分布式ROM 6w x 64d的COE文件如何编写

我试图找到一个很好的解释如何写一个COE文件，要么没有一个好的写，或者我没有找到它。
对此表单进行搜索并未对分布式ROM的COE文件做出任何好的解释。
我有一个6个地址输入（64个存储单元）由6个数据输出分布式ROM实例由核心生成，使用ISE 4.4为S6部分。
我能找到的几乎所有内容如下：
但是没有任何解释这个向量是什么以及如何编写它。
我想它与初选中的init字段类似，但是，嘿，让我们确定，因为它不在文档中。
**************************************************
****************
************分布式内存.COE文件示例***********
**************************************************
****************
;
分布式内存v2.0和。的示例内存初始化文件
;
后来。
;
;
此.COE文件与分布式内存核心的v1.0不兼容。
;
;
该示例指定深度= 32的内存的初始化值，
;
和宽度= 16。
在这种情况下，值以十六进制指定
;
格式。
memory_initialization_radix = 16;
memory_initialization_vector = 23f4 0721 11ff ABe1 0001 1 0A 0
23f4 0721 11ff ABe1 0001 1 0A 0
23f4 721 11ff ABe1 0001 1 A 0
23f4 721 11ff ABe1 0001 1 A 0;这在PG036中并没有好多少：
memory_initialization_vector。
每行存储器元件用二进制，十进制或十六进制数定义，具有等效二进制值，表示沿行宽度的单个存储器元件是否设置为“1”或“0”。每行存储器初始化
由逗号或空格分隔，直到记忆的深度。
不允许使用负值。示例COE文件：;
16x32分布式ROMmemory_initialization_radix = 16的样本初始化文件; memory_initialization_vector = 23f4 0721 11ff ABe1 0001 1 0A 023f4 0721 11ff ABe1 0001 1 0A 023f4 721 11ff ABe1 0001 1 A 023f4 721 11ff ABe1 0001 1 A 0;
地址值在哪里？
输出值在哪里？
底部2个条目似乎缺少位。
伦纳德



以下为原文

I'm trying to find a good explanation of how to write a COE file, either there is not a good write up or I'm not finding it. Doing a search on this form has not produced any good explanation on COE files for distributed ROM.

I have a 6 address inputs (64 memory locations) by 6 data outputs distributed ROM instance generated by core gen using ISE 4.4 for a S6 part. Pretty much all I can find is below:

But there is not any explanation what the vector is and how to write it. I imagine it is similar to the init field in the primatives but , hey lets be sure since it's not in the documentation.

******************************************************************************  Example of Distributed Memory .COE file  *****************************************************************************; Sample memory initialization file for Distributed Memory v2.0 and ; later.; ; This .COE file is NOT compatible with v1.0 of Distributed Memory Core.; ; The example specifies initialization values for a memory of depth= 32, ; and width=16. In this case, values are specified in hexadecimal; format.memory_initialization_radix = 16;memory_initialization_vector = 23f4 0721 11ff ABe1 0001 1 0A 0 23f4 0721 11ff ABe1 0001 1 0A 0 23f4 721 11ff ABe1 0001 1 A 0 23f4 721 11ff ABe1 0001 1 A 0;

This is in PG036 which is not much better:

memory_initialization_vector. Each row of memory elements are defined with a
binary, decimal, or hexadecimal number having an equivalent binary value that
represents whether an individual memory element along the width of the row is set to
‘1’ or ‘0.’ Each row of memory initialization is separated by a comma or white space, up
to the depth of the memory. Negative values are not allowed.
An example COE file:
; Sample Initialization file for a 16x32 distributed ROM
memory_initialization_radix = 16;
memory_initialization_vector =
23f4 0721 11ff ABe1 0001 1 0A 0
23f4 0721 11ff ABe1 0001 1 0A 0
23f4 721 11ff ABe1 0001 1 A 0
23f4 721 11ff ABe1 0001 1 A 0;

where are the address values? where are the output values? The bottom 2 entries seem to be missing bits.

Leonard