问题
In my 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
回答1:
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.
回答2:
Since you're lazy, like most programmers, here's a link to a simple example that you can use to parse some arbitrary code using Lua. From there, it should be simple to create your expression parser.
回答3:
This is for Lua users that are looking for a Lua equivalent of "eval".
The magic word used to be loadstring but it is now, since Lua 5.2, an upgraded version of load.
i=0
f = load("i = i + 1") -- f is a function
f() ; print(i) -- will produce 1
f() ; print(i) -- will produce 2
Another example, that delivers a value :
f=load('return 2+3')
print(f()) -- print 5
As a quick-and-dirty way to do, you can consider the following equivalent of eval(s), where s is a string to evaluate :
load(s)()
As always, eval mechanisms should be avoided when possible since they are expensive and produce a code difficult to read. I personally use this mechanism with LuaTex/LuaLatex to make math operations in Latex.
回答4:
The Lua documentation contains a section titled The Application Programming Interface which describes how to call Lua from your C program. The documentation for Lua is very good and you may even be able to find an example of what you want to do in there.
It's a big world in there, so whether you choose your own parsing solution or an embeddable interpreter like Lua, you're going to have some work to do!
回答5:
function calc(operation)
return load("return " .. operation)()
end
来源：https://stackoverflow.com/questions/1156572/evaluating-mathematical-expressions-using-lua