#include<iostream.h>
#include<stdio.h>
void main()
{
    struct STUDENT
      {
         char stu_name[20];
         char stu_sex;
         int stu_age;
     }student;
     gets(student.stu_name);
     cin>>(student.stu_sex);
}

#include
#include
struct Swimmingclub
{
     int memnumber;
     char memname[20];
     char memtype[4];
}
void main()
{
   Swimmingclub per1.per2;
   cout<<"Member Number:";
   cin>>per1.memnumber;
   cout<<"Member Name:";
   cin>>per1.memname;
   strcpy(per1.memtype."HIG");
   per1.memtype="HIG";
   per2=per1;
   cout<<"Member Number:"<
   cout<<"Member Name:"<
   cout<<"Member Type:"<
}

C^^

#define preprocessor directive is used to define a macro with some value/expression, which is substituted during compilation of program. Unlike variable, it does not occupy memory. e.g.
         #define Max 10
         void main()
         {
    int arr[Max];
         }
Const It is used in declaration of variable, it occupies memory to store a constant value, which once initialised cannot be changed
e.g. const int x=10;

typedef keyword is used to assign alternative names to existing data types.
e.g. typedef int in;
Now, whenever we want to declare a variable of types integer, rather than using int, we can use 'in' keyword in place.
e.g. void main()
       {
          typedef int in;
                         //in is an alternetive name to int 
          in a=10;
          cout<<a;  // a is of type integer

Syntax error is that when statements are wrongly written violating rules of the programming language.
e.g. Max + 2 = DMax is a syntax error as an expression cannot appear on the left side of an assignment operator.
Runtime error occurs in a program during its execution. Program execution halts when such an error is encountered.
e.g. division by zero, stack overflow, etc.

Type casting It is used by the programmer to convert value of one type to another type. It is forced type conversion. It is done by putting the target datatype in parentheses before the data to be converted.
e.g. float X=(double)15/6);

Automatic type conversion It is automatically done by the compiler itself wherever required. It is implicit type conversion
e.g. float X=3;

#include<iostream.h>
#include<conio.h>
#include<process.h>
void main()
{
    int j;
    cout<<"Enter the bit value 0, 1 "<<"\n";
    cin>>j;
    switch(j)
    {
        case 0;
          cout<<"Bit value is 0"<<"\n";
           break;
       case 1;
          cout<<"Bit value is 1"<<"\n";
          break;
      default:
          exit(0);
   } 
    getch();
}
Output
Enter the bit value 0,1
0
Bit value is 0

#include<iostream.h>
#include<conio.h>
void main()
{
    int i, j;
    for(i=0; i<10; i++)
    {
          cout<<"\n Outer loop executing. i="<<i;        
           for(j=0; j<2; j++)
            {
                  cout<<"\n Inner loop executing. j="<<j;
                  if(i==3)
                       goto stop;
          }
    } 
 cout<<"\n Loop exited.  i="<<i;
stop;
                   //This message does not print 
 cout<<"\n Jumped to stop. i="<<i;
getch();
}
Output
Outer loop executing. i=0
Inner loop executing. j=0
Inner loop executing. j=1
Outer loop executing. i=1
Inner loop executing. j=0
Inner loop executing. j=1
Outer loop executing. i=2
Inner loop executing. j=0
Inner loop executing. j=1
Outer loop executing. i=3
Inner loop executing. j=0
Jumped to stop. i=3

#include<iostream.h>
#include<conio.h>
void main()
{
    int i=0;
    do
    {
          i++;
          cout<<"before the continue\n";
          continue;
           cout<<"after  the continue, should";
           cout<<"never print\n";
} while(i<3)
 cout<<"after the do loop\n";
getch();
}
Output
before the continue
before the continue
before the continue
after the do loop

#include<iostream.h>
#include<conio.h>
void main()
{
    int i;
    for(i=1; i<10; i++)
    {
          cout<<i<<" ";
          if(i==4)
           break;
}
getch();
}     //Loop exit after printing 1 through 4
Output
1 2 3 4 

#include<iostream.h>
#include<conio.h>
void main()
{
   int i,j,k,l;
   clrscr();
   cout<<"The armstrong numbers in between 1 and 200 are";
for(i=1;i<=200;i++)
{
    k=0,j=i;
    while(j!=0)
    {
     l=j;
     k=k+1*1*1;
     j=j/10;
}
if(k==i)
   cout<<"\n"<<i;
}
getch();
}

#include
#include
void main()
{
    int n, i;
    clrscr();
    cout<<"Enter the number to find the multiple";
    cin>>n;
    for(i=1;i<=10;i++)
        cout< getch();
}