VHDL:切片数组的各个部分 [英] VHDL: slice a various part of an array

问题描述

我有一个 STD_LOGIC_VECTOR(0 到 639).在传入信号上，我必须遍历这个向量并获得它的下 2 位.

I have a STD_LOGIC_VECTOR (0 to 639). On an incoming signal I have to iterate through this vector and get next 2 bits of it.

我正在尝试使用整数counter来做类似的事情:

I'm trying to make something like, with an integer counter:

counter := counter+1;
MyVar := Data((counter*2) to ((counter*2)+1));

我得到以下信息:

错误 (10394):module.vhd(227) 处的 VHDL 错误:范围的左边界必须是常数

Error (10394): VHDL error at module.vhd(227): left bound of range must be a constant

upd:@user1155120 提出以下建议:将向量的每一位写入 MyVar 的每一位对应位

upd: the following was suggested by @user1155120: writing every single bit of vector to every single corresponding bit of MyVar

MyVar(0) := Data(counter * 2);
MyVar(1) := Data(counter * 2 + 1);

只要我使用 2 位 MyVar 就可以正常工作，但是如果我想使用 16-32-80 位变量怎么办?问题避免了，但没有解决.

Works fine as long as I use the 2bit MyVar, but what if i want to use a 16-32-80bit variable? Problem avoided, but not solved.

推荐答案

谷歌搜索显示错误消息来自 Quartus II(参见 ID:10394).它提供的 LRM 参考并不是特别有用，这是对合成的限制，您无法为多路复用器定义可变宽度的字长.不够聪明，无法检测到引用计数器的两个边界.

Googling shows the error message is from Quartus II (See ID: 10394). The LRM reference it provides isn't particularly helpful, it's a limitation on synthesis that you can't define a variable width word size for a multiplexer. Not smart enough to detect both bounds are referenced to counter.

如果你分别为 MyVar 的每一位表达一个多路复用器会发生什么?(索引名称而不是切片名称，两个变量赋值给 MyVar(1) 和 MyVar(0)).

What happens if you express a multiplexer for each bit of MyVar separately? (Indexed name instead of slice name, two variable assignments to MyVar(1) and MyVar(0)).

MyVar(0) := Data(counter * 2);
MyVar(1) := Data(counter * 2 + 1);

这个线程需要帮助解决奇怪的错误"范围的左边界必须是常量 " 建议使用循环来分配目标切片范围的每一位，这也是在每个元素的基础上使用索引名称.

This thread need a help with strange error" left bound of the range must be a constant " suggests using a loop to assign each bit of the target slice range, which is also using an indexed name on a per element basis.

for i in MyVar'RANGE loop
    MyVar(i) := Data(counter * 2 + i);
end loop;

counter 只需要索引范围的一半，你总是把它乘以 2.

counter only needs half the index range, you're always multiplying it by 2.

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