如何返回不同类型的数组

本文关键字:同类型 数组 返回 何返回 | 更新日期: 2023-09-27 18:11:24

我在c#中遇到的高级问题是制作描述机器人控制网络数据包(Ethercat)的数据结构的单个副本,然后使用该单个数据结构从数据包集合中提取数据。

当尝试使用来自累积包的数据时,问题就出现了,因为通过指定类型的强制转换或调用函数存在数据结构的隐式重复。为了帮助解释这个目标,我编写了一个python程序来做我想做的事情,并希望帮助确定在c#中是否有可能做到这一点。

我在c#中遇到的挑战是单一的函数"get_vector",它返回一个变量数值类型的同构集合。此类型在包结构中定义,在python中可以使用而无需定义重新定义数据结构。

import struct
# description of the complete packet
class PACKET_STRUCTURE :
# given a field name and a list of packets, return a vector
# this is the function that seems impossible in C# because the type of what is returned changes
def get_vector(self, name, packet_list):
    # locate the packet definition by the name of the vector
    result = [x for x in self.packet_def if x.field_name == name]
    # without error checking, pos contains the location of the definition
    pos = result[0].position;
    # decode ALL the pacckets in the (encoded) packet list - returning a list of [time_sec, status, position
    # in C# this step is similar to  using Marshal.PtrToStructure to transform from byte[] to a struct
    decoded_packet_list = [struct.unpack(self.fmt_str, packet) for packet in packet_list];
    # from the list of decoded_packets, extract the desired field into its own list          
    vector = [decode[pos] for decode in decoded_packet_list]
    # in C# this is similar to:
    # var CS_vector = decode_packet_list.Select(item => item.field_name).ToArray();
    # so far in C# there is no duplication of the packet structure.
    # but after this point, assume I cast CS_vector to object and return it - 
    # to use the object, I've not figured out how to avoid casting it to some type of array
    # eg double[], int32[]                        
    return vector        
    def __init__(self):
        self.packet_def = list();
        self.fmt_str = "<";
        self.cnt = 0;
    # add description of single item to the structure
    def add(self, struct_def) :
        struct_def.position = len(self.packet_def);
        self.packet_def.append(struct_def);
        self.fmt_str += struct_def.type;                
    # create a simple packet based on a counter based on the defined structure
    def make_packet(self):
        vals = [self.cnt*10+x for x in range(0, len(self.packet_def))];
        self.cnt += 1;
        pk = apply(struct.pack, [self.fmt_str] + vals)
        # print len(pk), ["%c" % x for x in pk]
        return pk
    def get_names(self):
        return [packet_items.field_name for packet_items in self.packet_def];

# the description of a single field within the packet
class PACKET_ITEM  :
    def __init__(self, field_name, type):
        self.field_name = field_name
        self.type = type;
        # self.offset = 0;
        self.position = 0;

if __name__ == "__main__" :
    INT32 = "l";
    UINT16 = "H";
    FLOAT = "f";
    packet_def = PACKET_STRUCTURE();
    # create an example packet structure - which is arbituary and could be anything - it could even be read from a file
    # this definition is the ONLY defintion of the packet structure
    # changes here require NO changes elsewhere in the program
    packet_def.add(PACKET_ITEM("time_sec", FLOAT))
    packet_def.add(PACKET_ITEM ("status",UINT16))
    packet_def.add(PACKET_ITEM ("position",INT32))
    # create a list of packets
    pk_list = list()
    for cnt in range(0,10) :
        pk_list.append(packet_def.make_packet());
    ################################
    # get the vectors without replicating the structure
    # eg no int32[] position = (int32[])get_vector()
    name_list = packet_def.get_names();
    for name in name_list :
        vector = packet_def.get_vector(name, pk_list);
        print name, vector

如何返回不同类型的数组

答案是将数组存储在类型为List<dynamic>的集合中

从集合中返回元素的函数的返回类型也应该是动态的。

这篇文章试图澄清我没有理解的问题,下面是更完整的答案。