Xenomai Timer :Xenomai has two time sources: the sytem timer, which counts the number of nanoseconds since 1970, and a hardware dependent high resolution counter which counts the time since an unspecified point in time (usually the system boot time). This hardware dependent high resolution counter is called “tsc” on a PC, and gave its name to Xenomai native API calls.rt_timer_tsc returns the value of this hardware dependent high-resolution counter.
rt_timer_info returns the same thing in the tsc member of the RT_TIMER_INFO structure, and the value of the system timer at exactly the same time as when the high-resolution counter was read.

This allows to have a correspondence between the two time sources.

rt_alarm_inquire is not related to this and returns some information
about a given alarm. Now, if you allow me, a little advice for the implementation of a “timer library”: you could be tempted to create only one periodic alarm object with Xenomai, and to manage a timer list yourself. Don’t do this. Creating an alarm object for each timer library object make Xenomai aware of the existence of all your application timers, this has several

advantages:
– it gives you information about all your timers in /proc/xenomai
– it allows Xenomai to use its anticipation algorithm for all your timers
– if you are concerned about the scalability of Xenomai timers list
management, you can check the options in the “Scalability” menu of
Xenomai configuration menu (“Real-time subsystem” sub-menu of kernel
configuration menu).
more about timers

Xenomai POSIX skin supports two clocks:
CLOCK_REALTIME maps to the nucleus system clock, keeping time as the amount of time since the Epoch, with a resolution of one system clock tick.

CLOCK_MONOTONIC maps to an architecture-dependent high resolution counter, so is suitable for measuring short time intervals. However, when used for sleeping (with clock_nanosleep()), the CLOCK_MONOTONIC clock has a resolution of one system clock tick, like the CLOCK_REALTIME clock.[1]

Imagine you sell those metallic digits used to number houses, locker doors, hotel rooms, etc. You need to find how many of each digit to ship when your customer needs to number doors/houses:
1 to 100
51 to 300
1 to 2,000 with zeros to the left
The obvious solution is to do a loop from the first to the last number, convert the counter to a string with or without zeros to the left, extract each digit and use it as an index to increment an array of 10 integers.
I wonder if there is a better way to solve this, without having to loop through the entire integers range.
GitHub Link
Thanx to mathematician’s Post that helped me to understand the complexity of problem, i just understood the equation and converted it to code. Enjoy !

There are a row of houses, each house can be painted with three colors red, blue and green. The cost of painting each house with a certain color is different. You have to paint all the houses such that no two adjacent houses have the same color. You have to paint the houses with minimum cost. How would you do it?

Note: Painting house-1 with red costs different from painting house-2 with red. The costs are different for each house and each color.

Approach:
Dynamic Programming solution:
we can paint the i’th house with blue, red or green.

Enter No. of Houses to paint : 6
1 4 5
1 3 9
6 7 2
9 6 4
3 5 2
6 7 4

Minimum Cost to paint all houses are 20

#include <stdio.h>
#include <stdarg.h>
#define COLOR 3
#define MAX(a,b) ((a) > (b) ? a : b)
#define MIN(a,b) ((a) < (b) ? a : b)
#define R 0
#define B 1
#define G 2
#define Color 3


int min( int num, ... )
{
    va_list arguments;
    int min=0;
    int tmp=0;
    /* Initializing arguments to store all values after num */
    va_start ( arguments, num );
    /* Sum all the inputs; we still rely on the function caller to tell us how
     * many there are */
    min = va_arg ( arguments, int );
    //printf("\n1st|%d|\n",min);
    for ( int x = 1; x < num; x++ )
    {
		tmp = va_arg(arguments,int);
		min = MIN(min,tmp);
    }
    va_end ( arguments );
    // Cleans up the list
	//printf("\nMin is %d", min);
    return min;
}
typedef struct {
    int  color;
    int cost;
    //struct nodee *root;
}node;
int main()
{
	node cost[3];
	int house = 3;
	printf("Enter No. of Houses to paint : ");
	scanf("%d",&house);
	int color[house][COLOR];
//	int ji=0;
	for(int i=0; i<house;++i){
		for(int j=0; j<COLOR;++j){
				scanf("%d",&color[i][j]);
//				color[i][j]= ++ji;
			}

		}
	//cost[0].cost=color[0][R];cost[1].cost=color[0][B];cost[2].cost=color[0][G];
	for(int i= 0; i< Color ;i++){
			cost[i].cost=color[0][i];
			cost[i].color = i;
		} // Initial Submission of values, and colour
//	printf("\nA.cost\t%d(%d)\tB.cost\t%d(%d)\tC.cost\t%d(%d)\n",cost[0].cost,cost[0].color,cost[1].cost,cost[1].color,cost[2].cost,cost[2].color);
	//A.color = R; B.color = B; C.color = G;
	int tmp_B=0,tmp_G=0,tmp_R=0;
	//printf("\nR_cost\t%d\tB_cost\t%d\tG_cost\t%d\n",R_cost,B_cost,G_cost);
	for(int i=1; i<house;++i)
		{
			for(int j= 0; j< Color ;j++){
				if(cost[j].color == R){
						if(color[i][B] < color[i][G]){
							cost[j].cost = cost[j].cost + color[i][B];
							cost[j].color = B;
							//add_linked_list();
							//cost[j].root = malloc(sizeof(struct node));
							//cost[j].root->next = 0;
							//cost[j].root->x = color[i][B];
						}else{
								cost[j].cost = cost[j].cost + color[i][G];
								cost[j].color = G;
						}
					}else if(cost[j].color == B){
							if(color[i][R] < color[i][G]){
								cost[j].cost = cost[j].cost + color[i][R];
								cost[j].color = R;
							}else{
									cost[j].cost = cost[j].cost + color[i][G];
									cost[j].color = G;
								}
						}else{
								//Case for G
								if(color[i][R] < color[i][B]){
									cost[j].cost = cost[j].cost + color[i][R];
									cost[j].color = R;
									}else{
											cost[j].cost = cost[j].cost + color[i][B];
											cost[j].color = B;
										}
							}
//			printf("\nA.cost\t%d(%d)\tB.cost\t%d(%d)\tC.cost\t%d(%d)\n",cost[0].cost,cost[0].color,cost[1].cost,cost[1].color,cost[2].cost,cost[2].color);
			}
			/*
			if(A.color == )
			//R_cost = min(2,color[i][B],color[i][G])+R_cost;
			if(color[i][B] < color[i][G]){

				tmp_B = R_cost + color[i][B];
				}else{
						tmp_G = R_cost + color[i][G];
					}

			//B_cost = min(2,color[i][R],color[i][G])+B_cost;
			if(color[i][R] < color[i][G]){

				tmp_R = B_cost + color[i][R];
				}else{
						tmp_G = B_cost + color[i][G];
					}
			//G_cost = min(2,color[i][B],color[i][R])+G_cost;
			if(color[i][R] < color[i][B]){

				tmp_R = G_cost + color[i][R];
				}else{
						tmp_B = G_cost + color[i][B];
					}
			R_cost = tmp_R;
			B_cost = tmp_B;
			G_cost = tmp_G;
			//printf("\n G %d:%d",color[i][B],color[i][R]);
			*/
//			printf("\nA.cost\t%d(%d)\tB.cost\t%d(%d)\tC.cost\t%d(%d)\n",cost[0].cost,cost[0].color,cost[1].cost,cost[1].color,cost[2].cost,cost[2].color);
		}
	printf("\nMinimum Cost to paint all houses are %d\n",min(3,cost[0].cost,cost[1].cost,cost[2].cost));
}

Hello guys let's we have an syntax and related values let's see how this works. Evaluate the given syntax 
z = ++x + y-- - ++y - x-- - x-- - ++y - x--
where x = 7 y = -3

z = ((((((++x + y--) - ++y) - x--) - x--) - ++y) - x--)
so first(++x + y--) will be evaluated
 (++x + y--)
   8 + (-3)        x=8 y = -4
 z = (((((5 - ++y) - x--) - x--) - ++y) - x--)
 (5 - ++y)
  5 - ( -3)  x=8 y = -3

 z = ((((8 - x--) - x--) - ++y) - x--)
 (8 - x--)
  8 - 8      x =7 y = -3 
 z = (((0- x--) - ++y) - x--)
 (0- x--)
  0 - 7      x =6 y = -3
 z = (((-7 - ++y) - x--)
 (-7 - ++y)
  -7 - (-2)  x = 6 y = -2
 z = (-5 - x--)
 -5 -6       x =5 y -2
 z = -11

So answer is -11

Hello, Guys let’s have a fun, i am giving  you a code just tell me what will be the output of this code if you enter there parameters.

For Test case 1 -->Q
For Test case 2 -->QQ

Please tell why putchar() function is not working and if it is working why it is not taking any Char input?


#include <stdio.h>
main()
{
putc(getc(stdin),stdout);
putchar(getchar( ));
return 0;
}


Explain the output for these test cases.

let’s your ns2 is installed in /home/arun/ns/ns-allinone2.34
I suppose that ns-2.34 is installed with the compiler gcc-4.3.
download both files in  …//ns.2.34
1. LeachCode
2.Installing_file
and make sure that in leach-setup.sh please change the path in this file and run

arun@merom:~$ cd /home/arun/ns/ns-allinone-2.34/ns-2.34/
arun@merom:/home/arun/ns/ns-allinone-2.34/ns-2.34$ sudo bash leach-setup.sh
arun@merom:/home/arun/ns/ns-allinone-2.34/ns-2.34$./configure
arun@merom:/home/arun/ns/ns-allinone-2.34/ns-2.34$make clean
arun@merom:/home/arun/ns/ns-allinone-2.34/ns-2.34$make depend
arun@merom:/home/arun/ns/ns-allinone-2.34/ns-2.34$make
arun@merom:/home/arun/ns/ns-allinone-2.34/ns-2.34$make install

after running  “make ” you will get error

[trace/cmu-trace.cc: In member function ‘void CMUTrace::format(Packet*, const char*)’:
trace/cmu-trace.cc:1327: error: ‘format_rca’ was not declared in this scope

run again all commands it will work 🙂

for more please refer

http://forum.wsnlab.ir/pdf/LEACH.inNS2.wsnlab.ir.pdf

Hello , one more important stuff to understand

I’m sure that this post will be as interesting and informative to C virgins (i.e. beginners) as it will be to those who are well versed in C. So let me start with saying that extern keyword applies to C variables (data objects) and C functions. Basically extern keyword extends the visibility of the C variables and C functions. Probably that’s is the reason why it was named as extern.

Though (almost) everyone knows the meaning of declaration and definition of a variable/function yet for the sake of completeness of this post, I would like to clarify them. Declaration of a variable/function simply declares that the variable/function exists somewhere in the program but the memory is not allocated for them. But the declaration of a variable/function serves an important role. And that is the type of the variable/function. Therefore, when a variable is declared, the program knows the data type of that variable. In case of function declaration, the program knows what are the arguments to that functions, their data types, the order of arguments and the return type of the function. So that’s all about declaration. Coming to the definition, when we define a variable/function, apart from the role of declaration, it also allocates memory for that variable/function. Therefore, we can think of definition as a super set of declaration. (or declaration as a subset of definition). From this explanation, it should be obvious that a variable/function can be declared any number of times but it can be defined only once. (Remember the basic principle that you can’t have two locations of the same variable/function). So that’s all about declaration and definition.
Now coming back to our main objective: Understanding “extern” keyword in C. I’ve explained the role of declaration/definition because it’s mandatory to understand them to understand the “extern” keyword. Let us first take the easy case. Use of extern with C functions. By default, the declaration and definition of a C function have “extern” prepended with them. It means even though we don’t use extern with the declaration/definition of C functions, it is present there. For example, when we write.

    int foo(int arg1, char arg2);

There’s an extern present in the beginning which is hidden and the compiler treats it as below.

    extern int foo(int arg1, char arg2);

Same is the case with the definition of a C function (Definition of a C function means writing the body of the function). Therefore whenever we define a C function, an extern is present there in the beginning of the function definition. Since the declaration can be done any number of times and definition can be done only once, we can notice that declaration of a function can be added in several C/H files or in a single C/H file several times. But we notice the actual definition of the function only once (i.e. in one file only). And as the extern extends the visibility to the whole program, the functions can be used (called) anywhere in any of the files of the whole program provided the declaration of the function is known. (By knowing the declaration of the function, C compiler knows that the definition of the function exists and it goes ahead to compile the program). So that’s all about extern with C functions.
Now let us the take the second and final case i.e. use of extern with C variables. I feel that it more interesting and information than the previous case where extern is present by default with C functions. So let me ask the question, how would you declare a C variable without defining it? Many of you would see it trivial but it’s important question to understand extern with C variables. The answer goes as follows.

    extern int var;

Here, an integer type variable called var has been declared (remember no definition i.e. no memory allocation for var so far). And we can do this declaration as many times as needed. (remember that declaration can be done any number of times) So far so good. 
Now how would you define a variable. Now I agree that it is the most trivial question in programming and the answer is as follows.

    int var;

Here, an integer type variable called var has been declared as well as defined. (remember that definition is the super set of declaration). Here the memory for var is also allocated. Now here comes the surprise, when we declared/defined a C function, we saw that an extern was present by default. While defining a function, we can prepend it with extern without any issues. But it is not the case with C variables. If we put the presence of extern in variable as default then the memory for them will not be allocated ever, they will be declared only. Therefore, we put extern explicitly for C variables when we want to declare them without defining them. Also, as the extern extends the visibility to the whole program, by externing a variable we can use the variables anywhere in the program provided we know the declaration of them and the variable is defined somewhere.
Now let us try to understand extern with examples.
Example 1:

Analysis: This program is compiled successfully. Here var is defined (and declared implicitly) globally.
Example 2:

Analysis: This program is compiled successfully. Here var is declared only. Notice var is never used so no problems.
Example 3:

Analysis: This program throws error in compilation. Because var is declared but not defined anywhere. Essentially, the var isn’t allocated any memory. And the program is trying to change the value to 10 of a variable that doesn’t exist at all.
Example 4:

Analysis: Supposing that somefile.h has the definition of var. This program will be compiled successfully.
Example 5:

Analysis: Guess this program will work? Well, here comes another surprise from C standards. They say that..if a variable is only declared and an initializer is also provided with that declaration, then the memory for that variable will be allocated i.e. that variable will be considered as defined. Therefore, as per the C standard, this program will compile successfully and work.
So that was a preliminary look at “extern” keyword in C.
I’m sure that you want to have some take away from the reading of this post. And I would not disappoint you. 
In short, we can say
1. Declaration can be done any number of times but definition only once.
2. “extern” keyword is used to extend the visibility of variables/functions().
3. Since functions are visible through out the program by default. The use of extern is not needed in function declaration/definition. Its use is redundant.
4. When extern is used with a variable, it’s only declared not defined.
5. As an exception, when an extern variable is declared with initialization, it is taken as definition of the variable as well.

Copied from GeekforGeeks Thanx GeekforGeeks.

Given a string containing only digits, restore it by returning all possible valid IP address combinations.
For example:
Given “25525511135”,
return [“255.255.11.135”, “255.255.111.35”]{hint:recursion,backtrack}

Here is the Solution ……
it was taken me about 5 hours to solve see !!!

//cc Arun Kumar Gupta

import java.util.*;
public class AllIpAddress
{
public static void main (String [] args)
{
Scanner sc = new Scanner(System.in);
String input = sc.next();
int[] intArray = new int[input.length()];
int length = input.length();

for (int i = 0; i < input.length(); i++) {
intArray[i] = Character.digit(input.charAt(i), 10);
//System.out.println(intArray[i]);
}
int i = 0 , j = 1 , k = 2;
AllIpAddress ip = new AllIpAddress();
ip.Ipaddress(intArray , i , j , k , length-1);

}
void Ipaddress(int [] intArray , int i , int j , int k , int length)
{
try{
int where = 0 ,change = 0;
int p1 =createIP( intArray , -1 , i);
int p2 =createIP( intArray , i ,  j );
int p3 =createIP( intArray ,  j , k );
int p4 =createIP( intArray ,  k ,  length );
//System.out.println(“**********Garbage”+p1+”.”+p2+”.”+p3+”.”+p4);
//System.out.println(“i = “+i+”j = “+j+”k = “+k);
if((p1<=255)&&(p2<=255)&&(p3<=255)&&(p4<=255)&&(p4>0))
{
System.out.println(p1+”.”+p2+”.”+p3+”.”+p4);
}
if(p4 >255)
{
//if(k != j)
//++k;
where =4;
//System.out.println(“Problem at P4”);
}
else if(p3>255)
{
//if(j != i)
//++j;
//–k;
where = 3;
//System.out.println(“Problem at P3”);
}
else if(p2>255)
{
//++i;
where = 2;
//System.out.println(“Problem at P2”);
}
else if(p1>255)
{

//System.out.println(“Problem at P1”);
System.exit(0);
}

if((k == length) && (j+1 == k))
{
//System.out.println(“if((k == length) && (j+1 == k))”);
//System.out.println(i+”_”+j+”_”+k);
++i;
j = i+1;
k = j+1;
change = 1;
//System.out.println(i+”_”+j+”_”+k);

}
if((k == length) &&(change == 0) )
{
//System.out.println(“if((k == length) &&(change == 0) )”);
++j;
k = j+1;
change = 0;
}
else if(change == 0)
{
if( (k) < length )
++k;
}
/*if(where ==4)
{
++k;
}
if(where ==3)
{
++j;
–k;
}
if(where ==2)
{
++i;
–j;
}
/*if(k==length)
{
–k;
–j;
}*/
Ipaddress(intArray , i , j , k , length);
}
catch(ArrayIndexOutOfBoundsException e){
//System.out.println(“Exception thrown  :” + e);
}
}
static int createIP(int [] intArray , int start , int end)
{
//System.out.println(“I am at createIP()start = “+start+”\tend = “+end);
int val = 1 , fina = 0;
for(int e =end ; e>start ; –e)
{
fina = intArray[e]*val + fina;
val = val*10;
}
return fina;
}
}

• Octal constants are written with a leading zero – 015.
• Hexadecimal constants are written with a leading 0x – 0x1ae.
• Long constants are written with a trailing L – 890L.

Character constants are usually just the character enclosed in single quotes; ‘a’, ‘b’, ‘c’. Some characters can’t be represented in this way, so we use a 2 character sequence as follows.

'\n' newline
'\t' tab
'\\' backslash
'\'' single quote
'\0' null ( Usedautomatically to terminate character string )

Defining Constants

int const a = 1;
const int a =2;

• You can declare the const before or after the type. Choose one an stick to it.
• It is usual to initialise a const with a value as it cannot get a value any other way.

#define TRUE  1
#define FALSE  0
#define NAME_SIZE 20

void strcpy(char *buffer, char const *string)

The enum Data type

enum colors {RED, YELLOW, GREEN, BLUE};

enum colors {RED=1, YELLOW, GREEN=6, BLUE };

#include 

int main() {
  enum {RED=5, YELLOW, GREEN=4, BLUE};

  printf("RED = %d\n", RED);
  printf("YELLOW = %d\n", YELLOW);
  printf("GREEN = %d\n", GREEN);
  printf("BLUE = %d\n", BLUE);
  return 0;
}

This will produce following results

  RED = 5
  YELLOW = 6
  GREEN = 4
  BLUE = 5

Copied From Tutorial C

as we know by default the pointers are near for example int *p is a near pointer… size of near pointer is 2 bytes in case of 16 bit compiler…….. n we already know very well size varies compiler to compiler…… They only store the offset of the address the pointer is referencing. . An address consisting of only an offset has a range of 0 – 64K bytes…. i think there is no need to discuss near pointers anymore…. so come to the main point….. that is far and huge pointers……