左移特定号码.1s具体没有.时间

Question

我想留下一个特定的号码.具体的没有.时间.我正在尝试这样的事情

//Left shift 3 1's by 5. 
int a = 0; 
a = 0b(111 << 5) ;  //Error : unable to find numeric literal
                    // operator 'operator""b'        
std::cout << a; //Should be 224

关于如何解决上述问题的任何建议？理想情况下,我想要这样的东西

int a = 0;
int noOfOnes = 3;
a = 0b(noOfOnes << 5);

我不确定如何在C++中完成上述操作？

Answer 1

它很简单((1u << a) - 1u) << b,s a的数量在哪里1,b是偏移量.

例:

#include <iostream>
#include <bitset>

unsigned foo(unsigned size, unsigned offset)
{
    return ((1u << size) - 1u) << offset;
}

int main()
{
    unsigned x = foo(5, 3);
    std::cout << std::bitset<16>(x); // 0000000011111000
}

Try it live

这种方法会破坏a (or b) >= sizeof(unsigned) * CHAR_BIT(感谢@Deduplicator)

如果在切换到最大可用积分类型后仍然存在问题,您可以添加一些安全检查:

#include <iostream>
#include <bitset>
#include <climits>

unsigned foo(unsigned size, unsigned offset)
{
    unsigned x = 0; 
    if (offset >= sizeof x * CHAR_BIT)
        return 0;
    if (size < sizeof x * CHAR_BIT)
        x = 1u << size;
    return (x - 1u) << offset;
}

int main()
{
    unsigned x = foo(32, 3);
    std::cout << std::bitset<32>(x);
}