在Verilog中,非阻塞赋值(non-blocking assignment)通常用于时序逻辑的描述,其语法为<=。关于非阻塞赋值中的动态类型(dynamic type),需要明确的是,Verilog本身是一种静态类型语言,不支持像某些动态类型语言(如Python)那样的动态类型特性。 然而,在非阻塞赋值中,我们仍然可以处理不同类型的信号,但这些信号的类型在编...
非阻塞赋值(Non-blocking Assignment)是个伪需求(2) https://mp.weixin.qq.com/s/5NWvdK3T2X4dtyRqtNrBbg 13hope: 个人理解,Verilog本身只是“建模”语言。具体到阻塞/非阻塞,只规定了两种赋值语句的行为。所以无论怎么写,仿真器和综合器都不会报错。但是存在两个问题,所描述的行为是否有物理电路与之对应;...
https://mp.weixin.qq.com/s/mH84421WDGRb7cuU5FEFIQ Verilog的赋值很是复杂,包括: 1. Continuous assignment; 2. Procedural assignment: a. Blocking Assignment; b. No
Verilog里有连续赋值(Continuous assignment) ,过程赋值(Procedural assignment),还有过程连续赋值(Procedural Continuous assignment)。 过程赋值又有阻塞赋值和非阻塞赋值。 "=" 表示阻塞过程赋值(Blocking Procedural assignment), "<="表示非阻塞过程赋值(Non-blocking Procedural assignment)。 过程赋值的过程可以理解为两...
还有一种带repeat循环的Intra-Assignment Delays的过程赋值,例如: always @(IN) OUT <= repeat(8) @(posedge clk) IN; 可用于阻塞或非阻塞赋值。 为了能更好理解,代码加了一些$display,以显示语句执行的时间,据此统计次数 modulecomb_logic_assign16;bitin;bito1,o2,o3,o4,o5,o6;initialbeginin=0;o1=0;...
// referencedesigner.com // 4 bit ring counter example module four_bit_ring_counter ( input clock, input reset, output [3:0] q ); reg[3:0] a; always @(posedge clock) if (reset) a = 4'b0001; else begin a <= a<<1; // Notice the non blocking assignment a[0] <=a[3]; ...
So first the blocking assignment statements using =. `timescale 1ns/1ps module blocking; reg p,q,r ; initial begin p = #10 1'b1;// Executed at time t = 10 units q = #30 0'b0;// Executed at time t = 10 + 30 = 40 units r = #20 1'b1;// Executed at time t = 40 +...
A VHDL variable assignment is working similarly to blocking procedural assignments in Verilog. They also work for synthesis. The main difference is the process local variable scope. "Global variables" exist in VHDL but are rarely supported for synthesis. Don't confuse Verilog blocking...
ao68000.v line 2841 shows result assigned with a non-blocking assignment. a068000.v line 2852 shows result assigned with a blocking assignment. I don't know what that the Altera synthesizer does with mixed assignment like this but Xilinx...
as long as you are using the non-blocking assignment it behaves like Daixiwen says. The evaluation take s place at the end of the cycle and the last assignment wins, but maybe you can re-write your code in order to make it easier to read. E.g a defalut should be a "default" wi...