将整数除法存储在浮点值C ++中 [英] Store division of integers in float value C++

问题描述

我只是想问一下，如果我不像这样将浮点数存储在浮点数中时，我不强制转换浮点数，会发生数字明智的事情:

I just wanted to ask what happens number wise if i do not typecast integers to float when storing in a float variable like this:

int32 IntVar1 = 100
int32 IntVar2 = 200
float FloatVar = IntVar1/IntVar2;

目前我正在这样做:

int32 IntVar1 = 100
int32 IntVar2 = 200
float FloatVar = float(IntVar1)/float(IntVar2);

但是在我拥有的代码数量上，这看起来确实很困难.我曾考虑过将int变量更改为float，但是我想那会影响性能.而且由于整数值不应包含任何小数，因此感觉像是完全浪费.

But in the amount of code i have, this looks really retarded. I thought about changing my int variables to float, but i guess that would be a performance hit. And since the integer values are not supposed to hold any decimals, it feels like a complete waste.

所以我想知道，选项1是否有可能起作用?还是我必须进行类型转换或将变量更改为浮点数?(所有类型转换几乎都使代码不可读)

So i wonder, are there any way that option 1 could be working? Or do i have to typecast OR change variables to float? (All typecasting pretty much makes the code unreadable)

推荐答案

看到功能的魔力:

float div(int x, int y)
{
    return float(x) / float(y);
}

现在您可以说:

int32 IntVar1 = 100
int32 IntVar2 = 200
float FloatVar = div(IntVar1, IntVar2);

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