While I was unable to upgrade my laptop to 1 GB of memory, I was suffering from installing Sun Solaris 10 U3 + Oracle 10g R2 in my current book, I got a laptop for the competition, haha ...... Comfortable. The following is my Installation Record Based on Quick Installation officially provided by Oracle.Hardware:DELL D400CPU: PM 1.6 cache 2 MRAM: 512 M on board + 512 M for a single entry = 1024 MHDD: Fujitsu 40 GBLCD: 12 "1024*768LAN: 100 m lan + 54 M
UC official Developer Center: http://www.uc.cn/business/developer.shtml
U3 kernel customization Parameters:
Landscape: Forced landscape Screen
Portrait: Force portrait Screen
Usage:
Supported versions: 8.6 +
2. full-screen control)
Parameters:
Yes: force full screen
Usage:
Supported versions: 8.6 +
3. browsermode)
Description: The application mode is a comprehensive switch set for web applications and game developers. The meta tag is us
C: \ Keil \ C51 \ Inc \ cypress D: \ cypress suite USB 3.4.7 \ firmware \ Inc ?? 1. Found the suite3.4.7 folderFx2.hThe file matches the fx2.h file carried inside Keil (the same ). ?? 2. TwoFx2regs. hFile does not match. suite3.4.7 comes with a newer version than Keil U3. ?? 2.1. the name (long) of the automatic pointer 1 in the Suite version is different from that in the Keil version (short), but it is well compatible. The Compatibility proces
In Ubuntu, go to odroid U3 Board eMMC card Burning write Xubuntu image file, the main steps are as follows:
1. Download the image file to burn, as required, I downloaded the desktop version of the XUBUNTU-13.04-DESKTOP-ARMHF_ODROIDU2XXXX.IMG.XZ.
2. Unzip the file and enter the command under Terminal:
Xz-d XUBUNTU-13.04-DESKTOP-ARMHF_ODROIDU2_20130503.IMG.XZ
3. View the file system and mount points. After inserting the eMMC card, enter the command:
Events are the core concept of GUI applications. GUI applications perform their functions by executing events. The event loops, event distribution, and event interceptions designed around events are allGuiThe core part of the framework and the basis
I have been writing this article for a long time. I found this article in my mailbox today because I have left my home AIX company. Article So paste it here.Environment: AIX 5.3.Go to the Subversion website and download the related packages. 1. Install APR [/U3/up070427/. cldev/pkgs/APR]>./configure -- prefix =/U3/up070427/. cldev/httpd[/U3/up070427/. cldev/pkgs/
to 3 persons
User ID
User Name
U1
Xiaoming
U2
Datong
U3
Wuming
(6) Relationship between users and roles:
User ID
Role ID
Inheritance number
U1
R1
0
U2
R2
0
U2
R1
1
U3
R4
0
U3
R3
1
the route Method description and whether to turn on data permissions---> Create roles---> Bind a created route to a created role---> Add a user---> role to create your own signature---> each request carries a signature--- > Bring this signature and resource binding when creating a resource
See what permissions your individual has
// userid 查询 RoleDoc 然后把所有返回的 PathMethods 做聚合 // 通过 PathMethods 也可以结合 RouterDoc 查询具体的数据验证权限
View permissions granted by others
// userid 查询 SignDoc 然后对返回,做
Cardinal Spline
This is relatively simple, with one end point, one start point, and two control points.
Points between the end point and the start point are implemented by interpolation. interpolation functions:
P(U) =PK-1 (-S * u + 2 S * u) +PK [(2-s) u * u + (S-3) u * u + 1] +PK + 1 [(S-2) * u + (3-2 s) * u + S * u] +PK + 2 (S * u-s * u * U)
Code implementation:
void getInterpolation(point p1,point p2,point p3,point p4,float s,float u,point *result){ float u2=u*u; float
resulting from the generation of x into the equations; x0 is the initial value of the unknown vector. If you want to solve the following equation group:
F1 (U1,U2,U3) = 0
F2 (u1,u2,u3) = 0
f3 (u1,u2,u3) = 0Then func can be defined as follows:
def func (x):
u1,u2,u3 = x return
[F1 (U1,U2,
Article Title: tiptop automatic backup script backup. sh. Linux is a technology channel of the IT lab in China. Includes basic categories such as desktop applications, Linux system management, kernel research, embedded systems, and open source.
# Create a crontab Scheduler
Crontab-e
0 *** sh/u3/backup/tool/backup. sh # automatic backup at every day
0 6 *** sh/bin/ftpauto # automatically upload at every morning
# Backup. sh
LANG = en_US
Ans = 'date | a
"[Root @ localhost ~] # Tail-1/etc/passwdU1: x: 4323: 4323: UUU:/home/u1:/bin/csh[Root @ localhost ~] # Groups u1U1: u1 g1 g2
2. Create the following user, group, and group member relationships
Group named adminsUser u2, using admins as the affiliated groupUser u3 also uses admins as the affiliated groupUser u4 cannot log on to the system interactively. The default home directory is/test/u4, And the passwords of u2,
cautionLet's look at an obvious DFS:
Initialization
Starting from a point u, add this point to a set and set it to U. Go through all the points that are connected to him, put them in another set S1, and then go through the first DFS
First time DFS:
Select a point U1 from the S1, and this point must be connected to any point in the set U. Add the U1 in the collection S1 to the collection S2, and add the U1 to the collection U for the second time DFS
Two times
You can choose from the following methods:First, using Eclipse, right-click the project to export the jar.Second, using Eclipse, right-click the project to export the runnable jar.Third, use the Eclipse plugin fat jar to export the executable jar package.There is now a development scenario:First project: A1Second item: B2Item Three: U3Several conditions and relationships:1 U3 references third-party libraries Log4j.jar and common-Io.jar. 2B2 reference
If the measured amplification circuit of several transistors three pin to ground point U1.U2.U3 are the following values, try to determine whether they are silicon tube or germanium tube, is NPN or PnP tube. and determine three electrodes of 1. u1=2.5v. u2=6v. u3=1.8v 2. u1=-6v. U2=-3v. u3=-2.8v I would like to draw up a detailed answer procedure.
A: (authoritat
component), after 8 bit quantization, the uncompressed pixel occupies 3 bytes.
The following four pixels are: [Y0 U0 V0] [Y1 U1 V1] [Y2 U2 V2] [Y3 U3 V3]
The stream stored is: Y0 U0 V0 Y1 U1 V1 Y2 U2 V2 Y3 U3
(2) YUV 4:2:2
The sampling rate of each color-difference channel is half that of the luminance channel, so the chroma sampling rate in the horizontal direction is only half of the 4:4:4. For uncom
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