blocks|key|68595|text|模%25只对整数有意义，因为它被定义为整数除法中的余数。你不能用浮点数做整数除法。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|68596|entityMap^0|1|1|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

Modulus <code>%</code> only makes sense with integers because it is defined as the remainder from integer division. You can't do integer division with floats.

blocks|key|220464|text|模算子不适用于浮子。您可能希望使用fmod函数：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|220465|http://www.cplusplus.com/reference/cmath/fmod/|offset|length|220466|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|1A|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@$D|O|E|P|1|Q]]|A|$]]|$1|F|3|-4|5|6|7|R|8|@]|9|@]|A|$]]]|G|$H|$5|I|J|K|A|$L|C]]]]

Modulus operator doesn't work with float. You probably want to use the fmod function:

<a href="http://www.cplusplus.com/reference/cmath/fmod/" rel="nofollow">http://www.cplusplus.com/reference/cmath/fmod/</a>

blocks|key|220478|text|请注意，%25操作符不适用于float。相反，您需要为您的需求使用fmod()。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|220479|double+fmod(double+numerator,+double+denominator);|code-block|syntax|javascript|220480|entityMap^0|4|1|C|5|V|6|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|U|8|@]|D|@]|E|$]]]|L|$]]

Please note that <code>%</code> operator doesn't work with <code>float</code>. Instead you need to use <code>fmod()</code> for your requirement.

<pre><code>double fmod(double numerator, double denominator);
</code></pre>

blocks|key|540638|text|FMOD功能是您需要的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|540639|#include+<stdio.h>

int+main(void){
++++float+x,y;
++++double+z;


++++scanf("%25f%25f",&x,&y);

++++z+=+fmod(x,+5.0f);

++++if(x<=y+&&+z==0)
++++++++printf("%25.2f",y-x-0.50);
++++else+if+(x>y+%7C%7C+z!=0)
++++++++++++printf("%25.2f",y);
++++return+0;
+}|code-block|syntax|javascript|540640|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|$]]

FMOD function is what you need:

<pre><code>#include &lt;stdio.h&gt;

int main(void){
 float x,y;
 double z;


 scanf("%f%f",&amp;x,&amp;y);

 z = fmod(x, 5.0f);

 if(x&lt;=y &amp;&amp; z==0)
 printf("%.2f",y-x-0.50);
 else if (x&gt;y || z!=0)
 printf("%.2f",y);
 return 0;
 }
</code></pre>

blocks|key|540653|text|正如@amdixon所说：“模运算符%25只适用于整数之间。”|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|540654|%25操作符的操作数应该是整数类型。
C11dr§6.5.5+2|blockquote|540655|为了与float算法保持一致，建议使用fmodf()而不是%25或fmod()。|540656|fmodf()计算x/y的浮点余数.|540657|//+float+z=x%255.0f;
float+z+=+fmodf(x+,+5.0f);|code-block|syntax|javascript|540658|备注：|540659|fmod()和家庭的结果可以预计是准确的。Ref|540660|%25是剩余部分。称它为模运算符重载它的意思，当使用a%25b时，它通常不满足期望，a或b都是负值。请参阅What's+the+difference+between+“mod”+and+“remainder”?|540661|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/users/1370847/amdixon|1|https://stackoverflow.com/a/20929050/2410359|2|https://stackoverflow.com/q/13683563/2410359^0|I|1|2|8|0|0|0|1|0|3|5|J|7|T|1|V|6|0|0|7|9|3|0|0|0|0|6|L|3|1|0|0|1|O|3|12|1|14|1|1D|1G|2|0^^$0|@$1|2|3|4|5|6|7|19|8|@$9|1A|A|1B|B|C]]|D|@$9|1C|A|1D|1|1E]]|E|$]]|$1|F|3|G|5|H|7|1F|8|@$9|1G|A|1H|B|C]]|D|@]|E|$]]|$1|I|3|J|5|6|7|1I|8|@$9|1J|A|1K|B|C]|$9|1L|A|1M|B|C]|$9|1N|A|1O|B|C]|$9|1P|A|1Q|B|C]]|D|@]|E|$]]|$1|K|3|L|5|6|7|1R|8|@$9|1S|A|1T|B|C]|$9|1U|A|1V|B|C]]|D|@]|E|$]]|$1|M|3|N|5|O|7|1W|8|@]|D|@]|E|$P|Q]]|$1|R|3|S|5|6|7|1X|8|@]|D|@]|E|$]]|$1|T|3|U|5|6|7|1Y|8|@$9|1Z|A|20|B|C]]|D|@$9|21|A|22|1|23]]|E|$]]|$1|V|3|W|5|6|7|24|8|@$9|25|A|26|B|C]|$9|27|A|28|B|C]|$9|29|A|2A|B|C]|$9|2B|A|2C|B|C]]|D|@$9|2D|A|2E|1|2F]]|E|$]]|$1|X|3|-4|5|6|7|2G|8|@]|D|@]|E|$]]]|Y|$Z|$5|10|11|12|E|$13|14]]|15|$5|10|11|12|E|$13|16]]|17|$5|10|11|12|E|$13|18]]]]

As <a href="https://stackoverflow.com/users/1370847/amdixon">@amdixon</a> commented: "modulus operator <code>%</code> only applies between integers."

<blockquote>
 The operands of the <code>%</code> operator shall have integer type. 
 C11dr §6.5.5 2
</blockquote>

In keeping with <code>float</code> arithmetic, suggest using <code>fmodf()</code> instead of <code>%</code> or <code>fmod()</code>. 
<code>fmodf()</code> computes the floating-point remainder of <code>x/y</code>.

<pre><code>// float z=x%5.0f;
float z = fmodf(x , 5.0f);
</code></pre>

Notes: 

The result of <code>fmod()</code> and family can be expected to be exact. <a href="https://stackoverflow.com/a/20929050/2410359">Ref</a>

The <code>%</code> is the remainder. Calling it the modulus operator overloads its meaning that often does not meet expectations when with <code>a%b</code>, either <code>a</code> or <code>b</code> is negative. See <a href="https://stackoverflow.com/q/13683563/2410359">What&#39;s the difference between “mod” and “remainder”?</a>

I was trying to run a program but it shows an error as:

<blockquote>
 Invalid binary operator float to int
</blockquote>

When I tried making it <code>float</code> it says:

<blockquote>
 Invalid binary operator float to float
</blockquote>

The problem is with <code>%</code> operator And its operands.

Please tell me what to do?

<pre><code>#include &lt;stdio.h&gt;
int main()
{
 float x,y;
 scanf("%f%f",&amp;x,&amp;y);
 float z=x%5.0f;
 if(x&lt;=y &amp;&amp; z==0)
 printf("%.2f",y-x-0.50);
 else if (x&gt;y || z!=0)
 printf("%.2f",y);
 return 0;
 }
</code></pre>

Error while running program with certain operator and operands in C

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我试图运行一个程序，但它显示了一个错误：无效二进制运算符浮点到int当我试着做float时，它说：无效二进制操作符浮点到浮动问题在于%算子及其操作数。请告诉我该怎么做？#include <stdio.h>int main(){ float x,y; scanf("%f%f",&x,&y); float z=x%5.0f; if(x<=y && z==0) pr

问在C中使用某些运算符和操作数运行程序时出错
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