函数传参涉及类型转换时，const与非const参数的差异是什么？

Hey fellow C++ learner! Let's unpack the critical points you're reading from Bjarne Stroustrup's classic—this distinction between initialization and assignment in parameter passing is super important for writing correct, idiomatic C++.

First, the chapter opens with a core rule that sets the foundation for how function arguments work:

参数传递的语义（semantics）与初始化（initialization）的语义完全相同。会检查参数类型，并在必要时进行隐式参数类型转换。

To put this in plain terms: when you pass an argument to a function, it’s exactly like creating a new variable (the function parameter) and initializing it with that argument. If the types don’t match but are compatible, the same implicit conversions you’d get during variable initialization kick in—think converting an int to a double, or a derived class pointer to a base class pointer.

Then Stroustrup draws a key contrast that trips up many programmers:

请注意，参数传递的语义与赋值（assignment）的语义不同。这对于const参数、引用参数以及……（原文未完成）而言十分重要。

Let me fill in that unfinished bit with why this distinction matters for those cases:

• Reference parameters: Initialization binds the reference directly to the original argument. If this were assignment instead, you’d be trying to overwrite the reference’s target (which isn’t allowed for const refs, and would behave entirely differently for non-const ones). This is why you can pass temporary values to const T& parameters—initialization allows it, but assignment never would.
• const parameters: Initialization lets you pass non-const arguments to a const parameter (since you’re just creating a read-only "view" of the value), but assignment to a const variable is invalid. The initialization semantics here keep the language safe and consistent.

内容的提问来源于stack exchange，提问作者Laschet Jain