C语言 错误：取浮点模数时，二进制% 的操作数无效
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error: invalid operands to binary % when taking modulus of float
提问by mjs
Its been a fairly long week, so forgive me if I am being thick.
这是一个相当长的一周，所以如果我很胖，请原谅我。
I have some code like:
我有一些代码，如：
float someFloat = 0;
//....do some stuff to someFloat
//....
if( someFloat % 1)
{
//take some action
}
I get a compiler error : error: invalid operands to binary %
我收到编译器错误： error: invalid operands to binary %
Assuming the compiler isnt on drugs, whats wrong with this?
假设编译器没有吸毒，这有什么问题？
EDIT: As an aside, what I actually wanted to do was detect non-integer value and round up. What I should have been doing was calling roundf (and I guess checking if the return is less than the operand and then incrementing if so, to take care that we have rounded up)
编辑：顺便说一句，我真正想做的是检测非整数值并四舍五入。我应该做的是调用 roundf （我想检查返回值是否小于操作数，如果是，则增加，以注意我们已经四舍五入）
回答by Paul R
%is an integer operator - use fmodor fmodffor doubles or floats.
%是整数运算符 - 将fmod或fmodf用于双精度或浮点数。
Alternatively if you expect your float to represent integer values then convert it to an intfirst, e.g.:
或者，如果您希望浮点数表示整数值，则将其转换为int第一个，例如：
if ((int)someFloat % 2 == 1) // if f is an odd integer value
{
...
}
回答by Jon
回答by Sohil Omer
%this only works with integers use fmod for floating point or double values**
%这仅适用于整数使用 fmod 为浮点或双精度值**
double fmod(double x, double y)
x -- This is the floating point value with the division numerator i.e. x.
y -- This is the floating point value with the division denominator i.e. y.
