Programming Concepts

Thinks of a situation where you need to do some work just once in a function irrespective of how many times that function is invoked then u can have a static variable which will keep a track of it. eg int main () { static int once = 0; if (!once) { once++; /*code need to be executed once*/ } /*reset of the code*/ }

throgh a socket programming bidirectional commumication.

OSI is Open System Interface Model. It is a concept which is used to make data transfer in steps. There are 7 layers and each layer interact with its counterpart. earlier a product made by any abc co was able to communicate with the same company product. to communicate with different co's products it was created. as it's name shows "OPEN System Interface"

Desktop application runs on a single machine. It could be a part of network. But Application and database resides on same machine. And only one user can access it at a time.Enterprise application is generally a multi-tier application. Many a concurrent users can access data and application at a time.

In GUI mode i.e winCVS, On menubar,there is checkout option.In that u can click and select the module that you want to check out.

In UNIX mode,you have to execute the following comand

>CVS checkout -r

Review if more formal than walkthrough .Walkthrough is done with a purpose of gain understanding of system and learning.

In laymans term walk-through is nothing but running thru the requirements while review is making an analysis of the progress and if the case may be a change has to be recommended

Please share your experience.

class singleten{ static int count; singleten() { if (count==1) { exit(0); } count++; }}singleten::count =0;void main(){singleten s1;singleten s2;}program is terminated as soon as second object is created.

There will be a lot of changes in fact. First of all, all the variables the function was suing needs to be pushed into the stack. (we need all these variables when we return from the other function). The Program counter will be pushed to stack too... (When the "return" of second function occurs we should know where to 'return' to)These are the basics which are bound to happen when a function calls another.

Buffer is something that is used to hold the data temporarly.

Ex-

If we are watching any thing online,then buffering of data takes place i.e for sometime data get stored in buffer then processed.

No! we can not run any simple program with out using any interface such as operatin systen or System software. but in some cases like Machine level languge or firmware run without any software because it directlly communicate with hardware and makes the hardware active so other program can run using his platform.

Raster data is a set of horizontal lines composed of individual pixels, used to form an image on a CRT or other screen.Raster data makes use of matrix of square areas to define where features are located. These squares are typically of uniform size, and their size determines the detail that can be maintained in the dataset.Raster data represent square areas, they describe interiors rather than boundaries. Vector data use X and Y coordinates to define the locations of points, lines, and areas (polygons) that correspond to map features, vector data tend to define centers and edges of features.They are excellent for capturing and storing spatial details

#include <stdio.h>

#include <conio.h>

#include <string.h>

void main()

{

char *str;

int i,len;

//not using any temp variable and assume we can use only string array and length

printf("Enter String : ");

scanf("%s",str);

len=strlen(str)-1;

for(i=0;i<strlen(str)/2;i++)

{

str[i]+=str[len];

str[len]=str[i]-str[len];

str[i]=str[i]-str[len--];

}

printf("Reverse String is : %s",str);

getch();

}

Please share your experience.

Cannot execute 13,000 FTS faster. First need to analyse the severity and priority of the functionality wise and then the selected test alone have to execute

Please share your views.

use the formulae

X(i,j)=Base+w[n(i-1)+(j-1)]

where m=7 ,n =9 ,i=2 ,j=3

hence 3000+8*[9(2-1)+(3-1)]

=3000+8*(9+2)

=3000+8*11=3088