來源:洪士灝教授:改行做軟體與自學新專業
某位臉友來信談到轉行自學的問題,由於是常見問題,在徵得他本人同意之下,我公開回答他的提問,也希望諸位先進給與指教。
Q. 臉友提問:
我是您 Blog & FB 的追蹤者,長時間閱讀老師所寫的文章,而老師也時常點出業界的弊病及提出改善的方法,台灣須從高資本低技術轉型到低資本高技術產業,已是刻不容緩的事情。
本身研究所畢業之後從事顯示器背光模組機構設計 (兩年),近年來因技術被大陸追趕,不斷被搶單,而高層只想賺可以立竿見影的 easy money,不願意長期投資先端的研發。因為硬體開發成本高的特性,底下的工程師縱使再有想法也無力影響公司改變,最後是失望離開。
因想要轉職&政府補助,在資策會進修過 java web 設計,目前做銀行AP的軟體撰寫。但是從硬體產業跳到軟體產業,發現若不從事深度技術的研發,依然是在做代工。若是想往系統底層技術發展,勢必須在系統軟體鑽研,而根據老師你近日的文章表示,此領域若不能全心投入很難有顯著的成果。
雖然本身透過下班之餘自學,但以非相關背景進入此領域還是感受到困難。由於您在平行運算&異質系統有深刻的研究,希望可以請教老師有關學習&如何踏入這行的建議。
A....
04 March, 2016
01 March, 2016
WINDOWS 中的 SHELLCODE 定位與緩衝區溢出
StandardThis is not just another paper describing basics of buffer overflows. There are lots of publications about this topic; therefore it does not make any sense to describe it again. If you are familiar with exploiting buffer overflows on Windows platform, do not think that this article has nothing to offer you in this article. It shows some interesting methods, which can be used during writing an exploit...
X86 LINUX SHELLCODE 設計解密
StandardIn our previous paper, Buffer Overflows Demystified, we told you that there will be more papers on these subjects. We kept our promise. Here is the second paper from the same series. The paper is about the fundamentals of shellcode design and totally Linux 2.2 on IA-32 specific. The base principles apply to all architectures, whereas the details might obviously not.
在前文「緩衝區溢出解密」中,筆者保證會撰寫更多有關這個主題的文章,當前所見為此系列的第二篇。...
09 June, 2013
Bitwise 的各種運算
StandardCode:
// ap8,Bitwise 的各種運算
public class ap8
{
public static void main(String args[])
{
int i=13;
int j=14;
int k=i+j;
short l=27;
String x = "0000000000000000000000000000000000";
String y = "1111111111111111111111111111111111";
String z = "0101010101010100101010101010101010";
String w = "1010101010101011010101010101010101";
String pad = "--------------------------------";
...
運算式真假值
StandardCode:
// ap7,運算式真假值
public class ap7
{
public static void main(String args[])
{
int a=126;
int b=-2*a;
boolean c=true;
boolean d=!c; //false
System.out.println("a=" + a);
System.out.println("b=-2*a=" + b);
System.out.println("c=" + c);
System.out.println("d=!c=" + d);
System.out.print("((a>b)?c:d)="+((a>b)?c:d)+"\n"); //true
System.out.print("((ab)?d:c)+"\n");...
Subscribe to:
Posts (Atom)