为什么Dart BigInt溢出?

Question

文档中明确表示

第三种整数表示，bigint，保存任意大的有符号整数。

是的，如果我做了

BigInt.parse('1401361620244348303075010764053798750806699051384896657186984262080729392169468490123546840166223286924678557627464612464171446377618435568706501027067406794991226499183297227160622976110894228304766631654633074535516079503916674556805715032127374387087475009709090')

那就行了

但是当我创建一个BigInt.from(0)并反复添加一些东西来转换这个base64编码的数字

EwnG/GyyyZR6cgLrLuY+cvvRMlNqIr0GgyqWYmrpvsWwNbVcRQ7FWJFuGWFON81W7FbX0wMyjRV7WsMmk0zisj2baRl3v3Y1LPA8ncXU9vVfqCyeXVmUgv1T9wi1k41Zjr6h7WTjZJvyQC4YpaYpZdOJcuYm8yVOlfUKJ10lm2p9yxPJLtStvwJFZy4uCF2p/sfDATIv9Vyny3Ewx/B85Ae+eg2nlRcDmZdu5ByoqOfEYaU6H1fzzHvUSUBZvHv9zBLQ6PrLG6DhYhXzxol3zpbV02NGq3WfeBLhfl4DOUiVEDi0HSLw3xyJU+rw8rS1hoQeYcyogZ8p0I3BiNRs1Q==`, 

转换成BigInt类型，它很快就会溢出:

...
543848068542839930
1120308820846263418
3426151830059957370
-5797220206794818438
-5797220206794818438
-5797220206794818438
...

和卡在-5797220206794818438。为什么是这样，我如何将base64编码的数字转换为BigInt ?我很困惑。

下面是将base64转换为BigInt的代码:

/// Bytes are assumed to be BigEndian.
BigInt bytesToBigInt(List<int> bytes) {
BigInt result = BigInt.from(0);
int e = 0;
for (int i in bytes.reversed) {
for (int c = 0; c < 8; c++) {
int include = i & 1;
i = i >> 1;
e++;
if (include == 1) {
final val = pow(2, e);
result += BigInt.from(include * val);
print(' $result');
}
}
}
return result;
}

Answer 1

您的特定问题是由于使用pow(2, e)(一旦e变大就容易溢出)而不是BigInt.pow(2, e)引起的。

此外，您的bytesToBigInt实现似乎比必要的更复杂。正如我在评论中指出的那样，如果您正在处理字节列表，那么处理每个位都是浪费的，并且BigInt直接支持按位操作符，因此可以简单地从字节序列创建BigInt:

/// Creates a [BigInt] from a sequence of [bytes], which is assumed to be in
/// big-endian order.
BigInt bytesToBigInt(Iterable<int> bytes) {
BigInt result = BigInt.zero;
for (int byte in bytes) {
result = (result << 8) | BigInt.from(byte);
}
return result;
}

或:

BigInt bytesToBigInt(Iterable<int> bytes) => bytes.fold(
BigInt.zero,
(resultSoFar, byte) => (resultSoFar << 8) | BigInt.from(byte),
);