原创 systemverilog中阻塞和非阻塞事件以及同步
一、SV中非阻塞事件 module test; event ev1, ev2; //belong to logic function partalways@(ev1) $display("ev1 triggered at %0ts",
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原创 Chapter 3 – Top Block
In a normalproject, the development of the DUT is done separately from the development ofthe testbench, so there are t
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原创 Chapter 2 – Defining the verification environment
Beforeunderstanding UVM, we need to understand verification. Right now, wehave a DUT and we will have to interact with
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原创 Chapter 5 – Driver
The driver is a block whose role is tointeract with the DUT. The driver pulls transactions from the sequencer and sends
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原创 Chapter 6 – Monitor
The monitor is aself-contained model that observes the communication of the DUT with thetestbench. At most, it should o
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原创 Chapter 7 – Agent
We have bothmonitors, the sequencer and the driver, so the next step is to connect them up.This is a job for the agent.
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原创 vi/vim 註釋字體顏色的設置
在vi中被註釋掉的字體顏色默認爲藍色，無法看清，通過設置用戶根目錄下.vimrc文件，可以更改顏色的配置。 通過hi命令來修改註釋的顏色，使用方法： 設置關鍵字爲藍色非粗體 ：hi Identifier ctermfg =blue c
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原创 Chapter 11 – Running the simulation
To run thesimulation, we simply execute the provided Makefile in the GitHub repository: $ make -f Makefile.vcs The te
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原创 Chapter 10 – Test
At last, we needto create one more block: the test. This block will derive from the uvm_test classand it will have two
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原创 Chapter 4 – Sequences and sequencers
The first step in verifying a RTL design is defining what kind of data should be sent to the DUT. While the driver deal
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原创 消息隊列（Message Queue）基本概念
轉載：http://kb.cnblogs.com/page/537914/ 背景 之前做日誌收集模塊時，用到flume。另外也有的方案，集成kafaka來提升系統可擴展性，其中涉及到消息隊列當時自己並不清楚爲什麼要使用消息隊列。
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原创 Chapter 8 – Scoreboard
The scoreboardis a crucial element in a self-checking environment, it verifies the properoperation of a design at a fun
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原创 systemverilog中$value$plusargs的用法
package pa;...... endpackage program pr;import pa::*;......;initial begin.
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原创 SystemVerilog中有關class類的基礎知識
1、class中的變量、宏定義等稱爲類的屬性，函數和任務稱爲類的方法 2、聲明對象時可以指定input/output/inout/ref 3、複製對
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原创 C++中淺拷貝、深拷貝、對象的複製、對象的賦值
一、概念字面理解 1、淺拷貝：發生對象複製時，只是對對像張數據成員進行簡單的賦值。涉及到動態分配問題，如果按淺拷貝進行復制，不做特殊處理，複製完成後，兩個對象中涉及動態分配空間的變量，不管他們各自的空間，及空間中的內容都是完全一樣的，當對
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