blocks|key|941928|text|在C%2B%2B中，运算符(用于POD类型)始终作用于相同类型的对象。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|941929|因此，如果它们不相同，则一个将被提升以匹配另一个。|941930|运算结果的类型与操作数相同(转换后)。|941931|if:
either+is++++++long+double+++++++other+is+promoted+>++++++long+double
either+is+++++++++++double+++++++other+is+promoted+>+++++++++++double
either+is+++++++++++float++++++++other+is+promoted+>+++++++++++float
either+is+long+long+unsigned+int+other+is+promoted+>+long+long+unsigned+int
either+is+long+long++++++++++int+other+is+promoted+>+long+long++++++++++int
either+is+long++++++unsigned+int+other+is+promoted+>+long++++++unsigned+int
either+is+long+++++++++++++++int+other+is+promoted+>+long+++++++++++++++int
either+is+++++++++++unsigned+int+other+is+promoted+>+++++++++++unsigned+int
either+is++++++++++++++++++++int+other+is+promoted+>++++++++++++++++++++int

Otherwise:
both+operands+are+promoted+to+int|code-block|syntax|javascript|941932|请注意。操作的最小大小是int。因此，在操作完成之前，short/char会被提升为int。|offset|length|style|CODE|941933|在您的所有表达式中，在执行操作之前，int都被提升为float。操作的结果是一个float。|941934|int+%2B+float+=>++float+%2B+float+=+float
int+*+float+=>++float+*+float+=+float
float+*+int+=>++float+*+float+=+float
int+/+float+=>++float+/+float+=+float
float+/+int+=>++float+/+float+=+float
int+/+int+++++++++++++++++++++=+int
int+%5E+float+=>++<compiler+error>|941935|entityMap^0|0|0|0|0|C|3|R|5|X|4|16|3|0|I|3|Q|5|14|5|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|X|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|Y|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|Z|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|10|8|@$M|11|N|12|O|P]|$M|13|N|14|O|P]|$M|15|N|16|O|P]|$M|17|N|18|O|P]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|19|8|@$M|1A|N|1B|O|P]|$M|1C|N|1D|O|P]|$M|1E|N|1F|O|P]]|9|@]|A|$]]|$1|S|3|T|5|H|7|1G|8|@]|9|@]|A|$I|J]]|$1|U|3|-4|5|6|7|1H|8|@]|9|@]|A|$]]]|V|$]]

In C++ operators (for POD types) always act on objects of the same type. 
Thus if they are not the same one will be promoted to match the other. 
The type of the result of the operation is the same as operands (after conversion).
<pre><code>if:
either is long double other is promoted &gt; long double
either is double other is promoted &gt; double
either is float other is promoted &gt; float
either is long long unsigned int other is promoted &gt; long long unsigned int
either is long long int other is promoted &gt; long long int
either is long unsigned int other is promoted &gt; long unsigned int
either is long int other is promoted &gt; long int
either is unsigned int other is promoted &gt; unsigned int
either is int other is promoted &gt; int

Otherwise:
both operands are promoted to int
</code></pre>
Note. The minimum size of operations is <code>int</code>. So <code>short</code>/<code>char</code> are promoted to <code>int</code> before the operation is done.
In all your expressions the <code>int</code> is promoted to a <code>float</code> before the operation is performed. The result of the operation is a <code>float</code>.
<pre><code>int + float =&gt; float + float = float
int * float =&gt; float * float = float
float * int =&gt; float * float = float
int / float =&gt; float / float = float
float / int =&gt; float / float = float
int / int = int
int ^ float =&gt; &lt;compiler error&gt;
</code></pre>

blocks|key|725908|text|如果不包括无符号类型，则会有一个有序的层次结构:符号字符、短整型、整型、长整型、长整型、浮点型、双精度、长整型双精度。首先，上述代码中在int之前的任何内容都将被转换为int。然后，在二元运算中，排名较低的类型将被转换为较高的类型，并且结果将是较高的类型。(您会注意到，在层次结构中，每当涉及浮点和整型时，整型都会转换为浮点型。)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|725909|Unsigned使事情变得有点复杂:它扰乱了排名，并且排名的一部分变成了实现定义的。因此，最好不要在同一个表达式中混合使用有符号和无符号。(大多数C%2B%2B专家似乎都避免使用unsigned，除非涉及到按位操作。至少，这是Stroustrup推荐的。)|725910|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

If you exclude the unsigned types, there is an ordered
hierarchy: signed char, short, int, long, long long, float,
double, long double. First, anything coming before int in the
above will be converted to int. Then, in a binary operation,
the lower ranked type will be converted to the higher, and the
results will be the type of the higher. (You'll note that, from
the hierarchy, anytime a floating point and an integral type are
involved, the integral type will be converted to the floating
point type.)

Unsigned complicates things a bit: it perturbs the ranking, and
parts of the ranking become implementation defined. Because of
this, it's best to not mix signed and unsigned in the same
expression. (Most C++ experts seem to avoid unsigned unless
bitwise operations are involved. That is, at least, what
Stroustrup recommends.)

blocks|key|942033|text|警告！|type|unstyled|depth|inlineStyleRanges|entityRanges|data|942034|转换是从左到右进行的。|942035|试试这个：|942036|int+i+=+3,+j+=+2;
double+k+=+33;
cout+<<+k+*+j+/+i+<<+endl;+//+prints+22
cout+<<+j+/+i+*+k+<<+endl;+//+prints+0|code-block|syntax|javascript|942037|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|O|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|P|8|@]|9|@]|A|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Caveat!

The conversions occur from left to right.

Try this:

<pre><code>int i = 3, j = 2;
double k = 33;
cout &lt;&lt; k * j / i &lt;&lt; endl; // prints 22
cout &lt;&lt; j / i * k &lt;&lt; endl; // prints 0
</code></pre>

I want to be better about knowing when I should cast. What are the implicit type conversion rules in C++ when adding, multiplying, etc. For example,

<pre><code>int + float = ?
int * float = ?
float * int = ?
int / float = ?
float / int = ?
int / int = ?
int ^ float = ?
</code></pre>

et cetera...

Will the expression always be evaluated as the more precise type? Do the rules differ for Java? 
Please correct me if I have worded this question inaccurately.

Implicit type conversion rules in C++ operators

Java

我想更好地知道我应该在什么时候选角。在进行加法、乘法等操作时，C++中的隐式类型转换规则是什么。例如，int + float = ?int * float = ?float * int = ?int / float = ?float / int = ?int / int = ?int ^ float = ?等等..。表...

问C++运算符中的隐式类型转换规则
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