blocks|key|4837654|text|还有更多的运算符。您可以找到C%2B%2B和C的排序表。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|4837655|但是的，你会发现它尊重这一点。(虽然我不确定这是不是你所说的…)|4837656|entityMap|0|LINK|mutability|MUTABLE|url|http://www.cppreference.com/wiki/operator_precedence|1|http://www.difranco.net/compsci/C_Operator_Precedence_Table.htm^0|E|3|0|I|1|1|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@$A|Q|B|R|1|S]|$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|6|7|W|8|@]|9|@]|C|$]]|$1|F|3|-4|5|6|7|X|8|@]|9|@]|C|$]]]|G|$H|$5|I|J|K|C|$L|M]]|N|$5|I|J|K|C|$L|O]]]]

There are far more operators than that. You can find a precedence tables for <a href="http://www.cppreference.com/wiki/operator_precedence" rel="nofollow">C++</a> and <a href="http://www.difranco.net/compsci/C_Operator_Precedence_Table.htm" rel="nofollow">C</a>.

But yes, you'll find it respects that. (Though I'm not sure it's exactly what you've said...)

blocks|key|1230053|text|这个问题有两个答案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1230054|其一是C%2B%2B确实遵循标准的数学优先规则，您称之为BODMAS。您可以看到C%2B%2B关联其所有运算符here的顺序。|offset|length|1230055|但是，如果操作中涉及的任何表达式都有副作用，则不能保证C%2B%2B会按照标准的数学顺序对它们进行计算。然而，这是一个比较高级的话题。|1230056|entityMap|0|LINK|mutability|MUTABLE|url|http://www.cppreference.com/wiki/operator_precedence^0|0|1B|4|0|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Q|8|@]|9|@$D|R|E|S|1|T]]|A|$]]|$1|F|3|G|5|6|7|U|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|I|$J|$5|K|L|M|A|$N|O]]]]

There are two answers to this question.

One is that C++ does follow standard mathematical precedence rules, which you refer to as BODMAS. You can see the order in which C++ associates all its operators <a href="http://www.cppreference.com/wiki/operator_precedence" rel="nofollow">here</a>.

However, if any of the expressions involved in the operation have side effects, then C++ is not guaranteed to evaluate them in what one might consider to be standard mathematical order. That's sort of an advanced topic, however.

blocks|key|1232564|text|其他人为您提供了运算符优先列表的链接。这些都很好。但是，如果您需要查看运算符优先表来确定您的代码告诉计算机要做什么，请同情您代码的维护者(包括未来的您)，只需使用括号。从长远来看，这会让你的意图更加清晰，并节省时间和心痛。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1232565|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|$]]

Other people have given you links to operator precedence lists. These are well and good. However, if you need to look at an operator precedence table to determine what your code tells computers to do, please take pity on your code's maintainers (including future you) and just use parentheses. It makes your intention much clearer and saves time and heartache in the long run.

blocks|key|1232548|text|C%2B%2B：http://msdn.microsoft.com/en-us/library/126fe14k.aspx|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1232549|C#：http://msdn.microsoft.com/en-us/library/aa691323(VS.71).aspx|1232550|entityMap|0|LINK|mutability|MUTABLE|url|http://msdn.microsoft.com/en-us/library/126fe14k.aspx|1|http://msdn.microsoft.com/en-us/library/aa691323(VS.71).aspx^0|4|1H|0|0|3|1O|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]]]]

C++: <a href="http://msdn.microsoft.com/en-us/library/126fe14k.aspx" rel="nofollow">http://msdn.microsoft.com/en-us/library/126fe14k.aspx</a>

C#: <a href="http://msdn.microsoft.com/en-us/library/aa691323(VS.71).aspx" rel="nofollow">http://msdn.microsoft.com/en-us/library/aa691323(VS.71).aspx</a>

blocks|key|3426326|text|C在算术表达式中以由以下运算符优先规则确定的精确顺序应用运算符，这些规则通常与代数中的规则相同：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3426327|首先计算括号对中包含的表达式中的运算符。据说括号处于“最高优先级”。在使用嵌套或嵌入括号的情况下，例如((a%2Bb)%2Bc)|3426328|首先应用最里面的一对括号中的运算符。|3426329|接下来应用乘法、除法和余数运算。如果表达式包含多个乘法、除法和余数运算，则计算将从左到右进行。乘法、除法和余数被认为处于相同的优先级。接下来计算加法和减法运算。如果表达式包含多个加法和减法运算，则计算将从左到右进行。加法和减法也具有相同的优先级，其优先级低于乘法、除法和余数运算的优先级。最后计算赋值运算符(=)。|3426330|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|L|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|M|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|I|$]]

C applies the operators in arithmetic expressions in a precise sequence determined by the following rules of operator precedence, which are generally the same as those in algebra:

Operators in expressions contained within pairs of parentheses are evaluated first. Parentheses are said to be at the “highest level of precedence.” In cases of nested, or embedded, parentheses, such as
 ( ( a + b ) + c ) 
the operators in the innermost pair of parentheses are applied first.

Multiplication, division and remainder operations are applied next. If an expression contains several multiplication, division and remainder operations, evaluation proceeds from left to right. Multiplication, division and remainder are said to be on the same level of precedence.
Addition and subtraction operations are evaluated next. If an expression contains several addition and subtraction operations, evaluation proceeds from left to right. Addition and subtraction also have the same level of precedence, which is lower than the precedence of the multiplication, division and remainder operations.
The assignment operator (=) is evaluated last.

blocks|key|3426360|text|我发现了一个不跟在“BODMAS”后面的表达式。这是我的c程序供你参考。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3426361|#include+<stdio.h>
int+main()+{
++int+a+=+6;
++int+b+=+4;
++int+c+=+2;
++int+result;
++result+=+a+-+b+%2B+c;+//+4
++printf("%25d+\n",+result);

++result+=+a+%2B+b+/+c;+//+8
++printf("%25d+\n",+result);

++result+=+(a+%2B+b)+/+c;+//+5
++printf("%25d+\n",+result);

++return+0;
}|code-block|syntax|javascript|3426362|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I found an expression that doesn’t follow “BODMAS”. Here is my c program for your reference 

<pre><code>#include &lt;stdio.h&gt;
int main() {
 int a = 6;
 int b = 4;
 int c = 2;
 int result;
 result = a - b + c; // 4
 printf("%d \n", result);

 result = a + b / c; // 8
 printf("%d \n", result);

 result = (a + b) / c; // 5
 printf("%d \n", result);

 return 0;
}
</code></pre>

When a expression is evaluated in C/C++, does it follow BODMAS [Bracket open Division Multiply Addition Substraction] rule? If not then how they are evaluated?

EDIT: More clearly, If the following expression is evaluated according to BODMAS rule,

(5 + 3)/8*9

First what is in brackets is processed.

8/8*9.

Then Division is done.

1*9

And then multiplication and so on.

Expression evaluation in C/C++ doesnt follow BODMAS rule?

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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当一个表达式在C/C++中求值时，它是否遵循BODMAS方括号中的开除乘加减法规则？如果不是，那么如何对它们进行评估？编辑:更清楚地说，如果根据BODMAS规则计算以下表达式，(5 + 3)/8*9首先，处理括号中的内容。8/8*9。然后除法就完成了。1*9然后是乘法，诸如此类。

问C/C++中的表达式计算不遵循BODMAS规则？
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回答 6

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问C/C++中的表达式计算不遵循BODMAS规则？
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