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为什么在 C++ 中混合 size_t 和 unsigned int 的模除法 go 错误

[英]Why does modulo division go wrong for mix of size_t and unsigned int in C++

Given a program给定一个程序

#include <iostream>
using namespace std;

int main()
{
     const size_t DoW = 7;
     const unsigned int DAYS_OF_WEEK = static_cast<unsigned int> (DoW);
     unsigned int dayOfFirstDay = 0;
     unsigned int _firstDayOfWeek = 1;
     unsigned int diff = (DAYS_OF_WEEK+ (dayOfFirstDay - _firstDayOfWeek) ) % DAYS_OF_WEEK;
     cout << "diff = ("  << DAYS_OF_WEEK << " + (" << dayOfFirstDay << " - " << _firstDayOfWeek << ")) %" << DAYS_OF_WEEK
         << " = " << diff << endl;
     return 0;
}

The output of that program is该程序的 output 是

diff = (7 + (0 - 1)) %7 = 6

which is expected.这是预期的。 But a modified program without static_cast但是没有static_cast的修改程序

#include <iostream>
using namespace std;

int main()
{
     const size_t DAYS_OF_WEEK = 7;
     unsigned int dayOfFirstDay = 0;
     unsigned int _firstDayOfWeek = 1;
     unsigned int diff = (DAYS_OF_WEEK+ (dayOfFirstDay - _firstDayOfWeek) ) % DAYS_OF_WEEK;
     cout << "diff = ("  << DAYS_OF_WEEK << " + (" << dayOfFirstDay << " - " << _firstDayOfWeek << ")) %" << DAYS_OF_WEEK
         << " = " << diff << endl;
     return 0;
}

outputs输出

diff = (7 + (0 - 1)) %7 = 3

which is not expected.这是意料之中的。 Why?为什么?

(Both programs are compiled with g++ 9.3.0 on Ubuntu 64 Bit) (这两个程序都是在 Ubuntu 64 位上使用g++ 9.3.0 编译的)

It seems on your platform size_t is 64-bit, and unsigned int is 32-bit.在您的平台上似乎size_t是 64 位,而unsigned int是 32 位。

There is no integral promotion to 64-bits 1 .没有对 64 位1整体提升 This is the danger of mixing 64-bit operands in expressions.这就是在表达式中混合 64 位操作数的危险。

So a 32-bit wraparound of -1 remains as 4294967295 when converted to 64 bits.因此,当转换为 64 位时,-1 的 32 位环绕保持为 4294967295。

And we get 7 + 4294967295 (performed in 64 bits) = 4294967302 (no wraparound).我们得到 7 + 4294967295(以 64 位执行)= 4294967302(无环绕)。

4294967302 % 7 = 3 4294967302 % 7 = 3


1 Except for systems where ( unsigned ) int itself is 64 bits, which is currently unlikely. 1除了 ( unsigned ) int本身是 64 位的系统,这目前不太可能。

Such result can happen when size_t has more width than unsigned int .size_t的宽度大于unsigned int时,可能会发生这种结果。

The subtraction of unsigned int and unsigned int wraps around and results in unsigned int . unsigned intunsigned int的减法环绕并导致unsigned int 0 - 1 results in -1 , and it may become 0xffffffff when unsigned int is 4-byte long. 0 - 1结果为-1 ,当unsigned int为 4 字节长时,它可能变为0xffffffff

Then, adding that with another unsigned int will result in unsigned int , so the result looks like normal subtraction and addition.然后,将其与另一个unsigned int相加将导致unsigned int ,因此结果看起来像正常的减法和加法。

On the other hand, adding with size_t will have it calculate in size_t domain, so truncation doesn't happen and the value 7 + 0xffffffff will be divided instead of 7 - 1 .另一方面,与size_t相加将使其在size_t域中计算,因此不会发生截断,并且值7 + 0xffffffff将被除而不是7 - 1

Here is an example code to check the values before division:这是一个示例代码,用于在除法之前检查值:

#include <iostream>
#include <ios>

int main()
{
     const size_t DoW = 7;
     const unsigned int DAYS_OF_WEEK = static_cast<unsigned int> (DoW);
     unsigned int dayOfFirstDay = 0;
     unsigned int _firstDayOfWeek = 1;
     size_t to_add = dayOfFirstDay - _firstDayOfWeek;
     size_t diff_uint = DAYS_OF_WEEK+ (dayOfFirstDay - _firstDayOfWeek);
     size_t diff_sizet = DoW+ (dayOfFirstDay - _firstDayOfWeek);
     std::cout << "sizeof(unsigned int) = " << sizeof(unsigned int) << '\n';
     std::cout << "sizeof(size_t) = " << sizeof(size_t) << '\n';
     std::cout << std::hex;
     std::cout << "to add     : 0x" << to_add << '\n';
     std::cout << "diff_uint  : 0x" << diff_uint << '\n';
     std::cout << "diff_sizet : 0x" << diff_sizet << '\n';
     return 0;
}

Here is an example of output :这是output 的示例

sizeof(unsigned int) = 4
sizeof(size_t) = 8
to add     : 0xffffffff
diff_uint  : 0x6
diff_sizet : 0x100000006

dayOfFirstDay - _firstDayOfWeek is an unsigned int . dayOfFirstDay - _firstDayOfWeek是一个unsigned int As _firstDayOfWeek is greater than dayOfFirstDay the value is an underflow and wrap around and becomes max value of unsigned int .由于_firstDayOfWeek大于dayOfFirstDay ,该值是下溢并环绕并成为unsigned int的最大值。 Let's call this value max_uint .我们称这个值为max_uint

On the other hand DAYS_OF_WEEK is a size_t which is probably a wider type than unsigned int .另一方面DAYS_OF_WEEKsize_t ,它可能是比unsigned int更宽的类型。 This means that DAYS_OF_WEEK + max_uint is not overflowing.这意味着DAYS_OF_WEEK + max_uint没有溢出。 So you end-up computing max_uint % 7 .所以你最终计算max_uint % 7 But max_uint % 7 is not equal to -1 ...但是max_uint % 7等于-1 ...

Try it with less obfuscation:尝试减少混淆:

#include <stdio.h>
#include <stddef.h>

int main() {
  printf("0u - 1u = %u\n", 0u - 1u);
  printf("7u + (0u - 1u) = %u\n", 7u + (0u - 1u));
  printf("7zu + (0u - 1u) = %zu\n", size_t{7} + (0u - 1u));
}

The output I get: output 我得到:

0u - 1u = 4294967295
7u + (0u - 1u) = 6
7zu + (0u - 1u) = 4294967302

As you can see, 0u - 1u results in a wraparound.如您所见, 0u - 1u导致环绕。 Adding this huge number to an unsigned int results in another wraparound.将这个巨大的数字添加到unsigned int会导致另一个环绕。 Adding it to a size_t doesn't as the entire value is representable.将其添加到size_t并不是因为整个值是可表示的。 For that reason, you get different results after the modulus operator.出于这个原因,在模运算符之后会得到不同的结果。

In the initializer expression of this declaration在此声明的初始化表达式中

unsigned int diff = (DAYS_OF_WEEK+ (dayOfFirstDay - _firstDayOfWeek) ) % DAYS_OF_WEEK;

the sub-expression (dayOfFirstDay - _firstDayOfWeek) is equal to the maximum value of the type unsigned int .子表达式(dayOfFirstDay - _firstDayOfWeek)等于unsigned int类型的最大值。

Thus in this sub-expression when DAYS_OF_WEEK has the type unsigned int因此,当DAYS_OF_WEEK的类型为unsigned int时,在此子表达式中

(DAYS_OF_WEEK+ (dayOfFirstDay - _firstDayOfWeek) )

an overflow occurs.发生溢出。

When DAYS_OF_WEEK has the type size_t when neither overflow occurs.DAYS_OF_WEEK的类型为size_t且均未发生溢出时。

This is the reason of different results.这就是结果不同的原因。

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