一个变量如何确定它的值而不在systemverilog中初始化它? [英] how does a variable to be decided it's value without it's initialize in systemverilog?

问题描述

现在我正在尝试挖掘 systemverilog 如下

Now I'm trying to digging the systemverilog as the below

denaliCdn_ahbTransaction burst1;

task sendTransfers;
      
burst1= new;       
burst1.Direction = DENALI_CDN_AHB_DIRECTION_WRITE;
burst1.FirstAddress = 32'h4020;//16416 M3 and M0 to S1
burst1.Kind = DENALI_CDN_AHB_BURSTKIND_INCR4;
burst1.Size = DENALI_CDN_AHB_TRANSFERSIZE_HALFWORD;
burst1.Data = new [8];

foreach (burst1.Data[ii])
burst1.Data[ii] = ii;

void'(activeMaster1.transAdd(burst1,0));
....
endtask

特别是，从这里开始，如何在未初始化的情况下确定 ii 的值?

Especially, from here , how does ii be decided it's value without initialized ?

foreach (burst1.Data[ii])
burst1.Data[ii] = ii;

一个变量如何在 systemverilog 中没有初始化的情况下确定它的值?

how does a variable to be decided it's value without it's initialize in systemverilog?

推荐答案

您的问题有一个通用的答案和一个特定的答案.

There is a general answer and a specific answer to your question.

一般的答案是 System-Verilog 中的所有类型都初始化为特定值:

The general answer is that all types in System-Verilog initialise to specific values:

int    0
real   0.0
string ""
logic  1'bx

但是，您问题的具体答案是您具体标识的代码

However, the specific answer to your question is that the code you specifically identify

foreach (burst1.Data[ii])
  burst1.Data[ii] = ii;

实际上是在执行初始化而不是依赖它.动态数组burst1.data由8个元素组成

is actually performing initialisation rather than relying on it. The dynamic array burst1.data is constructed with 8 elements in the line

burst1.Data = new [8];

线

foreach (burst1.Data[ii])

遍历该数组，循环 8 次，每个元素一次.变量 ii 取数组中每个元素的索引值，即 0 到 7.行

iterates over that array, looping 8 times, once for each element. The variable ii takes the value of the index of each element in the array, ie 0 to 7. The line

  burst1.Data[ii] = ii;

将数组的元素 ii 设置为值 ii，从而初始化数组.

Sets element ii of the array to the value ii and so is initialising the array.

这篇关于一个变量如何确定它的值而不在systemverilog中初始化它?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！