C++ 不能在双打上使用模数？
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Can't use modulus on doubles?
提问by Bhaxy
I have a program in C++ (compiled using g++). I'm trying to apply two doubles as operands to the modulus function, but I get the following error:
我有一个 C++ 程序（使用 g++ 编译）。我正在尝试将两个双精度数作为操作数应用于模数函数，但出现以下错误：
error: invalid operands of types 'double' and 'double' to binary 'operator%'
错误：“double”和“double”类型的无效操作数转换为二进制“operator%”
Here's the code:
这是代码：
int main() {
double x = 6.3;
double y = 2;
double z = x % y;
}
回答by Mysticial
The %operator is for integers. You're looking for the fmod()function.
该%运算符用于整数。您正在寻找fmod()功能。
#include <cmath>
int main()
{
double x = 6.3;
double y = 2.0;
double z = std::fmod(x,y);
}
回答by MSN
fmod(x, y)is the function you use.
fmod(x, y)是你使用的函数。
回答by Sceptical Jule
Use fmod()from <cmath>. If you do not want to include the C header file:
fmod()从使用<cmath>。如果不想包含 C 头文件：
template<typename T, typename U>
constexpr double dmod (T x, U mod)
{
return !mod ? x : x - mod * static_cast<long long>(x / mod);
}
//Usage:
double z = dmod<double, unsigned int>(14.3, 4);
double z = dmod<long, float>(14, 4.6);
//This also works:
double z = dmod(14.7, 0.3);
double z = dmod(14.7, 0);
double z = dmod(0, 0.3f);
double z = dmod(myFirstVariable, someOtherVariable);
回答by Mystical
You can implement your own modulus functionto do that for you:
您可以实现自己的模数函数来为您做到这一点：
double dmod(double x, double y) {
return x - (int)(x/y) * y;
}
Then you can simply use dmod(6.3, 2)to get the remainder, 0.3.
然后您可以简单地使用dmod(6.3, 2)来获取余数，0.3。
