c++ - AWK 中的大整数是否只有 53 位？

Question

很奇怪，我发现在awk中，大整数看起来只有53位。这是我的例子:

function bits2str(bits,data, mask)
{
    if (bits == 0)
        return "0"

    mask = 1
    for (; bits != 0; bits = rshift(bits, 1))
        data = (and(bits, mask) ? "1" : "0") data

    while ((length(data) % 8) != 0)
        data = "0" data

    return data
}

BEGIN{
    print 32,"\tlshift 48:\t", lshift(32,48), "\t", bits2str(lshift(32,48))
    print 429,"\tlshift 48:\t", lshift(429,48), "\t", bits2str(lshift(429,48))
}

输出是:

32  lshift 48:   0   0
429     lshift 48:   3659174697238528    00001101000000000000000000000000000000000000000000000000

但是在c++中，它的输出是:

32 lshift 48: 9007199254740992
429 lshift 48: 120752765008871424

比较两个输出后，我发现awk的只有53位，然后我研究了gawk的源代码(从文件builtin.c, gawk 4.1.1, http://ftp.gnu.org/gnu/gawk/中的3021行开始)，但我没有发现对int的特殊操作。那么，是什么原因造成的呢？为什么会这样？

Answer 1

在 AWK 中，所有数字都以 float 形式存储。

来自 Bitwise function :

For all of these functions, first the double precision floating-point value is converted to the widest C unsigned integer type, then the bitwise operation is performed. If the result cannot be represented exactly as a C double, leading nonzero bits are removed one by one until it can be represented exactly. The result is then converted back into a C double.

假设使用 IEEE-754，double 只能表示最大为 2⁵³ 的整数。