有一个Lua字符串replace()函数可以比gsub()更快地进行替换吗? [英] Is there a Lua string replace() function for faster replacements than gsub()?

问题描述

我看到Lua字符串函数的列表，并且看到.gsub()，用于全局搜索和替换:

I see a list of Lua string functions and I see the .gsub(), for global search and replace: http://www.gammon.com.au/scripts/doc.php?general=lua_string

所有lua字符串函数:

All lua string functions :

static const luaL_Reg strlib[] = {
  {"byte", str_byte},
  {"char", str_char},
  {"dump", str_dump},
  {"find", str_find},
  {"format", str_format},
  {"gfind", gfind_nodef},
  {"gmatch", gmatch},
  {"gsub", str_gsub},
  {"len", str_len},
  {"lower", str_lower},
  {"match", str_match},
  {"rep", str_rep},
  {"reverse", str_reverse},
  {"sub", str_sub},
  {"upper", str_upper},
  {NULL, NULL}
};

为什么没有简单，快速，简单的(非正则表达式)字符串替换功能? .gsub()如此高效以至没有好处吗?

Why is there no simple, fast, litteral (non-regex) string replace function? Is .gsub() so efficient that there is no benefit?

我发现这是2006年写的，但似乎并未包含在内: http://lua-users .org/wiki/StringReplace

I found this written in 2006 but it does not seem like it's included: http://lua-users.org/wiki/StringReplace

推荐答案

这可能是因为 gsub 能够完全执行replace函数的功能，而Lua的

This is likely because gsub is capable of doing exactly what a replace function would do, and Lua's design goals include that of a small, generally uncomplicated standard library. There's no need for a redundancy like this to be baked right into the language.

作为外部示例，Ruby编程语言在其标准库中同时提供了String#gsub和String#replace.由于这样的决定，Ruby是开箱即用的语言.

As an outside example, the Ruby programming language provides both String#gsub and String#replace in its standard library. Ruby is a much, much larger language out of the box because of decisions like this.

但是，Lua以它是一种非常容易扩展的语言而自豪.您显示的链接显示了整体编译Lua时如何将函数烘焙到标准库中.您也可以将其拼凑在一起以创建模块.

However, Lua prides itself on being a very easy language to extend. The link you've shown shows how to bake the function into the standard library when compiling Lua as a whole. You could also piece it together to create a module.

将需要的结果快速修补在一起(请注意，我们需要lmemfind函数="nofollow"> lstrlib.c ):

Quickly patching together the parts we need results in (note we need the lmemfind function from lstrlib.c):

#include <lua.h>
#include <lauxlib.h>
#include <string.h>

static const char *lmemfind
(const char *s1, size_t l1, const char *s2, size_t l2) {
    if (l2 == 0)
        return s1;  /* empty strings are everywhere */
    else if (l2 > l1)
        return NULL;  /* avoids a negative 'l1' */

    const char *init;  /* to search for a '*s2' inside 's1' */
    l2--;  /* 1st char will be checked by 'memchr' */
    l1 = l1-l2;  /* 's2' cannot be found after that */

    while (l1 > 0 && (init = (const char *) memchr(s1, *s2, l1)) != NULL) {
        init++;   /* 1st char is already checked */

        if (memcmp(init, s2+1, l2) == 0)
            return init-1;
        else {  /* correct 'l1' and 's1' to try again */
            l1 -= init-s1;
            s1 = init;
        }
    }

    return NULL;  /* not found */
}

static int str_replace(lua_State *L) {
    size_t l1, l2, l3;
    const char *src = luaL_checklstring(L, 1, &l1);
    const char *p = luaL_checklstring(L, 2, &l2);
    const char *p2 = luaL_checklstring(L, 3, &l3);
    const char *s2;
    int n = 0;
    int init = 0;

    luaL_Buffer b;
    luaL_buffinit(L, &b);

    while (1) {
        s2 = lmemfind(src+init, l1-init, p, l2);
        if (s2) {
            luaL_addlstring(&b, src+init, s2-(src+init));
            luaL_addlstring(&b, p2, l3);
            init = init + (s2-(src+init)) + l2;
            n++;
        } else {
            luaL_addlstring(&b, src+init, l1-init);
            break;
        }
    }

    luaL_pushresult(&b);
    lua_pushnumber(L, (lua_Number) n);  /* number of substitutions */
    return 2;
}

int luaopen_strrep (lua_State *L) {
    lua_pushcfunction(L, str_replace);
    return 1;
}

我们可以使用适当的链接(cc -shared，cc -bundle等)将其编译为共享对象，并像使用require的任何其他模块一样将其加载到Lua中.

We can compile this into a shared object with the proper linkage (cc -shared, cc -bundle, etc...), and load it into Lua like any other module with require.

local replace = require 'strrep'

print(replace('hello world', 'hello', 'yellow')) -- yellow world, 1.0

此答案是对以上评论的正式重构.

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