如何在 Swift 中从八进制转换为十进制

Question

您好，我正在尝试将 swift 中的八进制数转换为十进制数。最简单的方法是什么？

Answer 1

从八进制到十进制

这里有一个特定的 Int 初始值设定项

let octal = 10
if let decimal = Int(String(octal), radix: 8) {
    print(decimal) // 8
}

从十进制到八进制

let decimal = 8
if let octal = Int(String(decimal, radix: 8)) {
    print(octal) // 10
}

Note 1: Please pay attention: parenthesis are different in the 2 code snippets.

Note 2: Int initializer can fail for string representations of number with more exotic radixes. Please read the comment by @AMomchilov below.

Answer 2

您可以轻松地将八进制转换为十进制。 Swift 原生支持八进制语法。八进制数前必须写“0o”

let number = 0o10
print(number) // it prints the number 8 in decimal

Integer Literals

Integer literals represent integer values of unspecified precision. By default, integer literals are expressed in decimal; you can specify an alternate base using a prefix. Binary literals begin with 0b, octal literals begin with 0o, and hexadecimal literals begin with 0x.

这里是 documentation's reference.

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