C++ 一个 double 是否会将等式中的每个 int 提升为两倍？

Does one double promote every int in the equation to double?

提问by Alan Turing

Does the presence of one floating-point data type (e.g. double) ensure that all +, -, *, /, %, etc math operations assume double operands?

一种浮点数据类型（例如double）的存在是否确保所有 +、-、*、/、% 等数学运算都采用双操作数？

If the story is more complicated than that, is there a resource that describes these rules? Should I not ask such questions and always explicitly cast intto doublewhen the result of the equation is double. Here are some equations I'm thinking about. I purposefully did not compile and run then on my system, since this is the type of thing that could be compiler dependent.

如果故事比这更复杂，是否有描述这些规则的资源？我是否应该不问这样的问题并始终明确地强制int转换double为等式的结果是double. 这是我正在考虑的一些方程式。我故意没有在我的系统上编译和运行，因为这是可能依赖于编译器的类型。

int a(1), b(2), c(3);
double d(4.);
double result1 = a + b/d + c; // equal to 4 or to 4.5?
double result2 = (a + b)/d + c; // equal to 3 or to 3.75?    
double result3 = a/b + d; // equal to 4 or to 4.5?

回答by Slix

I purposefully did not compile and run then on my system, since this is the type of thing that could be compiler dependent.

我故意没有在我的系统上编译和运行，因为这是可能依赖于编译器的类型。

This is notcompiler dependent. C++ clearly defines the order of these operationsand how they are converted.

这与编译器无关。C++ 清楚地定义了这些操作的顺序以及它们是如何转换的。

How the conversion happens is dependent on the order of operations.

转换如何发生取决于操作顺序。

double result1 = a + b / d + c; // equal to 4 or to 4.5?

In this example, the division happens first. Because this is an int divided by a double, the compiler handles this by converting the int into a double. Thus, the result of b / dis a double.

在这个例子中，除法首先发生。因为这是一个 int 除以 double，编译器通过将 int 转换为 double 来处理这个问题。因此，结果b / d是双倍。

The next thing that C++ does is add ato the result of b / d. This is an int added to a double, so it converts the int to a double and adds, resulting in a double. The same thing happens with c.

C++ 做的下一件事是添加a到b / d. 这是一个加到双精度的整数，因此它将整数转换为双精度并相加，从而产生双精度。同样的事情发生在c.

double result3 = a / b + d; // equal to 4 or to 4.5?

In this example, division is handled first. aand bare both ints, so no conversion is done. The result of a / bis of type int and is 0.

在本例中，首先处理除法。a并且b都是整数，所以没有进行转换。的结果a / b是 int 类型且为 0。

Then, the result of this is added to d. This is an int plus a double, so C++ converts the int to a double, and the result is a double.

然后，将其结果添加到d. 这是一个int加一个double，所以C++把int转换成double，结果是double。

Even though a double is present in this expression, a / bis evaluated first, and the double means nothing until execution reaches the double. Therefore, integer division occurs.

即使此表达式中存在双精度值，a / b也会先求值，并且双精度值在执行达到双精度值之前没有任何意义。因此，发生整数除法。

I find promotion and conversion rules pretty complex. Usually integer-like numbers (short, int, long) are promoted to floating-point equivalents (float, double). But things are complicated by size differences and sign.

我发现促销和转换规则非常复杂。通常类似整数的数字（short、int、long）被提升为等价的浮点数（float、double）。但事情因尺寸差异和符号而变得复杂。

See this questionfor specifics about conversion.

有关转换的详细信息，请参阅此问题。

回答by Alan Turing

Does one doublepromote every intin the equation to double?

是否将等式中的double每一个都提升int为double？

No. Only the result of a single operation (with respect to precedence).

不。只有单个操作的结果（相对于优先级）。

double result1 = a + b/d + c; // equal to 4 or to 4.5?

4.5.

4.5.

double result2 = (a + b)/d + c; // equal to 3 or to 3.75?

3.75.

3.75。

double result3 = a/b + d; // equal to 4 or to 4.5?

4.

4.

回答by Ramy Al Zuhouri

You must consider the precedence of every operator, you must think like a parser:

您必须考虑每个运算符的优先级，您必须像解析器一样思考：

double result1 = a + b/d + c; // equal to 4 or to 4.5?

That's like a + (b/d) +c because the '/' operator has the biggest precedence.Then it doesn't matter what of these 2 operations is made for first, because the floating point operand is in the middle, and it "infects" other operands and make them be double.So it's 4.5.

那就像a + (b/d) +c 因为'/'运算符的优先级最高。那么这2个操作中的哪个先做就无所谓了，因为浮点操作数在中间，它“感染”其他操作数并使它们成为双倍。所以它是4.5。

double result2 = (a + b)/d + c; // equal to 3 or to 3.75?  

Same here, it's like ((a+b)/d )+c, so a+b is 3, that 3 becomes a floating point number because gets promoted to double, because is the dividend of d, which is a double, so it's 0.75+3, that is 3.75.

这里也一样，就像 ((a+b)/d )+c，所以 a+b 是 3，那个 3 变成浮点数，因为被提升为 double，因为是 d 的被除数，它是一个 double，所以它是 0.75+3，即 3.75。

double result3 = a/b + d; // equal to 4 or to 4.5?

It's like (a/b)+d, so a/b is zero and d is 4, so it's 4. A parser makes all the operations in order of precedence, so you can exactly know what will be the result of the expression.

这就像 (a/b)+d，所以 a/b 是 0，d 是 4，所以它是 4。解析器按优先级顺序进行所有操作，因此您可以确切地知道表达式的结果是什么。

回答by Eric Postpischil

Generally, if one operand of a binary operator is floating point and the other is integer, the integer is converted to floating point, and the result is floating point.

通常，如果二元运算符的一个操作数是浮点数，另一个是整数，则将整数转换为浮点数，结果为浮点数。

In a compound expression, with multiple subexpressions, each operator is processed individually, using the precedence rules you probably know. Thus, in a*b + c*d, a*bis evaluated, and c*dis evaluated, and the results are added together. Whatever is in c*dhas no effect in a*band vice-versa.

在具有多个子表达式的复合表达式中，使用您可能知道的优先级规则单独处理每个运算符。因此，在a*b + c*d，a*b被评估，并且c*d被评估，且结果被加在一起。任何内容c*d都没有影响，a*b反之亦然。

C++ is complicated, of course, and user-defined operators may have other behaviors.

当然，C++ 很复杂，用户定义的运算符可能有其他行为。

The authoritative resource that defines the rules is the C++ standard. The standard is quite large and technical. You might prefer to examine the C standard first. See this answerfor links to the standards. Any good book on C or C++ should describe the default type conversions and expression evaluation.

定义规则的权威资源是 C++ 标准。该标准是相当大的和技术性的。您可能更愿意先检查 C 标准。有关标准的链接，请参阅此答案。任何关于 C 或 C++ 的好书都应该描述默认类型转换和表达式计算。