C Programming

Many ways, depending on the robot complexity, how it is being ran and other factors. Conventional 'robots' have motors, sensors, and a computer. Usually the computer runs a program that does all the programming, this can be done in ANY language that the programmer fancies the most. There are some robots called BEAM robots that aren't programmed using code or programming, but instead react as a neural network (like an organic brain) or reactions from the pre built electronics in the robot. This is a VERY open question

There is a lot of different ones, so it really depends on what kind of software you want to design.

#include<stdio.h>

#include<conio.h>

#include<string.h>

#include<stdlib.h>

struct object_code

{

int locctr;

char add[10];

}obcode[300];

void main()

{

char input[100][16],output[100][16],binary[20],address[20],stloc[10];

int len,bitmask,loc,tlen=0,tloc,textloc,i=0,location,j,k,count=0,start,n,num=0,inc=0;

FILE *fp1,*fp2;

clrscr();

fp1=fopen("linput.dat","r");

fp2=fopen("reloadout.dat","w");

printf("Enter the location where the program has to be loaded:");

scanf("%s",stloc);

start=atoi(stloc);

location=start;

tloc=start;

fscanf(fp1,"%s",input[i]);

while(strcmp(input[i],"T")!=0)

{

strcpy(output[i],input[i]);

i++;

fscanf(fp1,"%s",input[i]);

strcpy(output[i],input[i]);

}

itoa(start,output[2],10);

while(strcmp(input[i],"E")!=0)

{

strcpy(output[i],input[i]);

if(strcmp(input[i],"T")==0)

{

for(j=0;j<3;j++)

{

i++;

fscanf(fp1,"%s",input[i]);

strcpy(output[i],input[i]);

}

bitmask=atoi(output[i]);

itoa(bitmask,binary,2);

strcpy(output[i],NULL);

textloc=atoi(output[i-2]);

textloc=textloc+start;

itoa(textloc,output[i-2],10);

for(n=0;n<(textloc-(tloc+tlen));n++)

{

strcpy(obcode[inc].add,"xx");

obcode[inc++].locctr=location++;

}

tlen=atoi(output[i-1]);

tloc=textloc;

k=0;

}

else

{

if(binary[k]==1)

{

num=0;

len=strlen(output[i]);

strcpy(address,NULL);

for(j=2;j<len;j++)

{

address[num]=output[i][j];

output[i][j]='\0';

num++;

}

loc=atoi(address);

loc=loc+start;

itoa(loc,address,10);

strcat(output[i],address);

}

k++;

len=strlen(output[i]);

num=0;

for(n=0;n<len;n++)

{

obcode[inc].add[num++]=output[i][n];

if(num>1)

{

obcode[inc++].locctr=location++;

num=0;

}

}

}

i++;

fscanf(fp1,"%s",input[i]);

}

strcpy(output[i],input[i]);

i++;

fscanf(fp1,"%s",input[i]);

loc=atoi(input[i]);

loc=loc+start;

strcpy(output[i],itoa(loc,address,10));

count=0;

i=0;

n=0;

fprintf(fp2,"%d\t",obcode[n].locctr);

for(n=0;n<inc;n++)

{

fprintf(fp2,"%s",obcode[n].add);

i++;

if(i>3)

{

fprintf(fp2,"\t");

i=0;

count++;

}

if(count>3)

{

fprintf(fp2,"\n%d\t",obcode[n+1].locctr);

count=0;

}

}

getch();

}

The stack search method cannot be performed recursively.

//Perhaps this is what you want.

//@author Rosencrantz

import java.util.Scanner;

class TriangleFun

{

public static void main(String[] args)

{

Scanner scan = new Scanner(System.in);

boolean done = false;

while(true)

{

System.out.print("Calculate AREA or PERIMETER? Type DONE to quit.");

String input = scan.nextLine().trim();

if(input.equalsIgnoreCase("DONE"))

break;

if(input.equalsIgnoreCase("AREA"))

{

System.out.print("Enter the base length: ");

double base = scan.nextDouble();

System.out.print("Enter the height: ");

double height = scan.nextDouble();

System.out.println("Area is: " + (1/2.0) * base * height);

}

else

{

System.out.print("Enter length of side 1: ");

double length1 = scan.nextDouble();

System.out.print("Enter length of side 2: ");

double length2 = scan.nextDouble();

System.out.print("Enter length of side 3: ");

double length3 = scan.nextDouble();

System.out.println("Perimeter is: " + (length1 + length2 + length3));

}

scan.nextLine();

}

}

}

//A better edited version is in the links

Not possible.

Functions in C language may call themselves (ie can be recursive) without restrictions.

You would use neither Java nor C++, you would use Objective-C, in conjunction with the Apple iPod API (iPod Library Access).

False :}

You will need to traverse the list i times to get to the element you want deleted. Each time you go thru the list you will need to remember the forward pointer from the previous element because when you get to the element you want to delete the previous forward pointer needs to be pointed to the I + 1 element of the list.

You could check the last element's next() node to see if it's null. If it's not null, then you've pointed your end pointer to the first element, and the queue is full.

in some cases, aloop control variable does not have to be initialized.

while true do

--do stuff

end

You don't. You can include math.h though, and use its functions eg. sin, cos, tg, sqrt, etc.

A merchant statement is a written record prepared by the processor (usually once a month) which lists all the transactions for the account, including fees charged.

asdasdasdsa

Write a C program to Draw a RAINBOW and fill the suitable colors ...

All the members of the structure can be accessed at once,where as in an union only one member can be used at a time. Another important difference is in the size allocated to a structure and an union. for eg: struct example { int integer; float floating_numbers; } the size allocated here is sizeof(int)+sizeof(float); where as in an union union example { int integer; float floating_numbers; } size allocated is the size of the highest member. so size is=sizeof(float);

coz at the run time object is provided memory , that's why object is called run time entity .

In order to print the first 10 prime numbers you need a function to determine if a given value is prime or not. The following algorithm is the standard method of doing so:

1. Let n be the value.

2. If n < 2 then return false.

3. If n is even then return true if n is 2, otherwise return false.

4. Let divisor = 3.

5. If divisor is greater than the square root of n then return true.

6. If divisor is a factor of value then return false.

7. Let divisor = divisor + 2.

8. Go to step 5.

This algorithm can be efficiently implemented in C++ as follows:

bool is_prime (const unsigned n)

{

if (n<2) return false;

if (!(n&1)) return n==2;

const unsigned m = static_cast<unsigned>(std::sqrt (static_cast<double>(n)));

for (unsigned d=3; d<=m; d+=2)

if (!(n%d)) return false;

return true;

}

With this function defined, we can now go ahead and write the complete program:

#include<iostream>

bool is_prime (const unsigned n) {/*as above*/}

int main()

{

std::cout << "First 10 primes:\n";

unsigned primes = 0;

unsigned num = 0;

while (primes < 10)

{

if (is_prime (num))

{

std::cout << num << std::endl;

++primes;

}

++num;

}

}

A representation of an object or event is a model.