ans.

d*d will be replaced by 10+10*10+10

during runtime.

so answer is 10+100+10 = 120

400

This boils down to (10 +10 * 10 + 10)

so answer is 120 ... but if the same macro was rewritten as 
#define d (10 + 10)

then d * d = (10 + 10 ) * (10 + 10)
           = 20 * 20
           = 400....

Pure macro concept....

# include<stdio.h>
# include <conio.h>

void main()
{
  char input;
  printf("enter an input to get ascii of input");
  scanf("%c",&input);
  printf("\n ascii of input %c: is %d",input,input);
  getch();
}

#include<stdio.h>
#include<conio.h>
void main()
{
char n;
printf("enter the character:");
scanf("%c",&n);
printf("the ascii value %c is %d",n,n);
getch();
}

#include <stdio.h>
#include <stdlib.h>
#include <conio.h>

void main()
{
	char num;
        printf("Enter the number");
        scanf("%c",&num);
	printf("ASCII of %d is %d\n",atoi(&num),num);

}

#include <stdio.h>
#include <stdlib.h>
#include <conio.h>

void main()
{
	char num;
        printf("Enter the number");
        scanf("%c",&num);
	printf("ASCII of %d is %d\n",atoi(&num),num);

}

void main()
{
	int num;
        printf("Enter the number");
        scanf("%d",&num);
	printf("ASCII of %d is %d\n",num,itoa(&num));

}

no. because return keyword return only one value.that ia 0 
or 1.

in c a function can only written one value 
the return value could be of any data type

ya we can return two or more than two values..... it's
possible..
by using concept of POINTERS..... but no need of return
keyword at all..... 

instead of call by value in the function use call by
reference concept.... 
take the following program:

int fun(int *,int *);
void main()
{
int j=800,k=1000;
fun(&j,&k);
printf("%d",j,k);
getch();
}
int fun(int *q,int *w)
{
q=q/2;
w=w/2;
}
the output of the followiung is : 400 & 500.
how it's possible, i ll explain,

                      since we are calling by reference we
are sending the address of the two variables. so in fun.
definition we are catching it by pointers..... so that
pointer variable is holding the address of the two variables
in main fun. which is passed through address.... so in the
function we are changing the values of j & k.... so this
will change the value directly in the address of those two
variables j & k....... so implicitely two values are
returned wit out return keyword....

sorryt sorry i made a mistake... i

in the line    q=q/2 is wrong
correct as  *q=*q/2;
and          *w=*w/2;

since addresses cant be divided...

very sorry

yes we can return two or more values from a function using
return keyword...

Use structure...return type of the function will b 'struct' 
and v ll store value in struct type.

no.

Using the return statement u can only return one value at a 
time.
So you can either return the value of a variable like you 
can return an integer, or you can return pointer (which may 
contain more than one values), which is pointing to 
dynamically allocated location, Like in given below example:
//Returning two values from a function.

#include "stdio.h"
#include "malloc.h"

int *values()
{
	int *ptr;
	ptr = (int*)malloc(2);
	*ptr = 10;
	*(ptr+1) = 20;
	return ptr;
}

int main()
{
	int *ptr = values();
	printf("%d\n%d",*ptr,*(ptr+1));
	return 0;
}

macro: one line function;not having any return type and 
argument so it would not report error even if any 
misbehaves in fnction;not having return statement;
macros would be replaced by that functions/value where ever 
macro name presents;if we need any later change in funcions 
then modifing only at macro is neccssary, sicne it is 
replacing wherever macro name presents.

functions:we would recieve error if the function call and 
function prototypes are mismatch;xplicitly it should have 
one return statement if the retrun type of fucntion is 
aother than void;control from main function would jump to 
function defintion of called fucntion if that function is 
called rather than replacing the function;it may have moe 
than one line function.If we want later change i fucntion 
execution,then we need to modify on function prototype(if 
modified),fucntion defintion and whereever that function is 
called.

#define PRINT printf("%d",a+b)
void main()
{
 int a=8,b=7;
 PRINT;
 getch();
}

#include <stdio.h>

void putint(int x);

int main(void)
{
	puts("[output]");
	putint(13725);
	putchar('\n');	
	putint(5500);
	putchar('\n');
	return 0;
}

void putint(int x)
{
	if (x)
	{
		putint(x / 10);
		putchar('0' + x % 10);
	}
}

#include<stdio.h>
int main()
{
 int a=2,b=3,sum =0;
 while(a--){
 sum = sum+b;
 puts(sum);

}

#include <stdio.h>
#include <sys/types.h>
#include <sys/stat.h>
#include <unistd.h> 

int get_file_size ( const char* file_name )
{
	struct stat _file_info;
	int _n_bytes = 0;

	if ( 0 <= stat ( file_name, &_file_info ) ) {
		if ( S_ISREG ( _file_info.st_mode ) )
			_n_bytes = _file_info.st_size;
	}

	return ( _n_bytes );
}

char* read_content ( const char* file_name )
{
	FILE* _file_pointer = NULL;
	char* cp_file_content = NULL;
	int _n_bytes = 0;

	_n_bytes = get_file_size ( file_name );
	cp_file_content = (char*) malloc ( ( _n_bytes + 1 ) 
* sizeof ( char ) );
	if ( NULL != cp_file_content ) {
		_file_pointer = fopen ( file_name, "r" );
		if ( _file_pointer ) {
			if ( _n_bytes != fread ( 
cp_file_content, 1, _n_bytes, _file_pointer ) ) {
				printf ( "\n The File 
Name :%s, Read Problem", file_name );
				free ( cp_file_content );
				cp_file_content = NULL;
			} else {
				cp_file_content [ 
_n_bytes ] = '\0';
			}
			fclose ( _file_pointer );
		}
	}

	return ( cp_file_content ); 
}

int main ( int argc, char* argv [] )
{
	char* cp_file_content = NULL;
	int n_length = 0;
	int n_counter = 0;

	if ( argc <= 1 || argc > 3 ) {
		printf ( "\n Usage : comment_remover 
<file_name>" );
		exit ( 0 );
	}

	cp_file_content = read_content ( argv [ 1 ] );

	if ( NULL != cp_file_content ) {
		n_length = strlen ( cp_file_content );
		for ( n_counter = 0; n_counter < n_length; 
n_counter++ ) {
			if ( ( * ( cp_file_content + 
n_counter ) == '/' ) 
					&& ( * ( 
cp_file_content + n_counter + 1 ) == '*' ) ) {
				while ( * ( cp_file_content 
+ n_counter ) != '\0' ) { 
					n_counter++;
					if ( ( * ( 
cp_file_content + ( n_counter - 1 ) ) == '*' ) 
							&& 
( * ( cp_file_content + n_counter ) == '/' ) ) {
					
	n_counter++; // move away from / (slash)
						break;
					}
				}
			} else if ( ( * ( cp_file_content + 
n_counter ) == '/' ) 
					&& ( * ( 
cp_file_content + n_counter + 1 ) == '/' ) ) {
				while ( * ( cp_file_content 
+ n_counter ) != '\n' ) n_counter++;
			} 
			printf ( "%c", * ( cp_file_content 
+ n_counter ) );
		}
	}
}

#include<stdio.h>
int main(int argc,char *argv[])
{
	FILE *f;
	int flag;
	f = fopen(argv[1],"r");	
	while(!feof(f))
	{
		ch = fgetc(f),flag = 0;
		if(ch == '/')
		{
			ch  = fgetc(f);
			if(ch == '*')
			{
				flag = 1;
				while(1)
					if(fgetc(f) == '*' && fgetc(f) == '/')
						break;
			}
			else if(ch == '/')
			{
				flag = 1;
				while(fgetc(f)!= '/n');
			}
			else	
				printf("/");// if it s division operator
		}
		if(!flag )
			printf("%c",ch); 
	}
	fclose(f);
}
/*
Run d prog as
>./a.out file_name.cpp
*/

#include <stdio.h>

char* str_str ( char* cp_str, char* cp_pattern )
{
	char* cp_temp = NULL;
	int n_pattern_length = 0;

	n_pattern_length = strlen ( cp_pattern );

	while ( cp_str && *cp_str ) {
		if ( !strncmp ( cp_str, cp_pattern, ( 
n_pattern_length ) ) ) {
			cp_temp = cp_str;
			break;
		}
		else cp_str++;
	}
	
	return ( cp_temp );

}

int main ( int argc, char* argv [ ] )
{
	char array [] = {"Hello World"};
	char* cp_temp = NULL;

	cp_temp = str_str ( array, "lo " );
	if ( NULL != cp_temp ) {
		printf ("\n%s", cp_temp);
	} else printf ("\nReturned null");

	return ( 0 );

}

#include <iostream>
#include<fstream>
using namespace std;
int main()
{
int arr[26]={0},i;
ifstream fin;
ofstream fout;
fout.open("input.txt",ios::app);
fin.open("input.txt",ios::out);
char  c;
fin>>c;
while(!fin.eof())
{
if(isalpha(c))
{
tolower(c);
switch(c)
{
case 'a':
        arr[0]++;
        break;
case 'b':
      arr[1]++;break;
case 'c':
       arr[2]++;break;
case 'd':
       arr[3]++;break;
case 'e':
       arr[4]++;break;
case 'f':
       arr[5]++;break;
case 'g':
        arr[6]++;break;
case 'h':
      arr[7]++;break;
case 'i':
       arr[8]++;break;
case 'j':
        arr[9]++;break;
case 'k':
        arr[10]++;break;
case 'l':
        arr[11]++;break;
case 'm':
        arr[12]++;break;
case 'n':
        arr[13]++;break;
case 'o': 
         arr[14]++;break;
case 'p':  
         arr[15]++;break;
case 'q':
       arr[16]++;break;
case 'r':
       arr[17]++;break;
case 's':
       arr[18]++;break;
case 't':
       arr[19]++;break;
case 'u':
      arr[20]++;break;
case 'v':
       arr[21]++;break;
case 'w':
       arr[22]++;break;
case 'x':
       arr[23]++;break;
case 'y':
       arr[24]++;break;
case 'z':
       arr[25]++;break;
}
}
fin>>c;
}//endl of while.*/
for(i=0;i<26;i++)
fout<<"no of letter "<<static_cast<char>(i+65)<<" is
"<<arr[i]<<endl;
fin.close();
return 0;
}

#include <stdio.h>

int main()
{
FILE *file;
int alpha[26]={0};
const char fileName[]="abc.txt";
char ch;
file= fopen(fileName,"r");
if(file == NULL)
{
  printf("cannot open the %s",fileName);
exit(8);
}
do
{
 ch=fgetc(file);
char c = tolower(ch);
switch(c)
{
case'a': alpha[0]++;
         break;
case'b': alpha[1]++;
         break;
case'c': alpha[2]++;
         break;
case'd': alpha[3]++;
         break;
case'e': alpha[4]++;
         break;
case'f': alpha[5]++;
         break;
case'g': alpha[6]++;
         break;
case'h': alpha[7]++;
         break;
case'i': alpha[8]++;
         break;
case'j': alpha[9]++;
         break;
case'k': alpha[10]++;
         break;
case'l': alpha[11]++;
         break;
case'm': alpha[12]++;
         break;
case'n': alpha[13]++;
         break;
case'o': alpha[14]++;
         break;
case'p': alpha[15]++;
         break;
case'q': alpha[16]++;
         break;
case'r': alpha[17]++;
         break;
case's': alpha[18]++;
         break;
case't': alpha[19]++;
         break;
case'u': alpha[20]++;
         break;
case'v': alpha[21]++;
         break;
case'w': alpha[22]++;
         break;
case'x': alpha[23]++;
         break;
case'y': alpha[24]++;
         break;
case'z': alpha[25]++;
         break;
}

}while(ch != EOF);
int i;
for(i=0;i<26;i++)
{  
  printf("%d\n",alpha[i]);
}
return 0;
}

this is the correct answer

#include <stdio.h>
int count[26];
int main()
{	
	FILE *f;
	int i;
	char ch;
	f = fopen("capti.c","r");	
	while(!feof(f))
	{
		ch = fgetc(f);
		count[ch - 'a']++;	
	}
	for(i = 0;i < 26;i++)
		printf("count[%c] = %d\n",65+i,count[i]);
	fclose(f);
	return 0;
}
//asuming only lower characters are required

#include<stdio.h>
main()
{
int n,m;
}

#include <stdio.h>

int reverse_digits ( int number, int* digits )
{
	int n_digits = 0;
	int rev_number = 0;

	while ( number ) {
		rev_number *= 10;
		rev_number += number % 10;
		number /= 10;
		n_digits++;
	}

	*( digits ) = n_digits;
	return ( rev_number );
}

int palindrome ( int number, int digits )
{
	int flag = 1;

	while ( number ) {
		if ( ( number % 10 ) != ( number / (int) 
pow ( 10, digits - 1 ) ) && ( digits > 1 ) ) {
			flag = 0;
			break;
		}

		// removing first and last digits
		if ( 1 < ( digits -= 2 ) ) {
			number /= 10;
		       	number = number % (int) pow ( 10, 
digits );
		} else break;
	}

	return ( flag );
}

int main ( int argc, char* argv [] )
{
	int number = 123;
	int digits = 0;

	int rev_number = 0;

	rev_number = reverse_digits ( number, &digits );

	if ( !palindrome ( number, digits ) ) {
		printf ("\n NOT Palindrome, creating 
Palindrome");
		/** 
		 * if you need 12321 use digits - 1
		 * if you need 123321 use digits
		 */
		number *= (int) pow ( 10, ( digits - 1 ) );
		number += rev_number % (int) pow ( 10, ( 
digits - 1 ) );
	}

	printf ( "\n The Number: %d", number );

	return ( 0 );
}

i have used functions.... i got this logic.

#include<stdio.h>
#include<conio.h>
void logic1(int);
void logic2();
void logic3();
void logic4();
int a=0;
void main()
{
int m;
printf("enter the number of lines :");
scanf("%d",&m);
for(int i=1;i<=(m/2+1);i++)
{
printf("\n");
if(i%2!=0)
logic1(i);
else
logic2();
}
for(i=1;i<=m/2;i++)
{
printf("\n");
if(i%2==0)
logic3();
else
logic4();
}
getch();
}
void logic1(int p)
{
a=(m/2+2)-p;
for(int i=a;i>=1;i--)
printf("%d",i);
}
void logic2();
{
a--;
for(int i=1;i<=a;i++)
printf("%d",i);
}
void logic3()
{
a++;
for(int i=a;i>=1;i++)
printf("%d",i);
}
void logic4()
{
a++;
for(int i=1;i<=a;i++)
printf("%d",i);
}

this is the correct logic..... some syntax mistakes was done
befoe ... nowq it's correct.... thank you

#include<stdio.h>
#include<conio.h>
void logic1(int,int);
void logic2();
void logic3();
void logic4();
int a=0;
void main()
{
int m;
printf("enter the number of lines :");
scanf("%d",&m);
for(int i=1;i<=(m/2+1);i++)
{
printf("\n");
if(i%2!=0)
logic1(i,m);
else
logic2();
}
for(i=1;i<=m/2;i++)
{
printf("\n");
if(i%2!=0)
logic3();
else
logic4();
}
getch();
}
void logic1(int p,int m)
{
a=(m/2+2)-p;
for(int i=a;i>=1;i--)
printf("%d",i);
}
void logic2();
{
a--;
for(int i=1;i<=a;i++)
printf("%d",i);
}
void logic3()
{
a++;
for(int i=a;i>=1;i++)
printf("%d",i);
}
void logic4()
{
a++;
for(int i=1;i<=a;i++)
printf("%d",i);
}

#include<stdio.h>
#include<conio.h>

void main()
{
	for(int i=1; i<=2; i++)
		for(int j=(i==1)?1:2;j<=5;j++)
		{
			for(int k = (j%2==1 && i==1)? 5 + 1 -j: (j%2==0 && i==2)?
j:1; (j%2==1 && i==1)? k>=1: (j%2==0 && i==2)? k>=1:(j%2==0
&& i==1)?k<=5 + 1 -j:k<=j ;(j%2==1 && i==1)? k--: (j%2==0 &&
i==2)? k--:k++)
				printf("%d", k);
			printf("\n");
		}
}

DIFFERENT LOGIC

#include<stdio.h>
#include<conio.h>
void main()
{
int m,n,p;
printf("enter the limit value");
scanf("%d",&m);
n=m+1;
for(int i=1;i<2*m-1;i++)
{
   if(i<=m)
{
n--;
p=0;
}
else
{
n++
p=1;
}
if(i%2==p)
{
for(int j=1;j<n;j++)
printf("%d",j);
}
else
{
for(j=n;j>=1;j--)
printf("%d",j);
}
}
getch();
}

#include<stdio.h>

int main()
{
  int i,j,n=5;

 for(i=n;i>=1;i--)
 {
  if(i%2 == 0){
  for(j=1;j<=i;j++)
  printf("%d",j);
  printf("\n");
  }
  else
  {
   for(j=i;j>=1;j--)
   printf("%d",j);
   printf("\n");
  }
 }
for(i=2;i<=5;i++)
{
  if(i%2 != 0) {
  for(j=1;j<=i;j++)
  printf("%d",j);
  printf("\n");
  }
  else
  {
  for(j=i;j>=1;j--)
  printf("%d",j);
  printf("\n");
  }
}
}

#include <stdio.h>

int get_digits ( int number )
{
	int n_digits = 0;
	while ( number ) {
		n_digits++;
		number = number / 10;
	}

	return ( n_digits );
}

int main ( int argc, char* argv [] )
{
	int number = 789876;
	int digits = 0;
	int temp_number = 0;
	int final_number = 0;
	int counter = 0;

	int start_point = 2;
	int end_point = 5;

	digits = get_digits ( number );

	if ( ( start_point < end_point ) && ( end_point <= 
digits ) ) {

		temp_number = number;
		if ( start_point - 1 ) final_number = 
number / pow ( 10, ( digits - ( start_point - 1 ) ) );

		counter = digits;
		while ( temp_number ) {
			if ( ( counter <= ( end_point ) ) 
&& ( counter >= ( start_point ) ) ) {
				final_number *= 10;
				final_number += ( 
temp_number % 10 );
			}
			temp_number /= 10;
			counter--;
		}
		if ( digits - end_point ) {
		       	final_number *= pow ( 10, ( digits -
 end_point ) );
			final_number += number % (int) ( 
pow ( 10, ( digits - end_point ) ) );
		}
	}

	printf ( "\n      Number: %d", number );
	printf ( "\nFinal Number: %d", final_number );
	printf ( "\nS_Pos: %d, E_Pos: %d", start_point, 
end_point );

	return ( 0 );

}

#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cmath>
using namespace std;
int getdnum(int num)
{
int numd=0;
while(num!=0)
{
numd++;
num=num/10;
}
return numd;
}
int reversenum(int i,int j ,int d,int num)
{
int a=(num/(pow(10,d-i+1)));
int b=(num/(pow(10,d-j)));
int c=num%static_cast<int>(pow(10,d-j));
int n=0;
int k;
for(k=0;k<=(j-i);k++)
{
n+=(b%10)*(pow(10,j-i-k));
b=b/10;
}
n=a*pow(10,d-i+1)+c+n*pow(10,d-j);
return n;

}
int main()
{
int i,j,k,l,m;
cin>>i>>j>>k;
int d=getdnum(i);
m=reversenum(j,k,d,i);
cout<<m<<endl;
return 0;
}

#include <stdio.h>

void reverse ( int* ip_array, int st_pos, int ed_pos )
{
	if ( ( ip_array ) && ( st_pos < ed_pos ) ) {
		* ( ip_array + st_pos ) ^= * ( ip_array + 
ed_pos ) ^= * ( ip_array + st_pos ) ^= * ( ip_array + 
ed_pos );
		reverse ( ip_array, ++st_pos, --ed_pos );
	}
}

int main ( int argc, char* argv [] )
{
	int int_array [20];
	int number = 789876;
	int counter = 0;
	int nloop = 0;
	int start_pos = 2;
	int end_pos = 5;

	/* split the number into an array */
	while ( number ) {
		int_array [ counter++ ] = number % 10;
		number = number / 10;
	}

	/* reverse the splited array */
	reverse ( int_array, 0, counter - 1 );

	/* reverse for the particular position */
	if ( ( start_pos < end_pos ) && ( end_pos <= 
counter ) ) {
		reverse ( int_array, ( start_pos - 1 ), ( 
end_pos - 1 ) );
		
		number = 0;
		for ( nloop = 0; nloop < counter; nloop++ )
		{
			number *= 10;
			number += int_array [ nloop ];
		}

		printf ( "\n %d", number );
	}

	return ( 0 );
}