如何在结构中正确初始化指向double值的指针？

Question

我试图获取指向结构中的double值的指针，我可以将其传递给例程，但是在获取除0之外的任何值时都会遇到问题。以下适用于其他结构！我使用过的值（例如只包含二进制文件的值！），但在这种情况下不是：

Rebol []

speed1: make struct! [
    d [double]
] [0.0]
speed1*-struct: make struct! [i [int]] third speed1
speed1*: speed1*-struct/i

以下是评估：

>>  speed1: make struct! [
[      d [double]
[     ] [0.0]
>>  speed1*-struct: make struct! [i [int]] third speed1
>>  speed1*: speed1*-struct/i
== 0

这是二进制的工作等价物！结构：

>>  binary: make struct! [bytes [binary!]] reduce [head insert/dup copy #{} 0 8]
>>  binary*-struct: make struct! [i [int]] third binary
>>  binary*: binary*-struct/i
== 39654648

我可以看到third函数的区别：

>> third speed1
== #{0000000000000840}
>> third binary
== #{F8145D02}

但是，我不确定长度的差异在这种情况下意味着什么。我究竟做错了什么？是否有不同的方法将指针传递给十进制值？

Answer 1

这是一种获取指向double值的指针的方法：

speed1-struct: make struct! [s [struct! [d [double]]]] [0.0]
speed1*-struct: make struct! [i [int]] third speed1-struct
speed1*: speed1*-struct/i

显示它的评估返回一个指针（显示为int）：

>> speed1-struct: make struct! [s [struct! [d [double]]]] [0.0]
>> speed1*-struct: make struct! [i [int]] third speed1-struct
>> speed1*: speed1*-struct/i
== 37329176