使用C语言计算中缀表达式的最简单方法是什么? [英] What is easiest way to calculate an infix expression using C language?

问题描述

假设用户以字符串形式输入中缀表达式?什么是使用C语言评估该表达式结果的最简单的方法(我最简单的意思是简称 t)?

Suppose the user inputs an infix expression as a string? What could be the easiest ( By easiest I mean the shortest) way to evaluate the result of that expression using C language?

可能的方法是通过使用堆栈将其转换为后缀.但这是一个相当漫长的过程.有什么方法可以使用诸如 atoi() 或 eval() 之类的功能，从而使工作更轻松?

Probable ways are converting it to a postfix then by using stacks.But its rather a long process. Is there any way of using functions such as atoi() or eval() that could make the job easier?

推荐答案

C没有内置的"eval"功能，但是有提供此功能的库.

C doesn't have an "eval" function built-in, but there are libraries that provide it.

我强烈建议使用 TinyExpr .它是免费的开源C代码，可通过字符串实现数学评估.TinyExpr只是1 C文件，大约500行代码.我认为您不会找到一种更短或更简单的方法来真正完成(而不仅仅是玩具示例).

I would highly recommend using TinyExpr. It's free and open-source C code that implements math evaluation from a string. TinyExpr is only 1 C file, and it's about 500 lines of code. I don't think you'll find a shorter or easier way that is actually complete (and not just a toy example).

这里是使用它的完整示例，应该说明它的简单程度:

Here is a complete example of using it, which should demostrate how easy it is:

#include "tinyexpr.h"
#include <stdio.h>

int main(int argc, char *argv[])
{
    printf("%f\n", te_interp("5 * 5", 0)); //Prints 25
    return 0;
}

如果您想自己构建一个表达式求解器，我建议您查看 TinyExpr源代码作为起点.非常干净而且易于遵循.

If you want to build an expression solver yourself, I would recommend looking at the TinyExpr source-code as a starting point. It's pretty clean and easy to follow.

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