float / int隐式转换 [英] float/int implicit conversion

问题描述

我正在做乘法和除法 float s和 int s，我忘记了隐式转换规则而且问题中的单词对谷歌来说似乎太过于模糊了，而不是在这里问这个问题。）

如果我有两个 int s，但我想做浮点除法，我是否只需要施放一个或两个操作数？如果乘以一个 float 和一个 int ，那么答案是a 你不能分配给 int $ <$ c $> code>从 float 除以 int 的结果，反之亦然。

所以答案是：

lockquote

如果我有两个 int s，但我想做浮点除法...？

一次投射就足够了。

lockquote>

如果我乘以一个 float 和一个 int ，答案是 float ？

是的。

---

  float f = 1000f; 
 int i = 3; 
 
 f = i; // Ok 
 i = f; //错误
 
 f = i / f; // Ok 0.003 
 f = f / i; //好的333.3333（3）
 
 i = i / f; //错误
 i = f / i; //错误
  

I'm doing multiplication and division of floats and ints and I forget the implicit conversion rules (and the words in the question seem too vague to google more quickly than asking here).

If I have two ints, but I want to do floating-point division, do I need only to cast one or both of the operands? How about for multiplication — if I multiply a float and an int, is the answer a float?

解决方案

You can’t assign to an int result from division of a float by an int or vice-versa.

So the answers are:

If I have two ints, but I want to do floating point division…?

One cast is enough.

If I multiply a float and an int, is the answer a float?

Yes it is.

---

float f = 1000f;
int i = 3; 

f = i; // Ok
i = f; // Error

f = i/f; //Ok 0.003
f = f/i; //Ok 333.3333(3)

i = i/f; //Error
i = f/i; //Error

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