Question

我需要在C中执行一个真正的数学模数。对于我来说，允许模数参数的负数是有意义的，因为我的模块化计算会产生负中间结果，必须将其放回到最少残留系统中。但允许负面模块是没有意义的，因此我写了

printf("%u\n", mod(-3, 11));

然而，使用负数和正模块调用此类函数

产生输出

using System;

namespace BankAccounts  
{  
   class Account  
   {  
       protected Account(decimal balance)  
       { _Balance = balance; }  

       private decimal _Balance;
       public decimal Balance
       {
           get { return _Balance; }
       }

       public override string ToString()
       {
           return string.Format("Balance: {0:c}", _Balance);
       }
   }
}

我不明白为什么。你能解释一下吗？

编辑：我知道operator％与数学模数不同，我知道它是如何为正数和负数定义的。我问的是它会为不同的签名做些什么，而不是不同的标志。

Answer 1

启用clang的

-Wconversion可以明确指出您的错误：

prog.cc:3:15: warning: implicit conversion changes signedness: 'unsigned int' to 'int' [-Wsign-conversion]
    int r = x % m;
        ~   ~~^~~
prog.cc:3:13: warning: implicit conversion changes signedness: 'int' to 'unsigned int' [-Wsign-conversion]
    int r = x % m;
            ^ ~
prog.cc:4:21: warning: operand of ? changes signedness: 'int' to 'unsigned int' [-Wsign-conversion]
    return r >= 0 ? r : r + m;
    ~~~~~~          ^
prog.cc:4:25: warning: implicit conversion changes signedness: 'int' to 'unsigned int' [-Wsign-conversion]
    return r >= 0 ? r : r + m;
                        ^ ~
prog.cc:9:12: warning: implicit conversion changes signedness: 'unsigned int' to 'int' [-Wsign-conversion]
    return mod(-3, 11);
    ~~~~~~ ^~~~~~~~~~~

live example on wandbox

转换为unsigned int时，-3变为4294967293。

4294967293 % 11等于1。

Answer 2

见C11 6.5.5（乘法运算符）/ 3：

通常的算术转换是在操作数上执行的。

usual arithmetic conversions由6.3.1.8定义。相关部分是：

否则，如果具有无符号整数类型的操作数的等级大于或等于等于另一个操作数的类型的等级，然后是操作数有符号整数类型转换为带有unsigned的操作数的类型整数类型。

因此，在void PerformTransformation(Gdiplus::Bitmap* bitmap, LPCTSTR SaveFileName) { Gdiplus::BitmapData* bitmapData = new Gdiplus::BitmapData; UINT Width = bitmap->GetWidth(); UINT Height = bitmap->GetHeight(); Gdiplus::Rect rect(0, 0,Width,Height ); // Lock a 5x3 rectangular portion of the bitmap for reading. bitmap->LockBits(&rect, Gdiplus::ImageLockModeWrite, PixelFormat32bppARGB, bitmapData); byte* Pixels = (byte*)bitmapData->Scan0; INT stride_bytes_count = abs(bitmapData->Stride); UINT row_index, col_index; byte pixel[4]; for (col_index = 0; col_index < Width; ++col_index) { for (row_index = 0; row_index < Height; ++row_index) { unsigned int curColor = Pixels[row_index * stride_bytes_count / 4 + col_index]; int b = curColor & 0xff; int g = (curColor & 0xff00) >> 8; int r = (curColor & 0xff0000) >> 16; if ((r + 10) > 255) r = 255; else r += 10; if ((g + 10) > 255) g = 255; else g += 10; if ((b + 10) > 255) b = 255; else b += 10; pixel[0] = b; pixel[1] = g; pixel[2] = r; Pixels[row_index * stride_bytes_count / 4 + col_index] = *pixel; } } bitmap->UnlockBits(bitmapData); ::DeleteObject(bitmapData); CLSID pngClsid; GetEncoderClsid(L"image/png", &pngClsid); bitmap->Save(SaveFileName, &pngClsid, NULL); } };中，x % m首先转换为unsigned int。

要避免此行为，您可以使用x，但如果x % (int)m这会出现故障。如果您想支持m > INT_MAX以及否定m > INT_MAX，则必须使用稍微复杂的逻辑。

Answer 3

其他答案很好解释OP在unsigned操作没有产生预期结果之前将负值转换为%时遇到了麻烦。

以下是解决方案：一个采用更广泛的数学（可能并不总是可用）。第二个是小心构造，以避免任何未定义的行为，（UB），实现定义（ID）行为或仅使用int, unsigned数学的溢出。它不依赖于2的补码。

unsigned int mod_ref(int x, unsigned int m) {
  long long r = ((long long) x) % m;
  return (unsigned) (r >= 0 ? r : r + m);
}

unsigned int mod_c(int x, unsigned int m) {
  if (x >= 0) {
    return ((unsigned) x) % m;
  }
  unsigned negx_m1 = (unsigned) (-(x + 1));
  return m - 1 - negx_m1 % m;
}

测试驱动程序

#include <limits.h>
#include <stdio.h>
#include <stdlib.h>

void testm(int x, unsigned int m) {
  if (m) {
    unsigned r0 = mod_ref(x, m);
    unsigned r1 = mod_c(x, m);
    if (r0 != r1) {
      printf("%11d %10u --> %10u %10u\n", x, m, r0, r1);
    }
  }
}

int main() {
  int ti[] = {INT_MIN, INT_MIN + 1, INT_MIN / 2, -2, -1, 0, 1, 2, INT_MAX / 2,
      INT_MAX - 1, INT_MAX};
  unsigned tu[] = {0, 1, 2, UINT_MAX / 2, UINT_MAX - 1, UINT_MAX};
  for (unsigned i = 0; i < sizeof ti / sizeof *ti; i++) {
    for (unsigned u = 0; u < sizeof tu / sizeof *tu; u++) {
      testm(ti[i], tu[u]);
    }
  }
  for (unsigned i = 0; i < 1000u * 1000; i++) {
    int x = rand() % 100000000;
    if (rand() & 1)
      x = -x - 1;
    unsigned m = (unsigned) rand();
    if (rand() & 1)
      m += INT_MAX + 1u;
    testm(x, m);
  }
  puts("done");
  return 0;
}

signed int modulo unsigned int产生无意义的结果

3 个答案: