从二进制快速字符串中获取有符号整数 [英] Get signed integer from swift string of binary

问题描述

我当前正在尝试使用功能

I'm currently trying to use the function

Int(binaryString, radix: 2)

将二进制字符串转换为Int.但是，此函数似乎总是将二进制字符串转换为无符号整数.例如

to convert a string of binary to an Int. However, this function seems to always be converting the binary string into an unsigned integer. For example,

Int("1111111110011100", radix: 2)

如果它正在执行带符号的int转换，则我期望从它获得-100时返回65436.我并没有真正使用二进制文件，所以我想知道我应该在这里做什么? Swift3内建有一种代码有效的方式来对带符号的int做到这一点吗?我最初希望它能工作，因为它是一个Int构造函数(不是UInt).

returns 65436 when I'd expect to get -100 from it if it were doing an signed int conversion. I haven't really worked with binary much, so I was wondering what I should do here? Is there a code-efficient way built-into Swift3 that does this for signed ints? I had initially expected this to work because it's an Int constructor (not UInt).

推荐答案

在游戏中玩耍可以得到所需的结果，如下所示:

Playing around you can get the desired result as follows:

let binaryString = "1111111110011100"
print(Int(binaryString, radix: 2)!)
print(UInt16(binaryString, radix: 2)!)
print(Int16(bitPattern: UInt16(binaryString, radix: 2)!))

输出:

65436
65436
-100

65436
65436
-100

所需的结果来自使用UInt16的位模式创建带符号的Int16.

The desired result comes from creating a signed Int16 using the bit pattern of a UInt16.

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