我如何在 C++ 中做模数？
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How do I do modulus in C++?
提问by user161190
How do I perform a mod operation between two integers in C++?
如何在 C++ 中的两个整数之间执行模运算？
回答by wilhelmtell
C++ has the %operator, occasionally and misleadingly named "the modulus" operator. In particular the STL has the modulus<>functor in the <functional>header. That's not the mathematical modulus operator, mind you, because in modulus arithmetics a mod bby definition evaluates to a non-negative value for any value of aand any positive value of b. In C++ the sign of the result of a % bis implementation-definedif either of the arguments is negative. So, we would more appropriately name the %operator the remainderoperator.
C++ 有%运算符，偶尔会误导性地命名为“模数”运算符。特别是 STLmodulus<>在<functional>头文件中有函子。请注意，这不是数学模数运算符，因为a mod b根据定义，在模数算术中，对于 的任何值a和任何正值，计算结果为非负值b。在 C++ 中，如果任一参数为负，则结果的符号a % b是实现定义的。因此，我们将%运算符命名为余数运算符更合适。
That said, if you truly want the mathematical modulus operator then you can define a function to do just that:
也就是说，如果你真的想要数学模运算符，那么你可以定义一个函数来做到这一点：
template<typename V>
V mod(const V& a, const V& b)
{
return (a % b + b) % b;
}
So long as bis a positive value a call to the above function will yield a non-negative result.
只要b是正值，对上述函数的调用就会产生非负结果。
回答by Jim Lewis
As the other answers have stated, you can use the C++ % operator. But be aware that there's a wrinkle no one has mentioned yet: in the expression a % b, what if ais negative?
Should the result of this operation be positive or negative? The C++ standard leaves this up to the
implementation. So if you want to handle negative inputs portably, you should probably
do something like r = abs(a) % b, then fix up the sign of rto match your requirements.
正如其他答案所述，您可以使用 C++ % 运算符。但请注意，有一个问题还没有人提到：在表达式中a % b，如果 a是负数呢？这个操作的结果应该是正的还是负的？C++ 标准将这留给了实现。因此，如果您想轻松地处理负输入，您可能应该执行类似的操作r = abs(a) % b，然后修复 的符号r以符合您的要求。
回答by Giffyguy
Like this: x=y%z
像这样：x=y%z
回答by Klaim
Using the modulus %operator :
使用模%运算符：
int modulus_a_b = a % b;
回答by ErdeNese
if you use double variable, you should use;
如果你使用双变量，你应该使用;
double x;
double y;
double result = fmod(x, y);
