将int隐式转换为double

问题描述：

所以，我有点困惑。我认为这应该可行：
在这种情况下， a 和 b 都是整数（准确地说是计数器）。

So, I'm a tad confused. I was under the impression that this should work: In this case, both a and b are ints (Counters to be exact).

由于 a / b 的结果可能包含小数位，因此整数显然不起作用。
因此，我对一个新的double进行了延迟，并在其中执行了如下计算：

As the result of a / b may possibly contain decimal places, ints obviously won't work. Therefore, I delared a new double and performed the calculation inside it like this:

double texturefactor = ((a / b) * 10);

这不符合我的预期， a 的结果/ b 始终是我使用int进行计算以存储结果时得到的结果。
另一方面，它起作用：

This doesn't work as I expected, and the result of a / b is always that which I would get if I performed the calculation using an int to store the results. On the other hand, this works:

double calculate1 = a;
double calculate2 = b;
double texturefactor = ((calculate1 / calculate2) * 10);

也许是愚蠢的问题-
1.我确定这应该有效-我知道在某些情况下VS会抱怨我试图将一种类型隐式转换为另一种-这就是我要尝试的方法！为什么不呢，我错过了什么吗？ :)
2.我应该只将计数器a和b转换为双精度并为自己省去转换的麻烦吗？还是

Couple of perhaps stupid questions- 1. I'm sure this ought to work- I know that in certain situations VS will complain that I've tried to implicitly convert from one type to another- That's what I'm trying to do! Why doesn't it, and have I missed something? :) 2. Should I just convert the counters a and b to doubles and save myself the trouble of the conversion, or is that trouble?

答

从内部到外部对表达式（包括子表达式的类型）进行求值：

The expression, including the types of subexpressions, is evaluated from inside to outside:

double texturefactor = ((a / b) * 10);

当编译器分析 a / b 时，它不知道结果以后会转换为 double ，因此它只是将计算编译为整数除法。

When the compiler analyses a / b, it has no idea that the result will later on be converted to double, so it just compiles the computation as an integer division.

显式地将两个操作数之一强制转换为 double ，足以避免这种混乱：

Explicitly casting one of the two operands to double right there is enough to avoid that confusion:

double texturefactor = (((double)a / b) * 10);

将int隐式转换为double

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