c++ - 长整数的意外模数行为

Question

当操作数是(un)signed 或long 时，C++ 的模% 运算符会表现得很奇怪。

为什么 mod(signed int, unsigned int) 和 mod(signed long long int, unsigned long long int) 会产生不同的结果？如果我想要一个正确的(此处:11183)，我该怎么做？

注意:如果按照我的算法，我应该调用 uint64_t mod(int64_t, uint64_t)。

// Tip: 11183 is the "correct" expected result.

(                   int)(-3365) % (                   int)(15156) = -3365
(  signed           int)(-3365) % (  signed           int)(15156) = -3365
(unsigned           int)(-3365) % (unsigned           int)(15156) = 11183
(  signed           int)(-3365) % (unsigned           int)(15156) = 11183
(unsigned           int)(-3365) % (  signed           int)(15156) = 11183

(              long int)(-3365) % (              long int)(15156) = -3365
(  signed      long int)(-3365) % (  signed      long int)(15156) = -3365
(unsigned      long int)(-3365) % (unsigned      long int)(15156) = 2555
(  signed      long int)(-3365) % (unsigned      long int)(15156) = 2555
(unsigned      long int)(-3365) % (  signed      long int)(15156) = 2555

(         long long int)(-3365) % (         long long int)(15156) = -3365
(  signed long long int)(-3365) % (  signed long long int)(15156) = -3365
(unsigned long long int)(-3365) % (unsigned long long int)(15156) = 2555
(  signed long long int)(-3365) % (unsigned long long int)(15156) = 2555
(unsigned long long int)(-3365) % (  signed long long int)(15156) = 2555

(          int_fast16_t)(-3365) % (          int_fast16_t)(15156) = -3365
(         uint_fast16_t)(-3365) % (          int_fast16_t)(15156) = 2555
(          int_fast16_t)(-3365) % (         uint_fast16_t)(15156) = 2555
(         uint_fast16_t)(-3365) % (         uint_fast16_t)(15156) = 2555

(          int_fast32_t)(-3365) % (          int_fast32_t)(15156) = -3365
(         uint_fast32_t)(-3365) % (          int_fast32_t)(15156) = 2555
(          int_fast32_t)(-3365) % (         uint_fast32_t)(15156) = 2555
(         uint_fast32_t)(-3365) % (         uint_fast32_t)(15156) = 2555

(          int_fast64_t)(-3365) % (          int_fast64_t)(15156) = -3365
(         uint_fast64_t)(-3365) % (          int_fast64_t)(15156) = 2555
(          int_fast64_t)(-3365) % (         uint_fast64_t)(15156) = 2555
(         uint_fast64_t)(-3365) % (         uint_fast64_t)(15156) = 2555

PS:关于我的系统的更多信息:

Linux pc-gi-446 4.4.0-36-generic#55-Ubuntu SMP x86_64 x86_64 x86_64 GNU/Linux
g++ (Ubuntu 5.4.0-6ubuntu1~16.04.4) 5.4.0 20160609

Answer 1

尝试

unsigned int x = -3365; printf("%u", x);

它不会打印“3365”，但会打印 unsigned int 减去 3365 + 1 的最大值，即 UINT_MAX-3365+1，即 4294963931.

因此 (unsigned int)(-3365) % (unsigned int)(15156) 与 unsigned int z = 4294963931 % 15156 相同； printf("%u\n", z) 并给出 11183。