填上可变数目的1

What's the best way to fill a variable with an unknown (at compile time) number of ones? For example, let's say:

用未知(在编译时)的1个数填充变量的最好方法是什么?例如,让我们说:

int n = 5;
int b = fillwithones(5);

now b contains 11111 (in binary).

现在b包含11111(二进制)。

I can't just hard code int b = 31 because n is not known ahead of time (in my application).

我不能只对int b = 31进行硬编码，因为(在我的应用程序中)不提前知道n。

I could do something like this:

我可以这样做:

int b = pow(2, n) - 1

But using a pow seems very wasteful.

但是使用pow似乎非常浪费。

Thanks!

谢谢!

2 个解决方案

#1

You can use left shift and then subtract 1:

你可以用左移然后减去1:

unsigned int b = (1U << n) - 1U;

// Broken down into steps
//  1           = 00000001b
//  1 << 5      = 00100000b
// (1 << 5) - 1 = 00011111b

The reason this works is 1 shifted left n times is the same as 2ⁿ, as each sole bit position represents a power of 2.

这样做的原因是左移1次n乘以等于2n，因为每个单独的位代表2的幂。

#2

A funny way to get the highest bits as 1 and the lowest bits as zero is using this nice trick:

得到最高位为1，最低位为0的有趣方法是使用这个很好的技巧:

#include <limits.h>

...

int b = INT_MIN >> n;

This works because shift left operation on a negative number will mantain the sign of the operation, and since INT_MIN is 10000....0000 shifting it by n to the left will give you n bits to 1, but on the other side.

这个作品,因为左移位操作在一个负数将基本的符号操作,因为INT_MIN 10000 ....0000将它向左平移n个比特就会得到n个比特的1，但是在另一边。

#1