操作员优先超载
我正在尝试为简单的 SQL 命令创建一个 C++ 编译模型。例如，这可能是我必须能够处理的主函数的一部分：
CREATE_TABLE(books) [ // create("books");ovr[
COLUMN(c1) TYPE(string), // "title string",
COLUMN(c2) TYPE(string), // "author string",
COLUMN(num1) TYPE(int) // "price int"
];
因此，为了做到这一点，我必须重载“[]”和“，”运算符。这样做之后，我发现“，”重载器是在“[]”重载器之前执行的。而我的猜测是“[]”应该首先执行。发生这种情况有什么特殊原因吗？或者仅仅是因为找到“]”时执行“[]”？
I am trying to make a C++ compiled model for simple SQL commands. For example this could be a part of my main function which i must be able to handle :
CREATE_TABLE(books) [ // create("books");ovr[
COLUMN(c1) TYPE(string), // "title string",
COLUMN(c2) TYPE(string), // "author string",
COLUMN(num1) TYPE(int) // "price int"
];
So in order to do that i had to overload the "[]" and "," operators. After doing so, I figured out that the "," overloader is executed before the "[]" one. Whereas my guess would be that "[]" should be executed first. Is there any particular reason why this happens? Or is it simply because the "[]" is executed when "]" is found?
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评论（5）
您的表达式将被编译为类似以下内容，这可能会解释评估顺序：
或者
取决于您的
operator,()的定义方式（以及COLUMN和已定义TYPE宏）。Your expression would be compiled to something like the following, which might explain the evaluation order:
or
depending on how your
operator,()is defined (and maybe how theCOLUMNandTYPEmacros are defined).必须首先计算
[]内的表达式 - 它需要作为参数传入，这就是为什么您会看到首先调用operator,的原因。The expression inside the
[]has to be evaluated first - it needs to be passed in as the argument, which is why you see theoperator,being called first.因为为了调用
operator[]，它需要一个参数。它将括号内的内容视为带逗号的表达式，并使用operator,获取单个结果来调用operator[]。Because in order to call
operator[], it needs a single parameter. It treats what's inside the brackets as an expression with commas, and usesoperator,to get a single result with which to call youroperator[].由于
[]运算符采用单个参数，因此它会等待[]之间的整个表达式被求值，然后再对其自身进行求值。Because the
[]operator takes a single argument, it waits for the entire expression between the[]to be evaluated before it is evaluated itself.看起来您正在使用
operator,()的结果作为operatot[]()的参数，因此首先评估前者是合乎逻辑的，如下所示f(g(x))。此外，运行时不会“解析”任何内容。 C++ 是一种编译语言。
It looks like you are using the result of
operator,()as the argument ofoperatot[](), so it's only logical that the former is evaluated first, as inf(g(x)).Also, nothing is "parsed" at runtime. C++ is a compiled language.