我们如何在Dart 2中将int转换为BigInt? [英] How do we cast an int as a BigInt in Dart 2?

问题描述

在Dart 2(Dart VM版本:2.0.0-dev.50.0)中使用 BigInt 类，我遇到了以下问题:

Playing around with the BigInt class in Dart 2 (Dart VM version: 2.0.0-dev.50.0), I came across the following issues:

BigInt x = 5;

print(x % 3); // 2
print(x ~/ 3); // 1

到目前为止，很好，但是...

So far, so good, but...

当我尝试访问方法时，例如:

When I try to access the methods, for example:

print(x.pow(2));

我得到一个 NoSuchMethodError:类'int'没有实例方法'pow'异常.看来 x 被视为香草味的 int ...

I get a NoSuchMethodError: Class 'int' has no instance method 'pow' exception. It appears that x is being treated as a vanilla flavored int...

此外，如果我尝试更明确地实例化 x :

Further, if I try to instantiate x more explicitly:

var x = new BigInt.from(5);

我得到 NoSuchMethodError:类'int'没有实例获取方法'_used'.

我误会了文档?具体来说，我们如何将 int 转换为 5 或 ['a'，'b'，'c'].length BigInt 对象?

Am I misunderstanding the documentation? Specifically, how do we cast an int, like 5 or ['a', 'b', 'c'].length, as a BigInt object?

尽管这不能解释为什么 BigInt x = 5 以及 x％3 之类的表达式不会产生错误并按预期工作，但这似乎可以仅仅是由于 BigInt 和 int 不能很好地配合.例如，一旦 x 确实代表了一个 BigInt (与使用 BigInt x = 5 实例化的情况不同，此后 x 似乎代表常规的 int )，则表达式 x％3 不起作用，因为％操作期望使用右侧的BigInt ，而不是 3 .(表达式需要更改为笨重的 x％new BigInt.from(3)，以摆脱 NoSuchMethodError:类'int'没有实例getter'_used'例外.)

Although this doesn't explain why BigInt x = 5 and then such expressions as x % 3 do not generate errors and work as expected, this appears to simply be due to BigInts and ints not playing well together. For example, once x really does represent a BigInt (unlike if it's instantiated using BigInt x = 5, after which x appears to represent a regular int) then the expression x % 3 doesn't work because the % operation is expecting a BigInt on the right hand side, which 3 is not. (The expression needs to change to the clunky x % new BigInt.from(3) to get rid of the NoSuchMethodError: Class 'int' has no instance getter '_used' exception.)

正如我已经提到的，这并不能解释所有内容.允许这样的表达式，例如 BigInt x = 5 ，然后像预期的那样运行诸如 x％3 的表达式，这表明 BigInt s和 int 是 meant ，它们至少可以在某些设计师的论文上无缝地协同工作...

As I mentioned already, this does not explain everything. The fact that such expressions as BigInt x = 5 are allowed, and that expressions such as x % 3 then work as expected suggests that BigInts and ints are meant to work seamlessly together, at least on some designer's papers...

推荐答案

您正在Dart 1生产模式下运行Dart，而没有使用分析器测试代码.这意味着您不会进行类型检查，并且您的程序确实不是类型正确的.文字 5 不是 BigInt ，而是 int .您将其分配给类型为 BigInt 的变量，但是在Dart 1生产模式下，该变量类型将被忽略，从而成功". BigInt s和 int s不能无缝地一起工作，但是如果您实际上不检查类型，则不会知道使用的是错误的类型

You are running Dart in Dart 1 production mode and without testing your code with the analyzer. That means that you get no type checking, and your program is indeed not type-correct. The literal 5 is not a BigInt, it is an int. You are assigning it to a variable typed as BigInt, but in Dart 1 production mode, the variable type is ignored, so that "succeeds". BigInts and ints are not meant to work seamlessly together, but if you don't actually check the types, you won't know that you are using the wrong one.

您的修正 var x = new BigInt.from(5); 是正确的.Dart不会进行隐式转换，您必须显式创建一个 BigInt 来获取隐式转换.这也意味着您必须这样做:

Your fix, var x = new BigInt.from(5);, is correct. Dart does not do implicit conversion, you have to explicitly create a BigInt to get one. That also means that you have to do:

var x = new BigInt.from(5);
print(x % new BigInt.from(3)); // 2 

该代码应该起作用.因此，如果抛出，则表示出了问题.我在Dart 2.0.0-dev.49.0中尝试了该代码，并且可以正常工作.您正在使用哪个版本的Dart( dart --version )以及如何运行它?

That code should work. so if it throws, something is wrong. I tried the code in Dart 2.0.0-dev.49.0, and it worked. Which version of Dart are you using (dart --version) and how are you running it?

如果您将程序作为 dart --checked program.dart 运行，则当您尝试将 int 分配给 BigNum时，将出现类型错误.变量.

If you run your program as dart --checked program.dart then you will get the type error when you try to assign an int to a BigNum variable.

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