Initialising Memories
The file_name and memory_nameare memory_start and memory_finish are optional, it missed out they default to the start index of the named memory and the end of the named memory respectively.
Memories can be stored in a file in the format shown below, the address is specified as @< address>, where the address is in hexadecimal.
@003
00000011
00000100
00000101
00000110
00000111
00001000
00001001
With the above file it can be seen if the memory is large it would become very tedious to work out the address of a specific byte, so it is normally a good idea to use milestones along the memory file, so a larger file may look something like the following:
@003
00000011
00000100
00000101
@006
00000110
00000111
@008
00001000
00001001
or if the data is contiguous, omit the address entirely.
Now that a memory file exists to access it, it has to be initialised for memory reading. This can be done by the following.
module testmemory;
reg [7:0] memory [9:0];
integer index; initial begin
$readmemb("mem.dat", memory); for(index = 0; index < 10; index = index + 1)
$display("memory[%d] = %b", index[4:0], memory[index]);
end
endmodule // testmemory
with the file mem.data as
1000_0001
1000_0010
0000_0000
0000_0001
0000_0010
0000_0011
0000_0100
0000_0101
0000_0110
0000_0000
EXERCISE
Store the above data in a file and run the above programme
Consider and understand the following code (run it to check):
module fileDemo;
integer handle, channels, index, rand;
reg [7:0] memory [15:0];
initial begin
handle = $fopen("mem.dat");
channels = handle | 1;
$display("Generating contents of file mem.dat");
$fdisplay(channels, "@2");
for(index = 0; index < 14; index = index + 1) begin
rand = $random;
$fdisplay(channels, "%b", rand[12:5]);
end
$fclose(handle);
$readmemb("mem.dat", memory);
$display("\nContents of memory array");
for(index = 0; index < 16; index = index + 1)
$displayb(memory[index]);
end
endmodule // fileDemo
Initialising Memories的更多相关文章
- (转) Written Memories: Understanding, Deriving and Extending the LSTM
R2RT Written Memories: Understanding, Deriving and Extending the LSTM Tue 26 July 2016 When I was ...
- 2018 Multi-University Training Contest 4 Problem K. Expression in Memories 【模拟】
任意门:http://acm.hdu.edu.cn/showproblem.php?pid=6342 Problem K. Expression in Memories Time Limit: 200 ...
- HDU6342-2018ACM暑假多校联合训练4-1011-Problem K. Expression in Memories
Problem K. Expression in Memories Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262 ...
- PatentTips - Wear Leveling for Erasable Memories
BACKGROUND Erasable memories may have erasable elements that can become unreliable after a predeterm ...
- Multi-processor having shared memory, private cache memories, and invalidate queues having valid bits and flush bits for serializing transactions
Multi-processor systems are often implemented using a common system bus as the communication mechani ...
- 杭电多校第四场 Problem K. Expression in Memories 思维模拟
Problem K. Expression in Memories Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262 ...
- SDM439平台出现部分机型SD卡不能识别mmc1: error -110 whilst initialising SD card【学习笔记】
SDM439平台出现部分机型SD卡不能识别mmc1: error -110 whilst initialising SD card 打印了如下的log: - ::>[ after ms - :: ...
- SD卡报错“error -110 whilst initialising SD card”
目前开发遇到了某些SD卡和TI的SOC芯片的驱动不协调的地方,具体表现为: uboot 阶段初始化mmc dev 1 没有任何串口信息输出,无法读写mmc Kernel阶段报错”SD卡初始化失败 er ...
- About memories in ASIC FPGA
1. Write first | Read First | No Change区别在于:en & wr的时候,dout是什么,三种case对应于: dout = din; dout = mem ...
随机推荐
- Qt数据类型转换
把QString转换为double类型 方法1.QString str="123.45"; double val=str.toDouble(); //val=123.45 方法2. ...
- An Overview of Complex Event Processing
An Overview of Complex Event Processing 复杂事件处理技术概览(一) 翻译前言:我在理解复杂事件处理(CEP)方面一直有这样的困惑--为什么这种计算模式是有效的, ...
- java线程学生进实训室
Instructor: Dr. Simina FlutureCSCI 34 CSCI 34 CSCI 34CSCI 34 0 Summer 201 ummer 201 ummer 201ummer 2 ...
- 最简单的ADABOOST人脸检测程序。COPY执行,前提是你配置OpenCV周围环境
#include "cv.h" #include "highgui.h" #include "stdio.h" void main() { ...
- HDU1698_Just a Hook(线段树/成段更新)
解题报告 题意: 原本区间1到n都是1,区间成段改变成一个值,求最后区间1到n的和. 思路: 线段树成段更新,区间去和. #include <iostream> #include < ...
- 重新想象 Windows 8 Store Apps (19) - 动画: 线性动画, 关键帧动画, 缓动动画
原文:重新想象 Windows 8 Store Apps (19) - 动画: 线性动画, 关键帧动画, 缓动动画 [源码下载] 重新想象 Windows 8 Store Apps (19) - 动画 ...
- 解决mongodb设备mongod命令不是内部或外部的命令
1:安装 去mongodb的官网http://www.mongodb.org/downloads下载32bit的包 解压后会出现下面文件 在安装的盘C:下建立mongodb目录,拷贝bin目录到该目录 ...
- Light OJ Dynamic Programming
免费做一样新 1004 - Monkey Banana Problem 号码塔 1005 - Rooks 排列 1013 - Love Calculator LCS变形 dp[i][j][k]对于第一 ...
- java基础程序题
发现自己初学java时保存在word里的练习题,哈哈,放博客里面来作为纪念吧~~~ [程序1] 题目:古典问题:有一对兔子,从出生后第3个月起每个月都生一对兔子,小兔子长到第四个月后每个月又生一对兔 ...
- Linux curl使用简单介绍 (转)
Curl是Linux下一个很强大的http命令行工具,其功能十分强大. 1) 二话不说,先从这里开始吧! $ curl http://www.linuxidc.com 回车之后,www.linuxid ...