Question

我从后端到ArrayBuffer得到无符号的64位整数，由于IEEE 754我无法得到＆amp;将此序列保持为数字。有没有办法“字符串化”这个庞大的数字？

简而言之：

var buffer = new ArrayBuffer(8);
var raw = new Uint8Array(buffer);

raw[0] = 255;
raw[1] = 255;
raw[2] = 255;
raw[3] = 255;
raw[4] = 255;
raw[5] = 255;
raw[6] = 255;
raw[7] = 255;

var view = new DataView(buffer);
var result = view.getUint64(0); /// :-(

提前谢谢。

Answer 1

是的，你可以对Int64进行字符串化，但没有＆＃34;内置＆＃34;这样做的方式就像你说的那样，JavaScript数字是双精度浮点数。

<?xml version='1.0' encoding='UTF-8'?>
<X12_00501_837_P xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
    <TS837_2000A_Loop>
        <NM1_SubLoop_2>
            <TS837_2010AA_Loop>
                <NM1_BillingProviderName>
                    <NM103_BillingProviderLastorOrganizationalName>VNA of Cape Cod</NM103_BillingProviderLastorOrganizationalName>
                    <NM109_BillingProviderIdentifier>1487651915</NM109_BillingProviderIdentifier>
                </NM1_BillingProviderName>
                <N3_BillingProviderAddress>
                  <N301_BillingProviderAddressLine>8669 NORTHWEST 36TH ST </N301_BillingProviderAddressLine>
                </N3_BillingProviderAddress>
            </TS837_2010AA_Loop>
        </NM1_SubLoop_2>
    </TS837_2000A_Loop>
</X12_00501_837_P>

h2d来自this answer here。

＆＃13;

// ...
function readInt64(view){
    var small = view.getUint32(0).toString(16); // present as base 16 string
    var large = view.getUint32(1).toString(16);
    // in a power-of-2  base it's just a string concat to add them
    var bigNumber = large + small; 
    return h2d(bigNumber);
}

＆＃13;

Answer 2

下面的DataView.getUint64()实现将为您提供一个尽可能精确的数字值：

How to Read 64-bit Integer from an ArrayBuffer / DataView in JavaScript

Uint8Array（8）字符串化

2 个答案: