blocks|key|952238|text|他做整数除法，这意味着3/4=+0。将其中一个括号转换为浮动|type|unstyled|depth|inlineStyleRanges|entityRanges|data|952239|+(float)(a.y+-+b.y)+/+(a.x+-+b.x);|code-block|syntax|javascript|952240|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

he does an integer divide, which means 3 / 4 = 0. cast one of the brackets to float 

<pre><code> (float)(a.y - b.y) / (a.x - b.x);
</code></pre>

blocks|key|736158|text|如果(a.y+-+b.y)小于(a.x+-+b.x)，则m始终为零。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|736159|所以就像这样铸造它。|736160|float+m+=+((float)(a.y+-+b.y))+/+((float)(a.x+-+b.x));|code-block|syntax|javascript|736161|entityMap^0|S|1|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|R|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|S|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|N|$]]

if (a.y - b.y) is less than (a.x - b.x), <code>m</code> is always zero. 

so cast it like this.

<pre><code>float m = ((float)(a.y - b.y)) / ((float)(a.x - b.x));
</code></pre>

blocks|key|952346|text|因为(a.y+-+b.y)可能小于(a.x+-+b.x)，所以在您的代码中，转换是在除法运算之后完成的，所以结果是一个整数，所以是0。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|952347|应该在/操作之前强制转换为float|952348|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|$]]

Because (a.y - b.y) is probably less then (a.x - b.x) and in your code the casting is done after the divide operation so the result is an integer so 0.

You should cast to float before the / operation

blocks|key|736294|text|您需要使用cast。我看到了其他答案，它们确实有效，但由于标签是C%2B%2B，我建议您使用static_cast|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|736295|float+m+=+static_cast<+float+>(+a.y+-+b.y+)+/+static_cast<+float+>(+a.x+-+b.x+);|code-block|syntax|javascript|736296|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/1255015/2628257^0|W|3|16|B|16|B|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@$9|X|A|Y|1|Z]]|E|$]]|$1|F|3|G|5|H|7|10|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

You need to use cast. I see the other answers, and they will really work, but as the tag is <code>C++</code> I'd suggest you to use <a href="https://stackoverflow.com/a/1255015/2628257"><code>static_cast</code></a>:

<pre><code>float m = static_cast&lt float &gt( a.y - b.y ) / static_cast&lt float &gt( a.x - b.x );</code></pre>

blocks|key|952477|text|您应该知道，在计算包含整数的表达式时，每个计算阶段的临时结果也会四舍五入为整数。在赋值给float+m时，该值仅在整数运算后转换为支持实数的float类型。这意味着，例如，3/4在变为0.0之前就已经是"0“值了。您需要强制转换为浮动，以使其更早发生。您可以通过在a.y、b.y、a.x、b.x、a.y+-+b.y或a.x+-+b.x中的任何一个上使用语法float(value)来做到这一点:只要其中一个术语在除法之前是浮点数，那么什么时候完成都没有关系，例如|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|952478|float+m+=+float(a.y+-+b.y)+/+(a.x+-+b.x);+
float+m+=+(float(a.y)+-+b.y)+/+(a.x+-+b.x);+
...etc...|code-block|syntax|javascript|952479|entityMap^0|18|7|1Y|5|3O|3|3S|3|3W|3|40|3|44|9|4E|9|4Y|C|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]|$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|15|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|L|$]]

You should be aware that in evaluating an expression containing integers, the temporary results from each stage of evaluation are also rounded to be integers. In your assignment to <code>float m</code>, the value is only converted to the real-number capable <code>float</code> type after the integer arithmetic. This means that, for example, 3 / 4 would already be a "0" value before becoming 0.0. You need to force the conversion to float to happen earlier. You can do this by using the syntax <code>float(value)</code> on any of <code>a.y</code>, <code>b.y</code>, <code>a.x</code>, <code>b.x</code>, <code>a.y - b.y</code>, or <code>a.x - b.x</code>: it doesn't matter when it's done as long as one of the terms is a float before the division happens, e.g.

<pre><code>float m = float(a.y - b.y) / (a.x - b.x); 
float m = (float(a.y) - b.y) / (a.x - b.x); 
...etc...
</code></pre>

blocks|key|736442|text|您正在对整数执行计算，并将其结果赋给float。因此编译器隐式地将整型结果转换为浮点型|type|unstyled|depth|inlineStyleRanges|entityRanges|data|736443|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

You are performing calculations on integers and assigning its result to float. So compiler is implicitly converting your integer result into float

blocks|key|736500|text|执行整数除法时，除非一个或多个操作数为浮点数，否则结果始终为整数。只需将其中一个/两个操作数类型转换为浮点数，编译器就会进行转换。当您希望算术按其应有的方式执行时，可以使用类型转换，以便结果将是正确的数据类型。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|736501|float+m+=+static_cast<float>(a.y+-+b.y)+/+(a.x+-+b.x);|code-block|syntax|javascript|736502|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

When doing integer division, the result will always be a integer unless one or more of the operands are a float. Just type cast one/both of the operands to a float and the compiler will do the conversion. Type casting is used when you want the arithmetic to perform as it should so the result will be the correct data type.
<pre class="lang-cpp prettyprint-override"><code>float m = static_cast&lt;float&gt;(a.y - b.y) / (a.x - b.x);
</code></pre>

blocks|key|952651|text|通过与(1.0)相乘，可以同时转换分子和分母。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|952652|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

you can cast both numerator and denominator by multiplying with (1.0) .

blocks|key|952697|text|您可以使用ios操纵器fixed()。它将允许您打印浮点值。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|952698|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

You can use ios manipulators fixed(). It will allow you to print floating point values.

Why is m always = 0? The x and y members of someClass are integers. 

<pre><code>float getSlope(someClass a, someClass b)
{ 
 float m = (a.y - b.y) / (a.x - b.x);
 cout &lt;&lt; " m = " &lt;&lt; m &lt;&lt; "\n";
 return m;
}
</code></pre>

C++ int float casting

编译器

为什么m总是= 0？someClass的x和y成员是整数。float getSlope(someClass a, someClass b){ float m = (a.y - b.y) / (a.x - b.x); cout << " m = " << m << "\n"; re...

问C++整数浮点铸造
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