C++ Programming

//The following is very simple program that will find the largest of any 3 numbers

#include <iostream>

using namespace std;

int main()

{

int n1, n2, n3, largest;

cout <<"Enter any 3 numbers ";

cin >>n1;

cin >>n2;

cin >>n3;

if(n1 > n2 && n1 > n3)

largest= n1;

else if(n2 >n1 && n2 >n3)

largest= n2;

else

largest= n3;

return 0;

}

It depends on what version of VB you are using, and what version of SQL Server you're attempting to connect to. Also, there are several ways to connect from each version of VB to each version of SQL Server (ODBC, ADODB, DSN-less, etc.). Since your question does not provide enough specifics to answer adequately, I refer you to the "Connection Strings" link at the bottom of this page (or you may just type in www.connectionstrings.com in your browser).

Remember that a pointer is just a variable containing the memory address of another variable. A pointer to a pointer is no different, other than that the address contains the address of another pointer. You use the * indirection operator to get the value of the variable being pointed at (the address of the other pointer), and the ** indirection operator to get at the value pointed at by the other pointer.

The following example illustrates how to access the values of pointers to int via an array of pointers to those pointers.

The memory address and the value of every variable is displayed for the benefit of clarity.

#include <iostream>

using namespace std;

int main()

{

// Set up an array of pointers to pointers to int variables.

int X = 1, Y=2; // The actual variables.

int* pX = &X; // Pointers to those variables

int* pY = &Y;

int** pp[2]; // Array of pointers to those pointers.

pp[0] = &pX;

pp[1] = &pY;

// Print the address of all variables and their stored values:

cout << "Var\t&Address\tValue" << endl;

cout << "---\t--------\t-----" << endl;

cout << "X\t0x" << &X << "\t" << X << endl;

cout << "Y\t0x" << &Y << "\t" << Y << endl;

cout << "pX\t0x" << &pX << "\t0x" << pX << endl;

cout << "pY\t0x" << &pY << "\t0x" << pY << endl;

cout << "pp\t0x" << &pp << "\t0x" << pp << endl;

cout << endl;

cout << "Note that both &pp and pp return the same value: the address of the array." << endl;

cout << "pp is simply an alias for the memory allocated to the array itself, it is" << endl;

cout << "not a variable that contains a value. You must access the elements of the" << endl;

cout << "array to get at the actual values stored in the array." << endl;

cout << endl;

// Use the array elements to access the pointers and the values they point to:

cout << "Elem\t&Address\tValue\t\t*Value\t\t**Value" << endl;

cout << "----\t--------\t-----\t\t------\t\t-------" << endl;

cout << "pp[0]\t0x" << &pp[0] << "\t0x" << pp[0] << "\t0x" << *pp[0] << "\t" << **pp[0] << endl;

cout << "pp[1]\t0x" << &pp[1] << "\t0x" << pp[1] << "\t0x" << *pp[1] << "\t" << **pp[1] << endl;

cout << endl;

return( 0 );

}

A "get" or "getter" function is a function that retrieves information. They are called getters because they are often named with a "get_" prefix, such as:

date get_current_date ();

The "get_" prefix is not actually required in C++, it is only required in languages that do not support function overloading. This is because most get functions also have a corresponding set function and you cannot use the same name twice without overloading:

date get_current_date ();

void set_current_date (const date&);

Getters and setters are usually defined as class member functions. However, rather than referring to them informally as getters and setters we refer to them formally as accessors and mutators, respectively. This reflects the fact that a getter typically accesses information without changing it whereas a setter typically mutates or changes information. For example:

class X {

private:

int data;

public:

X (int value): data {value} {} // constructor

X& operator= (int value) { data = value; return *this; } // mutator

int value () const { return data; } // accessor

// ...

};

The accessor is declared const and returns by value. In other words, the caller receives a copy of the information, not the information itself, and the class is left unaffected by the access.

The mutator, on the other hand, is declared non-const because the class representation (the data member) will need to be modified.

Constructors are not mutators because constructors create information, they do not mutate existing information.

This snippet assumes you have two values, x and y and you want to find the greater.

int max (int x, int y)
{
if (x > y)
{
return x;
}

return y;
}

C was initially developed by Dennis Ritchie from 1969 to 1973. C++ was initially developed by Bjarn Stroustrup from 1979 (when it was known as C with Classes) to 1983 (when it was renamed C++). Both developers worked at Bell Labs at the time.

To find the GCD of three numbers, a, b and c, you need to find the GCD of a and b first, such that d = GCD(a, b). Then call GCD(d, c). Although you could simply call GCD(GCD(a, b), c), a more useful method is to use an array and iteratively call the GCD(a, b) function, such that a and b are the first two numbers in the first iteration, which becomes a in the next iteration, while b is the next number. The following program demonstarates this method.

Note that the GCD of two numbers can either be calculated recursively or iteratively. This program includes both options, depending on whether RECURSIVE is defined or not. In a working program you'd use one or the other, but the iterative approach is usually faster because it requires just one function call and no additional stack space.

The program will create 10 random arrays of integers of length 3 to 5 and process each in turn. Note that the more numbers in the array, the more likely the GCD will be 1.

#include<iostream>

#include<time.h>

#define RECURSIVE // comment out to use iterative method

#define ARRAY // comment out to use non-arrays

#ifdef RECURSIVE

// Returns the GCD of the two given integers (recursive method)

unsigned int gcd(unsigned int a, unsigned int b)

{

if(!a)

return(b);

if(!b)

return(a);

if(a==b)

return(a);

if(~a&1)

{

if(b&1)

return(gcd(a>>1,b));

else

return(gcd(a>>1,b>>1)<<1);

}

if(~b&1)

return(gcd(a,b>>1));

if(a>b)

return(gcd((a-b)>>1,b));

return(gcd((b-a)>>1,a));

}

#else

// Returns the GCD of the two given integers (iterative method)

unsigned int gcd(unsigned int a, unsigned int b)

{

if(!a)

return(b);

if(!b)

return(a);

int c;

for(c=0; ((a|b)&1)==0; ++c)

{

a>>=1;

b>>=1;

}

while((a&1)==0)

a>>=1;

do{

while((b&1)==0)

b>>=1;

if(a>b)

{

unsigned int t=a;

a=b;

b=t;

}

b-=a;

}while(b);

return(a<<c);

}

#endif RECURSIVE

// Returns the greatest common divisor in the given array

unsigned int gcd(const unsigned int n[], const unsigned int size)

{

if( size==0 )

return( 0 );

if( size==1 )

return( n[0] );

unsigned int hcf=gcd(n[0],n[1]);

for( unsigned int index=2; index<size; ++index )

hcf=gcd(hcf,n[index]);

return(hcf);

}

int main()

{

using std::cout;

using std::endl;

srand((unsigned) time(NULL));

for(unsigned int attempt=0; attempt<10; ++attempt)

{

unsigned int size=rand()%3+3;

unsigned int* num = new unsigned int[size];

unsigned int index=0;

while(index<size)

num[index++]=rand()%100;

unsigned int hcf=gcd(num,size);

cout<<"GCD(";

index=0;

cout<<num[index];

while(++index<size)

cout<<','<<num[index];

cout<<") = "<<hcf<<endl;

delete[]num;

}

cout<<endl;

}

1. Calculate the mean average of the N numbers. Suppose it is stored in the variable D.

2. Now calculate the differences between each number with D and square the value. As there are N numbers so store this difference in an array. Example arr[0] = (num0 - D)^2; arr[1] = (num1 - D)^2; arr[N-1] = (numN - D)^2;

3. Now sum array value and divide by N, suppose the value is stored in F. Now square root F. It is the standard deviation of your N number.

I hope you can write the code by yourself or follow the part:

suppose you will store the N numbers in an array num. Now:

int num[N+2], D = 0; //or declare the num array as float or double if there are any precision value

for(int i = 0; i < N; i++)

{

D += num[i];

}

D /= N;

int arr[N+2];

for(int i = 0; i < N; i++)

{

arr[i] = (num[i] - D)^2; //square the difference.

}

int F = 0; //if precision value is accepted then declare F as float or double not int.

for(int i = 0; i < N; i++)

{

F += arr[i];

}

F /= N;

F = sqrt(F); //use #include <cmath> in your header file list so that you can use sqrt() function or simply use #include <bits/stdc++.h>

cout<<F<<endl;

- Thanks

write an algorithm to print the factorial of a given number and then draw the flowchart.

This looks like someones homework, defiantly someone looking for the easy way. When it comes to programming, the more you do the better you get. Experience counts (making your own mistakes and learning from the mistake).

C++ is a general purpose programming language. It has imperative, object-oriented and generic programming features, while also providing the facilities for low level memory manipulation. Source: Wikipedia.

The most basic input/output operations can be performed with std::cin for input and std::cout for output. For example:

#include<iostream>

int main (void) {

int n;

std::cout << "Enter a number: ";

std::cin >> n;

std::cout << "The value " << n << " was entered!\n";

}

Use the following function:

/* returns the average of two real numbers */

double average (const double a, const double b) {

return a/2.0 + b/2.0;

}

Note that we do not use (a+b) / 2.0 because the expression a+b could overflow the range of a double. By dividing each value by 2 before summing we ensure the result can never overflow no matter how large a and b are.

Ideally, you should write separate functions to cater for float and long double arguments independently:

/* returns the average of two long doubles */

long double average_lng (const long double a, const long double b) { return a/2.0 + b/2.0;

}

/* returns the average of two floats */

float average_flt (const float a, const float b) {

return a/2.0F + b/2.0F;

}

For mixed-mode arithmetic, always use the highest precision argument. E.g., the average of a float and a double is a double, so use the function that returns a double, not a float. The float argument will be implicitly cast to a double.

I suppose that you think to synthesis reaction.

Ideally, never. Friend functions should only be employed when a function (whether an operator overload or not) requires private access to a class, and it is not otherwise possible to provide a public interface without unduly undermining the class encapsulation. However, as programmer, it is your responsibility to ensure all friend functions adhere to the same class rules (which you yourself define) as do the members of your class, even though friends are not regarded as being members of the class. Ultimately, if you have no control over the friend function implementation, then you must not allow that function to be a friend of your class, as this will seriously undermine the encapsulation.

This error message means that somewhere in your program you have used a pointer-varible containing NULL-value. (Within an actual OS it with stop the program immediately, but in MS-DOS it doesn't.)

method is a process

property is about it.

please check the answer from others also. because its only my guess.

SORRY

THANK YOU

The built-in (fundamental) data types are int, char, bool, float, double and pointer. All other integral types are either modified types or aliases (typedefs). User-defined types (including those provided by the standard library) are not considered part of the language.

SOURCE CODE:

#include

#include

void push(int st[],int data,int &top);

void disp(int st[],int &top);

int pop(int st[],int &top);

int flg=0;

int top=-1,tos=-1;

int st[50];

void push(int st[],int data,int &top)

{

if(top==50-1)

flg=0;

else

{

flg=1;

top++;

st[top]=data;

}

}

int pop(int st[],int &top)

{

int pe;

if(top==-1)

{

pe=0;

flg=0;

}

else

{

flg=1;

pe=st[top];

top--;

}

return(pe);

}

void disp(int st[],int &top)

{

int i;

if(top==-1)

{

printf("\nStack is Empty");

}

else

{

for(i=top;i>=0;i--)

printf("\t%d",st[i]);

}

}

void main()

{

int dt,opt;

int q=0;

clrscr();

printf("This Program Is Used to Perform PUSH & POP operations On Stack");

printf("\n\n\tMain Menu.........");

printf("\n\n1.Push");

printf("\n\n2.Pop");

printf("\n\n3.Exit");

do

{

printf("\n\n\tEnter Your Choice 1-3:");

scanf("%d",&opt);

switch(opt)

{

case 1:

printf("\nEnter the Element to be Push:");

scanf("%d",&dt);

push(st,dt,tos);

if(flg==1)

{

printf("\nAfter Inserting the Element, Stack is:\n\n");

disp(st,tos);

if(tos==50-1)

printf("\nStack is Now Full");

}

else

printf("\nStack Overflow Insertion Not Possible");

break;

case 2:

dt=pop(st,tos);

if(flg==1)

{

printf("\n\tData Deleted From the Stack is:%d\n",dt);

printf("\n\tAfter Deleting the Element from the stack is:\n\n");

disp(st,tos);

}

else

printf("\nStack Empty,Deletio Not Possible:");

break;

case 3:

q=1;

break;

default:printf("\nWrong Choice Enter 1-3 Only");

}

}while(q!=1);

}

OUTPUT

Main Menu.........

1.push

2.pop

3.exit

Enter your choice 1-3:1

Enter the element to be push:4

After inserting the elements,stack is:

4

Enter your choice 1-3:1

Enter the element to be push:7

After inserting the elements,stack is:

7 4

Enter your choice 1-3:1

Enter the element to be push:4