Question

有什么区别：

1

BigIntegerList problem;
   LinkedList x;
   BigIntegerNode curr;

   problem = new BigIntegerList();
   //get value of first line stating #of bigInts to read in
   problem.n = LinkedList.readInteger(in);
   //read big ints from text file based on first number passed, n
   for(int i=0; i<problem.n;i++)
   {
     x = new LinkedList();
     x.readBigInteger(in);
     //case that the list is empty
     if(problem.n<1)
     {
       problem.start = new BigIntegerNode(x,null);
       problem.rear = problem.start;
     //list is not empty, add nodes to rear
     }else
     {
       curr = new BigIntegerNode(x,null);
       problem.rear.next = curr; -----> this is where i get a nullpointer....
       problem.rear = curr;
     }
   }
   return problem;

2

forever begin
    **@posedge(clk) begin**
        if(vif.sof == 1) begin
             //some code here
        end
    end
 end

与@（posedge clk）相关的begin..end会有所作为吗？

Answer 1

程序代码中的@和#等event_controls本身并不是声明;它们是后面的陈述的前缀。语句可以是一个简单的语句，如赋值，或者像begin / end或fork / join这样的块。只要允许单个语句，就允许使用块。

当你写@(posedge clk);时，它真的是@(posedge clk) null_statement;

我应该给你足够的信息来回答你的问题，但这是另一种变化：

forever 
        @posedge(clk)
        if(vif.sof == 1) begin
           //some code here
        end

现在，如果分号跟@(posedge clk)之后有分歧，则会有很大差异。

@（posedge clk）begin end ....和@（posedge clk）之间有什么区别;？

1 个答案: