c++ - C++ 11中的运算符模数变化？

标签 c++ c++11 language-lawyer modulo c++03

<分区>

Possible Duplicate:
C++ operator % guarantees

在 C++ 98/03 中

5.6-4

The binary / operator yields the quotient, and the binary % operator yields the remainder from the division of the first expression by the second. If the second operand of / or % is zero the behavior is undefined; otherwise (a/b)*b + a%b is equal to a. If both operands are nonnegative then the remainder is nonnegative; if not, the sign of the remainder is implementation-defined.

在 C++ 11 中:

5.6 -4

The binary / operator yields the quotient, and the binary % operator yields the remainder from the division of the first expression by the second. If the second operand of / or % is zero the behavior is undefined. For integral operands the / operator yields the algebraic quotient with any fractional part discarded;81 if the quotient a/b is representable in the type of the result, (a/b)*b + a%b is equal to a.

如您所见，缺少为符号位定义的实现，它会发生什么情况？

最佳答案

% 的行为在 C++11 中得到加强，现在已完全指定(除以 0 除外)。

向零截断和恒等式 (a/b)*b + a%b == a 的组合意味着 a%b 对于正数总是正数a 和 negative 表示负 a。

其数学原因如下:

设÷为数学除法，/为C++除法。

对于任何 a 和 b，我们有 a÷b = a/b + f(其中 f 是小数部分)，并且根据标准，我们还有 (a/b)*b + a%b == a.

a/b 已知会向 0 截断，因此我们知道如果 a÷b 为正，负是a÷b是负:

sign(f) == sign(a)*sign(b)

a÷b = a/b + f 可以重新排列为 a/b = a÷b - f。 a可以展开为(a÷b)*b:

(a/b)*b + a%b == a => (a÷b - f)*b+a%b == (a÷b)* b.

现在左侧也可以展开了: