Inline function

Inline functions are an imperative feature of C++ and are frequently used with classes. Inline functions are functions where the call is made to inline functions. The actual code then gets placed in the calling program. In a program when we call a function the program jumps to the address of the function and when it reaches the end of the function it comes back. This jumping actually involves much more and takes time. But when an inline function is called the compiler will replace the call with the function code. So in reality in that place there will be no function call, only the code of the function. No jumping needed to different addresses. The general format of inline function is as follows:

inline datatype function_name(arguments)

Example:

The following code creates and calls an inline function:

#include<iostream.h>

inline int average(int a, int b)
{
   return (a + b) / 2;
}

void main()

{
   int result = average(12, 14);

   cout << "The average of number 12, 14 is " << result << "\n";

   getch();

}

It is most important to know that the inline specifies is a request, not a command, to the compiler. If, for various reasons, the compiler is unable to fulfill the request, the function is compiled as a normal function.

What is macros in C

A macro is a fragment of code which has been given a name. When we use the name of macro in program, it is replaced by the contents of the macro. You may define any valid identifier as a macro, even if it is a c keyword. There are two types of macros. One is object-like macros and another is function-like macros. Object-like macros do not take parameters; function-like macros do. The generic syntax for declaring an identifier as a macro of each type is:

For object-like macros:

 #define <identifier> <replacement token list>

For example:

#include<stdio.h>

#include<conio.h>

#define sum a+b

void main()

{

Clrscr();

int a=5,b=10,c;

c=sum; //if we give the semicolon (;)when we define the macro; then we cannot give the semicolon when we write the name of identifier. If we give the semicolon when we define the macro, then we give the semicolon when we write the name of identifier in program

printf(“The sum of a+b is: %d”,c);

getch();}

For function-like macros:

#define <identifier> (<parameter list>) <replacement token list>

For example:   

#include<stdio.h>

#include<conio.h>

#define square(x) x*x;

Void main()

{

Clrscr();

int i=2,j;

j=square(i) //Here is no semicolon because I give the semicolon when I define the macros.

printf(“The value of j is: %d”,j);

getch();
}

 The #undef directive removes the definition of a macro.

Written by arnob;

Reverse an integer by using Recursion

#include<stdio.h>
#include<conio.h>
int re (int,int,int*,int);
void main()
{
clrscr();
int i,j=0,n[5],k=0,m;
printf("Enter a number: ");
scanf("%d",&i);
m=re(i,j,n,k);
for(j=0;j<m;j++)
printf("%d",n[j]);
getch();
}
re(int i,int j,int *n,int k)
{
if(i==0) return k;
j=i%10;
n[k]=j;
i=i/10;
re(i,j,n,k+1);

program written by arnob;

48. Writer a C program which store students name, id and gpa. and make a serach option that when we enter a name for search, if the name is there then show that students name,id and gpa. If the name is not there then show NOT FOUND!.



#include<stdio.h>
#include<conio.h>
#include<string.h>
struct student
{
char name[40];
char id[20];
char gpa[5];
}stu[100];
void main()
{
clrscr();
int i=0,j,n;
char name[40];
printf("How many students informatio: ");
scanf("%d",&n);
for(i=0;i<n;i++)
{
printf("\nEnter the name: ");
scanf("%s",stu[i].name);
printf("\nEnter the id: ");
scanf("%s",stu[i].id);
printf("\nEnter the gpa: ");
scanf("%s",stu[i].gpa);
clrscr();
}
printf("\nEnter the search name: ");
scanf("%s",name);
for(i=0;i<n;i++)
{
clrscr();
j=strcmp(name,stu[i].name);
if(j==0)
{
printf("\nName :");
puts(stu[i].name);
printf("\nId :");
puts(stu[i].id);
printf("\nGPA :");
puts(stu[i].gpa);
break;
}
}
if(j!=0)
printf("Not found !");
getch();
}

program written by arnob;

47. Find the hight and second hight number by using RECURSION

/* Find the hight number by using recursion */

#include<stdio.h>
#include<conio.h>
int main()
{
clrscr();
int n,a[100],i,j;
printf("How many nmber you want to input: ");
scanf("%d",&n);
for(i=0;i<n;i++)
{
printf("Enter the %d number",i+1);
scanf("%d",&a[i]);
}
int hight(int,int,int,int*);
j=hight(n,0,0,a);
printf("The hight number is %d",j);
return 0;
}
int hight(int n,int i,int m,int *a)
{
if(i>n) return m;
if(m<a[i])
m=a[i];
hight(n,i+1,m,a);
}

/* Find the second hight number by using recursion */

#include<stdio.h>
#include<conio.h>
int main()
{
clrscr();
int n,a[100],i;
printf("How many nmber you want to input: ");
scanf("%d",&n);
for(i=0;i<n;i++)
{
printf("Enter the %d number",i+1);
scanf("%d",&a[i]);
}
int hight(int,int,int,int,int*);
hight(n,0,0,0,a);
printf("The second hight number is %d",a[1]);
return 0;
}
int hight(int n,int i,int j,int k,int *a)
{
if(i>n-1) return 0;
for(j=i+1;j<n;j++)
if(a[i]<a[j])
{
k=a[i];
a[i]=a[j];
a[j]=k;
}
hight(n,i+1,j,k,a);
}

written by anrob.

46. Find the even number from a array by using RECURSION

Question: Get input a array frmon user and find the hight number. You can only use one array and can not use any globel variable.

Ans: If you can use more then one array then you can easyle find the hight number by only using for loop. But you can not use more then one array. So you can solve this problem by using recursion.

/*program for find the even number */

#include<stdio.h>

#include<conio.h>

int main()

{

clrscr();

int a[100];n,i;
printf("How many number you want to input: ");
scanf("%d",&n);
for(i=0;i<n;i++)
{
printf("Enter the %d number: ");
scanf("%d",&a[i]);
}

void chk_evn(int,int,int*,int*)

chk_evn(0,0,&n,a);

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

printf("%d",a[i]);

return 0;

}

void chk_evn (int i,int j,int *n,int *a)

{

if(i==n){*n=j;return n;}

if(a[i]%2==0)

a[j++]=a[i];

chk_evn(i+1;j,n,a);

}

program written by arnob

45. Passing 2-D Array to a Function by using pointer

There are two ways in which er can pass a 2-D array to a function by using pointer. These are illustrated in the following program.

/*Two ways of accessing a 2-D array*/

#include<alloc.h>

void main()

{

int a [3][4]={

               1,2,3,4,

               5,6,7,8,

               9,0,1,6

             };

clrscr();

display(a,3,4);

show(a,3,4);

}

display(int *q,int row, int col)

{

int i,j;

for(i=0;i<row;i++)

{

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

printf("%d",*(q+i*col+j));

printf("\n");

}

printf("\n");

}

show(int (*q)[4],int row, int clo)

{

int i,j;

int *p;

for(i=0;i<row;i++)

{

p=q+i;

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

printf("%d",*(p+j));

printf("\n");

}

printf("\n");

}

And here is the output...

1234

5678

9016

1234

5678

9016

In the display() finction we have collected the base address of the 2-D array being passed to it in an ordinary int pointer. Thenthrough the two for loops using the ex[ression *(q+i*col+j) we have reached the appropriate element in the array. Suppose i is equal to 2 and j is equal to 3, then we wish to reach the element give this element or not. The exprewssion *(q+i*col+j) becomes *(4001+2*4+3). This turns out to be *(4001+11). Since 4001 is address of an integer, *(4001+11) turns out to be *(4023). Value at this address is +. This is indeed same as a[2][3]. A more general formula for accessing each array element would be:

*(base address + row *no of columns +column no)

In the show( ) function we have defined q to be a pointer to an array of 4 integers through the declaration

int (*q)[4];

To begin with, q hods the base address of the xeroth 1-D array, i.e 4001. This address is then assigned to p, an int pointer, and then using this pointer all elements of the xeroth 1-D array are accessed. Next thime through the loop when i takes a value 1, the expression q+i fetches the address of the first 1-D arry. This is because, q is a pointer to zeroth 1-D array and adding 1 to it would given us the address of the next 1-D array. This address is once again assigned to p, and using it all elements of the next 1-D arraY are accessed. 

Arnob