C Programming

The enum keyword means enumeration.

The flowchart in deriving the sum of the square root of first N even numbers, You may follow the steps provided below:

# Draw the start symbol then a flow line connecting to item #2 # Draw the init box for the syntax: set variable_A=0, variable_Chk=0, variable_Sum=0, variable_Sqrt, then a flow line connecting to item #3 # Draw the input box and write variable_A then a flow line connecting to item #4 # Draw the decision box for 'Is variable_A not numeric?'. if yes, draw a flow line connecting to item#3 else draw a flow line connecting to item#5 # Draw the process box for the syntax: compute variable_Chk=variable_A / 2 then a flow line connecting to item #6 # Draw the decision box for 'Is variable_chk not whole number?. if yes, draw a flow line connecting to item #3 else draw a flow line connecting to item #7 # Draw the process box for the syntax: compute variable_Sum=variable_Sum + variable_A then a flow line connecting to item #8 # Draw the decision box for 'Do you want to add another number?'. if yes, draw a flow line connecting to item#3 else draw a flow line connecting to item #9 # Draw the process box for the syntax: compute variable_Sqrt=SQRT(variable_Sum) then a flow line connecting to item #10 # Draw the output box and write variable_Sum, variable_Sqrt then a flow line connecting to item #11 # Draw the end symbol.

Where:

variable_A contains a given N number, variable_Chk contains the quotient of variable_A / 2, variable_Sum contains the sum of N numbers and variable_Sqrt contains the result.

You need to check the day, month and year separately, to ensure they are within range. You also need to deal with leap days (29th February) which is every 4 years, but not the 100th unless it is the 400th. Finally, you need to deal with the change from the Julian to Gregorian calendars, which skips the 5th to 14th October, 1582.

bool checkdate(unsigned int d, unsigned int m, unsigned int y)

{

return( !(( d<1 d>31 m<1 m>12 ( y==1582 && m==10 && (d>4 && d<15 )))

( d==31 && ( m==2 m==4 m==6 m==9 m==11 ))

( m==2 && ( d>29 ( d==29 && ( y%4 ( !( y%100 ) && y%400 )))))));

}

Yes, that's the idea of an array.

Yes, that's the idea of an array.

Yes, that's the idea of an array.

Yes, that's the idea of an array.

There is no pause function as such, but you can easily roll your own:

#include <iostream>

#include <limits>

void Pause()

{

std::cout << "Press ENTER to continue...";

std::cin.ignore( std::numeric_limits<std::streamsize>::max(), '\n' );

}

int main()

{

Pause();

return( 0 );

}

There being only 7 symbols to consider (IVXLCDM), conversion is easily achieved in any number of ways. In general, numerals are formed from left to right, largest value to smallest. However, if a smaller value precedes a larger value, the smaller value is subtracted from the larger value (or is negated). This can lead to problems such as IVX. Reading left to right this would become 10 - ( 5 - 1 ) = 10 - 4 = 6. There's nothing wrong with this, but most people would accept 6 = VI, not IVX.

The problem is there has never been an official standard relating to how Roman numerals are formed. Decimal 1999 could be represented as MCMXCIX or MIM or MDCCCCLXXXXVIIII or even a mixed format like MCMXCVIIII. All are intrinsically correct. However, only the first example conforms to what many would consider to be the "unofficial" standard, whereby certain combinations are no longer permitted (such as IIII, IM and VX).

This standard has been incorporated into the following code.

#include <iostream>

using namespace std;

int main()

{

char roman[11];

int decimal[10];

memset( roman, 0, 11 );

memset( decimal, 0, 10 * sizeof( int ));

cout << endl;

cout << "Enter a Roman number (max. 10 chars from I, V, X, L, C, D or M): ";

cin.getline( roman, 11, '\n' );

strupr( roman ); // convert to uppercase for consistency

// check validity, including all invalid combinations

if( !strlen( roman )

( strspn( roman, "IVXLCDM") != strlen( roman ))

( strstr( roman, "IIII" ))

( strstr( roman, "XXXX" ))

( strstr( roman, "CCCC" ))

( strstr( roman, "MMMM"))

( strstr( roman, "IL" ))

( strstr( roman, "IC" ))

( strstr( roman, "ID" ))

( strstr( roman, "IM" ))

( strstr( roman, "XD" ))

( strstr( roman, "XM" ))

( strstr( roman, "VX" ))

( strstr( roman, "VL" ))

( strstr( roman, "VC" ))

( strstr( roman, "VD" ))

( strstr( roman, "VM" ))

( strstr( roman, "LC" ))

( strstr( roman, "LD" ))

( strstr( roman, "LM" ))

( strstr( roman, "DM" ))

( strstr( roman, "IIV" ))

( strstr( roman, "IIX" ))

( strstr( roman, "XXL" ))

( strstr( roman, "XXC" ))

( strstr( roman, "CCD" ))

( strstr( roman, "CCM" )))

{

cout << roman << " is not a valid roman number." << endl;

return( -1 );

}

// convert to decimal, in reverse order.

int c = 9, total = 0;

while( c >= 0 )

{

switch( roman[c] )

{

case('I'): decimal[c] = 1; break;

case('V'): decimal[c] = 5; break;

case('X'): decimal[c] = 10; break;

case('L'): decimal[c] = 50; break;

case('C'): decimal[c] = 100; break;

case('D'): decimal[c] = 500; break;

case('M'): decimal[c] = 1000; break;

}

if( c < 9 ) // subtraction required?

if( decimal[c] < decimal[c+1] )

decimal[c] *= (-1); // negate

// update total.

total += decimal[c--];

}

cout << "Roman " << roman << " is decimal " << total << endl;

return( 0 );

}

Calculate y in the following equation: y = 2 + 3 x 4 What answer did you get? Did you get 20? If you did, you messed up. You should have gotten 14. If you got 20, you did the operations in the order in which they appear, from left to right, which is a mistake. You added 2 and 3, to get 5, and then you multiplied by 4, to get 20. But you should have multiplied 3 by 4, first, to get 12, and added that product to 2, to get 14. That's because multiplication (and division) take precedent over addition (and subtraction). Now, try this: y = (2 + 3) x 4 Did you get 20? If so, you got the right answer. Do you know why?

//Program to check number is prime or not using recursive function

#include<stdio.h>

#include<stdlib.h>

void prime(int num,int count)

{

if(count<num)

{

if(num%count==0)

{

printf("%d is not Prime Number\n",num);

goto exit;

}

count += 1;

if(count<num)

{

prime(num,count);

}

}

if(num==count)

{

printf("%d is a Prime Number\n",num);

}

exit:

return 0;

}

int main()

{

system("cls");

int gvar;

printf("Enter the number = ");

scanf("%d",&gvar);

prime(gvar,2);

printf("\nashokakhil@gmail.com\n");

system("PAUSE");

return 0;

}

I think this can be another solution

#include<stdio.h>

#include<conio.h>

int prime(int);

main()

{

int i=1,r;

clrscr();

r=prime(i);

if(r==1)

printf("\n\n\tNo is prime ");

getch();

}

int prime(int i)

{

int n=1,ans,flag=1;

i++;

ans=n%i;

if(ans==0)

{

printf("\t\t\n\nNo is not prime");

flag=0;

return flag;

}

if((i!=n-1)&&(n!=1))

flag=prime(i);

return flag;

}

write a program that reads in the size of the side of square and then pints a hollow square of that size out of asterisks and blanks?

To address the American labor shortage

a for loop is defined with an boolean expression to indicate when it should terminate. A for each loop iterates once for each item in a collection.

for example, "for each (book in bookshelf)" will iterate once for each book on the bookshelf, providing access to the current book.

a for loop is defined like "for (int i = 0; i<10;i++)" meaning the loop will iterate as long as the condition is true (i < 10), and will increment on each loop.

Note: there is no 'for each' loop in C language, but there is a 'foreach' in PHP.

In C, structures are uninitialized by default. To initialize a structure you will typically zero the memory allocated to the structure and then set specific members to specific values. If all members are non-zero, you can simply set those members rather than zero the memory first.

In C++, structures are initialized via inline initializes and/or through the class constructor.

# include<stdio.h>

# include<conio.h>

void main()

{

clrscr();

int a,b;

printf ("Enter the first value:");

scanf ("%d",& a );

printf ("Enter the second value:");

scanf ("%d",& b );

printf ("\n\nBefor swaping the values ");

printf ("\nThe first value is %d",a);

printf ("\nThe second value is %d",b);

printf ("\n\nAfter swaping the values ");

printf ("\nThe first value is %d",b);

printf ("\nThe second value is %d",a);

}

/* write a program to print Days of Week using switch-case structure */

#include<stdio.h>

#include<conio.h>

void main()

{

int n;

clrscr();

printf("\n Enter Day of weak as Number 1 to 7 ");

scanf("%d",&n);

switch(n)

{

case 1:

printf("\n MONDAY ");

case 2:

printf("\n TUESDAY");

case 3:

printf("\n WEDNESDAY");

case 4:

printf("\n THURSDAY");

case 5:

printf("\n FRIDAY");

case 6:

printf("\n SATURDAY");

case 7:

printf("\n SUNDAY");

default :

printf("\n no operation is required");

}

getch();

}

it is not regular language .it is high level language

How to download and install Turbo C++ in Windows 7....

1.Download dosbox 0.74 from dosbox.com and save in E: drive(say).

2.Download Turbo C++ 3.0.

3.Create a folder Turbo in E: drive & save all contents of Turbo C++ 3.0.

4.Open dosbox.

5.Type as follows:-

mount e e:\ <Press enter>

e: <Press enter>

cd turbo <Press enter>

install install <Press enter>

6.Proceed as per requirements.

7.A folder TC appears in E: drive.

8.To repeat the point 5. commands whenever you open the dosbox, go to E:\Dosbox-0.74\Dosbox-0.74 Options

9.Scroll down to the end of the page and write as follows:-

mount E E:\

E:

CD TC

CD BIN

TC.EXE

10.Now open dosbox by clicking its icon on desktop and find to open Turbo C++ directly, but in a small dialob box.

•Much more portable than low level languages (can be transferred over different computers) •Many tutorials and manual for the languages

•Many prewritten and tested algorithms made (no need to "reinvent the wheel")

•Excellent for general purpose programming

From inside any loop statement, the continue; statement will skip any remaining statements and re-evaluate the loop's conditional expression. If that expression remains true, a new iteration of the loop begins, otherwise control passes to the statement that follows the loop. Note that in a for or while loop, the conditional expression is defined before the loop body but in a do loop it is defined after the loop body.

main refers to the entry point of an application. All programs must have a main() function that returns an integer to the calling program or script. The return value can be used for any purpose, but generally a non-zero negative value is used to indicate that an error occurred (zero meaning no error).

Programs written in a high level language might be slower than ones written in Assembly language; but it is not always so, it is very easy to write un-effective programs in Assembly.

getch(); is used for unbuffered input. e.x:

int main()

{

char num=0;

printf("Press a keyboard button: ");

num = getch(); //This brings in 1 character that the user pressed on the keyboard

printf("\nYou pressed: %c", num); //This prints the character you pressed

getchar(); // I am using getchar(); to stop the program from ending after pressing buttons

return 0;

}

My input will be within the ().

output:

Press a keyboard button: (v)

You pressed: v

EOP //End of program

I hope this has helped you!

No. The program counter must be stored in a dedicated register. The stack is in working memory and you cannot operate on working memory; all values must be moved into a register in order to operate upon them. It makes no sense to move a program counter in and out of memory unless performing a context switch and you can't use a stack for context switching; a priority queue must be used for this. Keep in mind that the address of the top of the stack has to be moved in and out of its register during a context switch. It doesn't make sense to load the stack register from a priority queue before you can determine where the program counter value is. It's easier to keep all state information in the same place in the priority queue where it belongs.