cinitializationexpressiondeclaration

Is initialization an assignment expression?

I found this paragraph in C17(6.7.9 #1):

initializer:
    assignment-expression
    { initializer-list }
    { initializer-list , }
initializer-list:
    designation{opt} initializer
    initializer-list , designation{opt} initializer
designation:
    designator-list =
designator-list:
    designator
    designator-list designator
designator:
    [ constant-expression ]
    . identifier

Does this mean that the initializer expands to an assignment expression and we can use it this way?:

#include <stdio.h>
int main(void)
{
    if((int n = 9) == 9){}
    
    return 0;
}

  1. If not, why not?

  2. Why is initialization not an expression?

  3. Why is initialization referred to as a declaration?

  4. Why is the assignment expression not an initializer?

  5. What is a declarator?

I'm sorry for such indiscreet questions.I'm new to C and I just want to know.

And if it's not difficult for you, could you provide the relevant paragraphs from the C Standard?

Thank you very much!

Solution

  • Does this mean that the initializer expands to an assignment expression and we can use it this way?

    It is actually the opposite: The grammar says you can have an assignment inside an initialization, not an initialization inside an assignment.

    If not, why not?

    In the grammar rules, the : in A: B means A can be a B.

    Why is initialization not an expression?

    In the grammar, initializer appears in init-declarator (that is, initializer appears on the right side of a : rule with init-declarator on the left), which appears in init-declarator-list, which appears in declaration. None of those appear in the grammar for expression, except a compound literal has initializers. A compound literal is something you can use in an expression, but it will not be declaring any variables.

    Why is initialization referred to as a declaration?

    Where do you see that?

    Why is the assignment expression not an initializer?

    An assignment expression can be an initializer. For example, in int a, b = a = 3;, the a = 3 is an assignment expression. Its value is 3, so b is initialized to 3.

    What is a declarator?

    A declarator is the part of a declaration that goes where the D is here: Type D;, excluding the = initializer that could be there with the D. That is, you could have int foo = 3;, but only the foo would be the declarator. (There can be additional parts to Type and can be multiple declarators, in a list, but I have used this simplified example for illustration.) That D can be a name (int foo;), an array declarator (int foo[3];), a pointer declarator (int *foo;), a function declarator (int foo(char, float);), or a parenthesized declarator (int (*foo); or int (foo);), and it can be combinations of those (int *foo[3];, int (*foo)[3]). (There can be additional parts of those declarators, such as int foo[const 3];, which I do not cover further here.)