C ++中64位整数的Bitwise（Bitshift）操作

I'm trying to deal with bitboards, which requires me to set a particular bit in a 64-bit unsigned integer. To set bit i, I perform a bitwise OR with the bitboard in question, with a left shifted number.

我正在尝试处理位板,这需要我在64位无符号整数中设置一个特定的位。要设置位i,我对所讨论的位板执行按位OR运算,左移数字。

#include <stdint.h>
uint64_t kings = 0ULL; // Also tried unsigned long long int before.
kings |= 1 << i;

It works fine from bit 0 to bit 31, but fails to work for bits 32 to 63. I suspect it's because the evaluation of the right side happens to be in a 32-bit integer. Therefore, I tried a temporary variable.

它从第0位到第31位工作正常,但不能用于第32位到第63位。我怀疑这是因为右侧的评估恰好是32位整数。因此,我尝试了一个临时变量。

uint64_t temp = 0ULL;
temp |= 1 << i;

Perhaps it still evaluates the right hand side as a 32-bit integer, or that it's some other problem I cannot figure out. To output the integer, I'm using std::bitset<64>. For example:

也许它仍然将右侧评估为32位整数,或者它是我无法弄清楚的其他问题。要输出整数,我使用的是std :: bitset <64>。例如:

uint64_t kings = 0ULL;
kings |= 1 << 3;
kings |= 1 << 59;

Expected decimal value: 576460752303423496

预期的十进制值:576460752303423496

Actual: 8

std::bitset<64> x(kings);
std::cout << x;

Bit value: 0000000000000000000000000000000000000000000000000000000000001000

位值:0000000000000000000000000000000000000000000000000000000000001000

Clearly, only kings |= 1 << 3; worked correctly.

显然,只有国王| = 1 << 3;工作正常。

In summary, what is the issue with bits 32 to 63 and how can I work around it?

总之,32到63位有什么问题,我该如何解决?

3 个解决方案

#1

You need to use 1LL as 64 bit value before you use shift operator << to get 64 bit result:

在使用shift operator < <获得64位结果之前,需要使用1ll作为64位值:< p>

#include <stdint.h>
uint64_t kings = 0ULL; 
kings |= 1ULL << i;

#2

You need to bitshift a 64 bit integer:

你需要bithift一个64位整数:

kings |= 1i64 << 59;

#3

what is the issue with bits 32 to 63?

32到63位有什么问题?

The literal 1 is of type int. The type of a shift operator's result is the type of its LHS (after usual arithmetic conversions have been performed on it). It appears to be 32 bits on your implementation, so shifting it by more than 31 bits yields undefined behavior.

文字1的类型为int。移位运算符的结果类型是其LHS的类型(在对其执行通常的算术转换之后)。它的实现似乎是32位,因此将其移位超过31位会产生未定义的行为。

Use a 64-bit integer as the left operand of the shift operator:

使用64位整数作为移位运算符的左操作数:

temp |= static_cast<uint64_t>(1) << i;

#1