为什么 Rust 既有按值调用又有按引用调用? [英] Why does Rust have both call by value and call by reference?

问题描述

有些语言，比如 Haskell，不区分值传递和引用传递.然后，编译器可以通过启发式方法大致选择最有效的调用约定.一个启发式示例适用于 Linux x64 ABI:如果参数的大小大于 16 字节，则传递一个指向堆栈的指针，否则传递寄存器中的值.

Some languages, like Haskell, make no distinction between pass-by-value and pass-by-reference. The compiler can then approximately choose the most efficient calling convention with a heuristic. One example heuristic would be for the Linux x64 ABI: if the size of parameter is greater than 16 bytes, pass a pointer to the stack otherwise pass the value in registers.

在 Rust 中保留按值传递和按引用传递(当然是不可变的)的概念并强制用户选择有什么好处?

What is the advantage of keeping both notions of pass-by-value and pass-by-reference (non-mutable of course) in Rust and forcing the user to choose?

如果看到值被修改，会不会是传值是传引用+复制的语法糖?

Could it be the case that pass-by-value is syntactic sugar for pass-by-reference + copy if the value is seen to be modified?

推荐答案

两件事:

1. Rust 将基于类似的启发式将某些按值传递调用转换为按引用传递.
2. 传值表示所有权转移，传引用表示借用.这些非常不同，并且与您所询问的 asm 级别的问题完全正交.

换句话说，在 Rust 中，这两种形式具有不同的语义.不过，这并不排除进行优化.

In other words, in Rust, these two forms have different semantics. That doesn't preclude also doing the optimization, though.

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