使用Lua评估数学表达式 [英] Evaluating Mathematical Expressions using Lua
问题描述
在我的上一个问题中，我一直在寻找一种在C中规避复杂数学表达式的方法其中的建议需要实现某种类型的解析器.
但是一个答案，建议使用Lua评估表达式.我对这种方法很感兴趣，但是对Lua一无所知.
在Lua有经验的人可以阐明一些想法吗?
我特别想知道的是 Lua提供了哪些API来评估以字符串形式传递的数学表达式?如果没有API可以做这样的事情，那么也许有人可以通过它来阐明链接的答案似乎是一种好方法:)
谢谢
我要评估的表达式类型得到了一些用户输入，例如
y = x ^ 2 + 1/x-cos(x)
计算y的x值范围
设置Lua解释器实例，并将其传递给要评估的表达式，然后返回调用该表达式的函数来评估该表达式很简单.您甚至可以让用户拥有变量...
这是我编写的示例代码，并已编辑为其他答案.无论如何，最好将其放在标记为Lua的问题上，因此我也在此处添加它.我对此进行了编译并在少数情况下进行了尝试，但是如果不注意错误处理等问题，那么在生产代码中当然不应信任它.所有通常的注意事项都适用于此.
我使用适用于Windows的Lua 中的Lua 5.1.4在Windows上进行了编译和测试.在其他平台上，您必须从通常的来源或从www.lua.org中找到Lua.
更新:该示例使用简单直接的技术将Lua API的全部功能和复杂性隐藏在尽可能简单的界面后面.它可能原样有用，但可以通过多种方式加以改进.
我鼓励读者研究更多易于生产的 lhf 获取，该代码利用API避免了一些快速而肮脏的情况我使用过的字符串操作.他的库还将数学库提升为全局名称空间，以便用户可以说sin(x)或2 * pi而不必说math.sin等.
LE的公共接口
这是文件le.h:
/* Public API for the LE library.
*/
int le_init();
int le_loadexpr(char *expr, char **pmsg);
double le_eval(int cookie, char **pmsg);
void le_unref(int cookie);
void le_setvar(char *name, double value);
double le_getvar(char *name);
使用LE的示例代码
这是文件t-le.c，演示了此库的简单用法.它采用单个命令行参数，将其作为表达式加载，并使用11步将全局变量x从0.0更改为1.0进行求值:
#include <stdio.h>
#include "le.h"
int main(int argc, char **argv)
{
int cookie;
int i;
char *msg = NULL;
if (!le_init()) {
printf("can't init LE\n");
return 1;
}
if (argc<2) {
printf("Usage: t-le \"expression\"\n");
return 1;
}
cookie = le_loadexpr(argv[1], &msg);
if (msg) {
printf("can't load: %s\n", msg);
free(msg);
return 1;
}
printf(" x %s\n"
"------ --------\n", argv[1]);
for (i=0; i<11; ++i) {
double x = i/10.;
double y;
le_setvar("x",x);
y = le_eval(cookie, &msg);
if (msg) {
printf("can't eval: %s\n", msg);
free(msg);
return 1;
}
printf("%6.2f %.3f\n", x,y);
}
}
这是t-le的一些输出:
E:...>t-le "math.sin(math.pi * x)" x math.sin(math.pi * x) ------ -------- 0.00 0.000 0.10 0.309 0.20 0.588 0.30 0.809 0.40 0.951 0.50 1.000 0.60 0.951 0.70 0.809 0.80 0.588 0.90 0.309 1.00 0.000 E:...>
LE的实现
这里是le.c，实现了Lua Expression评估程序:
#include <lua.h>
#include <lauxlib.h>
#include <stdlib.h>
#include <string.h>
static lua_State *L = NULL;
/* Initialize the LE library by creating a Lua state.
*
* The new Lua interpreter state has the "usual" standard libraries
* open.
*/
int le_init()
{
L = luaL_newstate();
if (L)
luaL_openlibs(L);
return !!L;
}
/* Load an expression, returning a cookie that can be used later to
* select this expression for evaluation by le_eval(). Note that
* le_unref() must eventually be called to free the expression.
*
* The cookie is a lua_ref() reference to a function that evaluates the
* expression when called. Any variables in the expression are assumed
* to refer to the global environment, which is _G in the interpreter.
* A refinement might be to isolate the function envioronment from the
* globals.
*
* The implementation rewrites the expr as "return "..expr so that the
* anonymous function actually produced by lua_load() looks like:
*
* function() return expr end
*
*
* If there is an error and the pmsg parameter is non-NULL, the char *
* it points to is filled with an error message. The message is
* allocated by strdup() so the caller is responsible for freeing the
* storage.
*
* Returns a valid cookie or the constant LUA_NOREF (-2).
*/
int le_loadexpr(char *expr, char **pmsg)
{
int err;
char *buf;
if (!L) {
if (pmsg)
*pmsg = strdup("LE library not initialized");
return LUA_NOREF;
}
buf = malloc(strlen(expr)+8);
if (!buf) {
if (pmsg)
*pmsg = strdup("Insufficient memory");
return LUA_NOREF;
}
strcpy(buf, "return ");
strcat(buf, expr);
err = luaL_loadstring(L,buf);
free(buf);
if (err) {
if (pmsg)
*pmsg = strdup(lua_tostring(L,-1));
lua_pop(L,1);
return LUA_NOREF;
}
if (pmsg)
*pmsg = NULL;
return luaL_ref(L, LUA_REGISTRYINDEX);
}
/* Evaluate the loaded expression.
*
* If there is an error and the pmsg parameter is non-NULL, the char *
* it points to is filled with an error message. The message is
* allocated by strdup() so the caller is responsible for freeing the
* storage.
*
* Returns the result or 0 on error.
*/
double le_eval(int cookie, char **pmsg)
{
int err;
double ret;
if (!L) {
if (pmsg)
*pmsg = strdup("LE library not initialized");
return 0;
}
lua_rawgeti(L, LUA_REGISTRYINDEX, cookie);
err = lua_pcall(L,0,1,0);
if (err) {
if (pmsg)
*pmsg = strdup(lua_tostring(L,-1));
lua_pop(L,1);
return 0;
}
if (pmsg)
*pmsg = NULL;
ret = (double)lua_tonumber(L,-1);
lua_pop(L,1);
return ret;
}
/* Free the loaded expression.
*/
void le_unref(int cookie)
{
if (!L)
return;
luaL_unref(L, LUA_REGISTRYINDEX, cookie);
}
/* Set a variable for use in an expression.
*/
void le_setvar(char *name, double value)
{
if (!L)
return;
lua_pushnumber(L,value);
lua_setglobal(L,name);
}
/* Retrieve the current value of a variable.
*/
double le_getvar(char *name)
{
double ret;
if (!L)
return 0;
lua_getglobal(L,name);
ret = (double)lua_tonumber(L,-1);
lua_pop(L,1);
return ret;
}
备注
上面的示例总共由189行代码组成，包括少量的注释，空白行和演示.对于知道如何计算一个变量的合理任意表达式并具有previous question I was looking for a way of evaulating complex mathematical expressions in C, most of the suggestions required implementing some type of parser.
However one answer, suggested using Lua for evaluating the expression. I am interested in this approach but I don't know anything about Lua.
Can some one with experience in Lua shed some light?
Specifically what I'd like to know is Which API if any does Lua provide that can evaluate mathematical expressions passed in as a string? If there is no API to do such a thing, may be some one can shed some light on the linked answer as it seemed like a good approach :)
Thanks
The type of expression I'd like to evaluate is given some user input such as
y = x^2 + 1/x - cos(x)
evaluate y for a range of values of x
It is straightforward to set up a Lua interpreter instance, and pass it expressions to be evaluated, getting back a function to call that evaluates the expression. You can even let the user have variables...
Here's the sample code I cooked up and edited into my other answer. It is probably better placed on a question tagged Lua in any case, so I'm adding it here as well. I compiled this and tried it for a few cases, but it certainly should not be trusted in production code without some attention to error handling and so forth. All the usual caveats apply here.
I compiled and tested this on Windows using Lua 5.1.4 from Lua for Windows. On other platforms, you'll have to find Lua from your usual source, or from www.lua.org.
Update: This sample uses simple and direct techniques to hide the full power and complexity of the Lua API behind as simple as possible an interface. It is probably useful as-is, but could be improved in a number of ways.
I would encourage readers to look into the much more production-ready ae library by lhf for code that takes advantage of the API to avoid some of the quick and dirty string manipulation I've used. His library also promotes the math library into the global name space so that the user can say sin(x) or 2 * pi without having to say math.sin and so forth.
Public interface to LE
Here is the file le.h:
/* Public API for the LE library.
*/
int le_init();
int le_loadexpr(char *expr, char **pmsg);
double le_eval(int cookie, char **pmsg);
void le_unref(int cookie);
void le_setvar(char *name, double value);
double le_getvar(char *name);
Sample code using LE
Here is the file t-le.c, demonstrating a simple use of this library. It takes its single command-line argument, loads it as an expression, and evaluates it with the global variable x changing from 0.0 to 1.0 in 11 steps:
#include <stdio.h>
#include "le.h"
int main(int argc, char **argv)
{
int cookie;
int i;
char *msg = NULL;
if (!le_init()) {
printf("can't init LE\n");
return 1;
}
if (argc<2) {
printf("Usage: t-le \"expression\"\n");
return 1;
}
cookie = le_loadexpr(argv[1], &msg);
if (msg) {
printf("can't load: %s\n", msg);
free(msg);
return 1;
}
printf(" x %s\n"
"------ --------\n", argv[1]);
for (i=0; i<11; ++i) {
double x = i/10.;
double y;
le_setvar("x",x);
y = le_eval(cookie, &msg);
if (msg) {
printf("can't eval: %s\n", msg);
free(msg);
return 1;
}
printf("%6.2f %.3f\n", x,y);
}
}
Here is some output from t-le:
E:...>t-le "math.sin(math.pi * x)" x math.sin(math.pi * x) ------ -------- 0.00 0.000 0.10 0.309 0.20 0.588 0.30 0.809 0.40 0.951 0.50 1.000 0.60 0.951 0.70 0.809 0.80 0.588 0.90 0.309 1.00 0.000 E:...>
Implementation of LE
Here is le.c, implementing the Lua Expression evaluator:
#include <lua.h>
#include <lauxlib.h>
#include <stdlib.h>
#include <string.h>
static lua_State *L = NULL;
/* Initialize the LE library by creating a Lua state.
*
* The new Lua interpreter state has the "usual" standard libraries
* open.
*/
int le_init()
{
L = luaL_newstate();
if (L)
luaL_openlibs(L);
return !!L;
}
/* Load an expression, returning a cookie that can be used later to
* select this expression for evaluation by le_eval(). Note that
* le_unref() must eventually be called to free the expression.
*
* The cookie is a lua_ref() reference to a function that evaluates the
* expression when called. Any variables in the expression are assumed
* to refer to the global environment, which is _G in the interpreter.
* A refinement might be to isolate the function envioronment from the
* globals.
*
* The implementation rewrites the expr as "return "..expr so that the
* anonymous function actually produced by lua_load() looks like:
*
* function() return expr end
*
*
* If there is an error and the pmsg parameter is non-NULL, the char *
* it points to is filled with an error message. The message is
* allocated by strdup() so the caller is responsible for freeing the
* storage.
*
* Returns a valid cookie or the constant LUA_NOREF (-2).
*/
int le_loadexpr(char *expr, char **pmsg)
{
int err;
char *buf;
if (!L) {
if (pmsg)
*pmsg = strdup("LE library not initialized");
return LUA_NOREF;
}
buf = malloc(strlen(expr)+8);
if (!buf) {
if (pmsg)
*pmsg = strdup("Insufficient memory");
return LUA_NOREF;
}
strcpy(buf, "return ");
strcat(buf, expr);
err = luaL_loadstring(L,buf);
free(buf);
if (err) {
if (pmsg)
*pmsg = strdup(lua_tostring(L,-1));
lua_pop(L,1);
return LUA_NOREF;
}
if (pmsg)
*pmsg = NULL;
return luaL_ref(L, LUA_REGISTRYINDEX);
}
/* Evaluate the loaded expression.
*
* If there is an error and the pmsg parameter is non-NULL, the char *
* it points to is filled with an error message. The message is
* allocated by strdup() so the caller is responsible for freeing the
* storage.
*
* Returns the result or 0 on error.
*/
double le_eval(int cookie, char **pmsg)
{
int err;
double ret;
if (!L) {
if (pmsg)
*pmsg = strdup("LE library not initialized");
return 0;
}
lua_rawgeti(L, LUA_REGISTRYINDEX, cookie);
err = lua_pcall(L,0,1,0);
if (err) {
if (pmsg)
*pmsg = strdup(lua_tostring(L,-1));
lua_pop(L,1);
return 0;
}
if (pmsg)
*pmsg = NULL;
ret = (double)lua_tonumber(L,-1);
lua_pop(L,1);
return ret;
}
/* Free the loaded expression.
*/
void le_unref(int cookie)
{
if (!L)
return;
luaL_unref(L, LUA_REGISTRYINDEX, cookie);
}
/* Set a variable for use in an expression.
*/
void le_setvar(char *name, double value)
{
if (!L)
return;
lua_pushnumber(L,value);
lua_setglobal(L,name);
}
/* Retrieve the current value of a variable.
*/
double le_getvar(char *name)
{
double ret;
if (!L)
return 0;
lua_getglobal(L,name);
ret = (double)lua_tonumber(L,-1);
lua_pop(L,1);
return ret;
}
Remarks
The above sample consists of 189 lines of code total, including a spattering of comments, blank lines, and the demonstration. Not bad for a quick function evaluator that knows how to evaluate reasonably arbitrary expressions of one variable, and has rich library of standard math functions at its beck and call.
You have a Turing-complete language underneath it all, and it would be an easy extension to allow the user to define complete functions as well as to evaluate simple expressions.
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