blocks|key|2424167|text|您需要解析字符串。C中没有eval()+(与大多数静态语言一样)，所以您需要编写自己的解析器或找到一些库来帮助。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2424168|由于最容易使用的解析器是C%2B%2B而不是C，所以我宁愿使用完全可嵌入的语言。我最喜欢的是Lua，如果不包括库，它可以非常轻量级。而且，它的语法比C的好，所以您的用户可能更喜欢它。|2424169|当然，Lua是一种成熟的编程语言，所以它可能不合适，或者它可以在其他方面有所帮助(使扩展应用程序变得更容易)。|2424170|entityMap|0|LINK|mutability|MUTABLE|url|http://lua.org/^0|D|6|0|16|3|0|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@$9|S|A|T|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|U|8|@]|D|@$9|V|A|W|1|X]]|E|$]]|$1|H|3|I|5|6|7|Y|8|@]|D|@]|E|$]]|$1|J|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|K|$L|$5|M|N|O|E|$P|Q]]]]

you need to parse the string. there's no <code>eval()</code> in C (as in most static languages), so you need to either write your own parser or find some library to help.

since most easy to use parsers are for C++ and not C, i'd rather use a full embeddable language. my absolute favorite is <a href="http://lua.org" rel="nofollow noreferrer">Lua</a>, which can be incredibly lightweight if you don't include the libraries. also, the syntax is nicer than C's, so your users might like it better.

of course, Lua is a full-blown programming language, so it might not be appropriate, or maybe it could help in other ways (to make it easier to extend your application).

blocks|key|980771|text|一种干净(可能不是短)的方法是构建一棵树，就像编译器一样。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|980772|例如，假设您有一个表达式"2%2B3“。“%2B”就是头。'2‘将是左子，'3’将是正确的孩子。|980773|因为每个表达式的计算结果都是一个值，所以这个树可以扩展为无限复杂的表达式:它只需要按照每个操作符的优先级排序。低优先级运算符(如'%2B‘位于顶部，而高优先级运算符(如'*')位于底部。然后，从下向上计算树上的表达式。|980774|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|I|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|J|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|G|$]]

One clean (possible not short) way to do it is to build a tree, like a compiler would.

For example, say you have the expression "2+3". The '+' would be the head. The '2' would be the left child and the '3' would be the right child. 

Since each expression evaluates to a value, this tree can be extended for infinitely complex expressions: it just needs to be sorted in order of precedence for each operator. Low precedence operators (like '+' go at the top, while high-precedence operators (like '*') go at the bottom. You would then evaluate the expressions on the tree from the bottom up.

blocks|key|898318|text|您需要构建一些脚本语言的解释器。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|898319|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

You need to build in interpreter of some scripting language.

blocks|key|898328|text|将字符串转换为作为操作数和运算符的令牌数组。将infix令牌数组转换为反向波兰符号数组。在方程在RPN中之后，您可以从堆栈中弹出令牌并对它们进行操作。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|898329|看看维基百科在反向波兰表示法上的文章。文中给出了如何进行转换和计算。|offset|length|898330|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Reverse_Polish_notation^0|0|7|7|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|O|8|@]|9|@$D|P|E|Q|1|R]]|A|$]]|$1|F|3|-4|5|6|7|S|8|@]|9|@]|A|$]]]|G|$H|$5|I|J|K|A|$L|M]]]]

Convert the string into an array of tokens which are the operands and operators.
Convert the infix token array to a Reverse Polish Notation array.
After the equation is in RPN, then you can pop tokens off the stack and operate on them.

Take a look at the Wikipedia article on <a href="http://en.wikipedia.org/wiki/Reverse_Polish_notation" rel="nofollow noreferrer">Reverse Polish Notation</a>. It shows how to do the conversion and the calculation.

blocks|key|898340|text|当然，最有教育意义的方法(甚至可能是最简单的方法，一旦您知道了)就是学习如何编写您自己的递归下降分析器。C中infix表达式的解析器不是很长。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|898341|这里有一个的一些优秀的博客文章由excellent对解析。(这是与你最相关的一个，但我强烈推荐所有这些。)它包含infix表达式解析器的源代码--诚然是用Python，而不是C，但是转换应该相当简单，在这个过程中您将学到很多东西。|898342|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Recursive_descent_parser|1|http://eli.thegreenplace.net/2009/03/14/some-problems-of-recursive-descent-parsers/^0|18|7|0|0|0|5|1|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@$A|Q|B|R|1|S]]|C|$]]|$1|D|3|E|5|6|7|T|8|@]|9|@$A|U|B|V|1|W]]|C|$]]|$1|F|3|-4|5|6|7|X|8|@]|9|@]|C|$]]]|G|$H|$5|I|J|K|C|$L|M]]|N|$5|I|J|K|C|$L|O]]]]

Certainly the most instructive way (and possibly even the easiest, once you know how) is to learn how to write your own <a href="http://en.wikipedia.org/wiki/Recursive_descent_parser" rel="noreferrer">recursive descent parser</a>. A parser for infix expressions in C isn't very long.

<a href="http://eli.thegreenplace.net/2009/03/14/some-problems-of-recursive-descent-parsers/" rel="noreferrer">Here's one</a> of a number of excellent blog posts by Eli Bendersky on parsing. (This one is the one that's most relevant to you, but I highly recommend all of them.) It contains source code for an infix expression parser -- admittedly in Python, not C, but the conversion should be fairly straightforward, and you'll learn a lot in the process.

blocks|key|898436|text|C没有内置的"eval“函数，但也有提供它的库。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|898437|我强烈建议使用TinyExpr。它是免费的、开源的C代码，可以从字符串中实现数学计算。TinyExpr只是一个C文件，大约有500行代码。我不认为你会找到一个更短或更容易的方式，实际上是完整的(而不仅仅是一个玩具的例子)。|offset|length|898438|下面是一个使用它的完整示例，它应该演示它有多容易：|898439|#include+"tinyexpr.h"
#include+<stdio.h>

int+main(int+argc,+char+*argv[])
{
++++printf("%25f\n",+te_interp("5+*+5",+0));+//Prints+25
++++return+0;
}|code-block|syntax|javascript|898440|如果您想自己构建表达式求解器，我建议您将TinyExpr源代码作为一个起点。它很干净，很容易被跟踪。|898441|entityMap|0|LINK|mutability|MUTABLE|url|https://codeplea.com/tinyexpr|1|https://github.com/codeplea/tinyexpr^0|0|7|8|0|0|0|0|K|B|1|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Z|8|@]|9|@$D|10|E|11|1|12]]|A|$]]|$1|F|3|G|5|6|7|13|8|@]|9|@]|A|$]]|$1|H|3|I|5|J|7|14|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|15|8|@]|9|@$D|16|E|17|1|18]]|A|$]]|$1|O|3|-4|5|6|7|19|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]|W|$5|R|S|T|A|$U|X]]]]

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

I would highly recommend using <a href="https://codeplea.com/tinyexpr" rel="noreferrer">TinyExpr</a>. 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).

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

<pre><code>#include "tinyexpr.h"
#include &lt;stdio.h&gt;

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

If you want to build an expression solver yourself, I would recommend looking at the <a href="https://github.com/codeplea/tinyexpr" rel="noreferrer">TinyExpr source-code</a> as a starting point. It's pretty clean and easy to follow.

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?

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?

What is easiest way to calculate an infix expression using C language?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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假设用户输入一个infix表达式为字符串？使用C语言计算表达式结果的最简单方法(，我指的是简写的t)是什么？可能的方法是将其转换为后缀，然后使用stacks.But，这是一个相当长的过程。是否有任何方法可以使用诸如、atoi()、或eval()之类的函数来简化工作？

问使用C语言计算infix表达式最简单的方法是什么？
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问使用C语言计算infix表达式最简单的方法是什么？
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