for why it says assignment-expression specifically:
The way the syntax of expressions and operator precedence is described in C, is by having each operator group's syntax point at the operator group with higher precedence than itself. For example look at the syntax for an assignment expression:
assignment-expression:
conditional-expression
It points at the conditional ?: operator which will be the next operator group with higher precedence than assignment operators.
And for the comma operator with lowest precedence of all, the syntax is actually:
expression:
assignment-expression
expression,assignment-expression
That is, the syntax for an expression in C. Now for some reason, C99 apparently didn't want to allow comma operators inside []. I don't know why - perhaps it would make qualified/static array parameter declarators too strange. So instead of using expression in the direct-declarator syntax, they picked the next operator group above the comma operator, the assignment operators.
