c - ANSI C 或 ISO C 是否指定 -5 % 10 应该是多少？

Question

我似乎记得 ANSI C 没有指定当模运算符的任一操作数为负时应该返回什么值(只是它应该是一致的)。是后来指定的，还是一直指定的，我记错了？

Answer 1

C89，不完全 (§3.3.5/6)。它可以是 -5 或 5，因为 -5/10 可以返回 0 或 -1(% 是根据涉及 / 的线性方程定义的， * 和 +):

When integers are divided and the division is inexact, if both operands are positive the result of the / operator is the largest integer less than the algebraic quotient and the result of the % operator is positive. If either operand is negative, whether the result of the / operator is the largest integer less than the algebraic quotient or the smallest integer greater than the algebraic quotient is implementation-defined, as is the sign of the result of the % operator. If the quotient a/b is representable, the expression (a/b)*b + a%b shall equal a.

C99，是的(§6.5.5/6)，结果必须是-5:

When integers are divided, the result of the / operator is the algebraic quotient with any fractional part discarded.⁸⁸⁾ If the quotient a/b is representable, the expression (a/b)*b + a%b shall equal a.

_{88) This is often called "truncation toward zero".}

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类似地，在 C++98 中，结果是实现定义的(§5.6/4)，遵循 C89 的定义，但提到首选向零舍入规则，

... If both operands are nonnegative then the remainder is nonnegative; if not, the sign of the remainder is implementation-defined⁷⁴⁾.

_{⁷⁴⁾ According to work underway toward the revision of ISO C, the preferred algorithm for integer division follows the rules defined in the ISO Fortran standard, ISO/IEC 1539:1991, in which the quotient is always rounded toward zero.}

事实上它成为了 C++0x 中的标准规则 (§5.6/4):

... For integral operands the / operator yields the algebraic quotient with any fractional part discarded;⁸² ...

_{82) This is often called truncation towards zero.}