Lua覆盖#的字符串

本文介绍了Lua覆盖#的字符串的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我正在尝试对Lua中的字符串实施我自己的长度方法. 我已经成功重写了len()方法的字符串，但是我不知道如何为#运算符执行此操作.

I'm trying to imlement my own length method for strings in Lua. I have successfully overriden len() method for string, but I have no idea how to do this for # operator.

orig_len = string.len function my_len(s) print(s) return orig_len(s) end string.len = my_len abc = 'abc'

如果我打电话:

print(abc:len())

它输出:

abc 3

但是

print(#abc)

仅输出"3"，这意味着它调用了原始长度函数，而不是我的.有没有办法让#调用我的length函数?

Outputs only '3' and that means it called original length function instead of mine. Is there a way to make # call my length function?

推荐答案

我正在尝试在Lua中对字符串使用自己的长度方法.

I'm trying to imlement my own length method for strings in Lua.

您不能在Lua上做到这一点.

You can't do this from Lua.

您需要修改Lua源，特别是虚拟机(lvm.c)，并更改其对操作码OP_LEN的处理.在Lua 5.2中，您需要更改luaV_objlen来检查元方法，然后才能获取字符串的实际长度:

You'd need to modify the Lua source, specifically the virtual machine (lvm.c) and change its handling of the opcode OP_LEN. In Lua 5.2 you'd need to change luaV_objlen to check the metamethod before getting the string's actual length:

case LUA_TSTRING: { tm = luaT_gettmbyobj(L, rb, TM_LEN); // <--- add this line if (!ttisnil(tm)) // <--- add this line break; // <--- add this line setnvalue(ra, cast_num(tsvalue(rb)->len)); return; }

但这似乎是运算符重载滥用，例如重载+是指除法之类的东西.

But this seems like operator overloading abuse, like overloading + to mean division or something.