长整数出现意外的模数行为 [英] Unexpected modulo behavior with long integers
问题描述
当操作数是(un)signed 或 long 时，C ++的模数％运算符的行为会很奇怪.
The modulo % operator of C++ behaves strangely when operands are (un)signed or long.
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(在这里:11183)怎么办?
Why would mod(signed int, unsigned int) and mod(signed long long int, unsigned long long int) produce different results? How do I do if I want the correct one (here: 11183)?
注意::如果我遵循自己的算法，则应调用 uint64_t mod(int64_t，uint64_t).
Note: If I follow my algorithm, I should call 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:有关我的系统的更多信息:
PS: More information about my system:
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
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
推荐答案
％不是取模.
另请参见"mod"和"remainder"有什么区别?，C答案也适用于此.
See also What's the difference between "mod" and "remainder"?, a C answer that applies here too.
在C ++中，％是除法的余数. -3365/15156 ->0.除法的其余部分为-3365.
In C++, % is the remainder of a division. -3365/15156 --> 0. The remainder of the division is -3365.
(int)(-3365) % (int)(15156) = -3365
以下内容一开始似乎是错误的，但是由于转换为 unsigned 数学，因此是正确的
(unsigned long int)(-3365) % (unsigned long int)(15156) = 2555
(ULONG_MAX + 1 -3365) % 15156 = 2555
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