口译员：简化了多少？

发布于 2024-09-01 07:43:13 字数 374 浏览 6 评论 0原文

在我的解释器中，如下代码

x=(y+4)*z
echo x

解析并“优化”为解释器执行的四个单个操作，几乎类似于汇编：

add 4 to y
multiply <last operation result> with z
set x to <last operation result>
echo x

在现代解释器（例如：CPython、Ruby、PHP）中，简化程度如何解释器运行的“操作码”有多简化？
当尝试使解释器的结构和命令更加复杂和高级时，我能否获得更好的性能？这肯定会困难得多，或者？

原文

In my interpreter, code like the following

x=(y+4)*z
echo x

parses and "optimizes" down to four single operations performed by the interpreter, pretty much assembly-like:

add 4 to y
multiply <last operation result> with z
set x to <last operation result>
echo x

In modern interpreters (for example: CPython, Ruby, PHP), how simplified are the "opcodes" for which are in end-effect run by the interpreter?
Could I achieve better performance when trying to keep the structures and commands for the interpreter more complex and high-level? That would be surely a lot harder, or?

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第几種人 2024-09-08 07:43:13

在Python的情况下，你可以让它告诉你给定函数的字节码 dis 模块。

from dis import dis
def foo():
    x=(y+4)*z
    print x

dis(foo)

给你：

2           0 LOAD_GLOBAL              0 (y)
            3 LOAD_CONST               1 (4)
            6 BINARY_ADD          
            7 LOAD_GLOBAL              1 (z)
           10 BINARY_MULTIPLY     
           11 STORE_FAST               0 (x)

3          14 LOAD_FAST                0 (x)
           17 PRINT_ITEM          
           18 PRINT_NEWLINE       
           19 LOAD_CONST               0 (None)
           22 RETURN_VALUE

其中一些是无关的（例如最后的 LOAD_CONST 和 RETURN_VALUE 用于 foo() 中隐式的 return None ），但 Python 似乎推送 y 和4 入栈，加，压入z，乘，然后写入x。然后按下 x 并打印

In Python's case, you can have it tell you the bytecode for a given function with the dis module.

from dis import dis
def foo():
    x=(y+4)*z
    print x

dis(foo)

gives you:

2           0 LOAD_GLOBAL              0 (y)
            3 LOAD_CONST               1 (4)
            6 BINARY_ADD          
            7 LOAD_GLOBAL              1 (z)
           10 BINARY_MULTIPLY     
           11 STORE_FAST               0 (x)

3          14 LOAD_FAST                0 (x)
           17 PRINT_ITEM          
           18 PRINT_NEWLINE       
           19 LOAD_CONST               0 (None)
           22 RETURN_VALUE

Some of that is extraneous (e.g. the LOAD_CONST and RETURN_VALUE at the end are for the implicit return None in foo()), but Python appears to push y and 4 onto the stack, add, push z, multiply, and write to x. Then it pushes x and prints

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