Question:How to convert large binary number to hexadecimal 

#include<stdio.h>
#define MAX 1000

int main(){

   char binaryNumber[MAX],hexaDecimal[MAX];
   int temp;
   long int i=0,j=0;
  
   printf("Enter any number any binary number: ");
   scanf("%s",binaryNumber);

   while(binaryNumber[i]){
      binaryNumber[i] = binaryNumber[i] -48;
      ++i;
   }

   --i;
   while(i-2>=0){
       temp =  binaryNumber[i-3] *8 + binaryNumber[i-2] *4 +  binaryNumber[i-1] *2 + binaryNumber[i] ;
       if(temp > 9)
            hexaDecimal[j++] = temp + 55;
       else
            hexaDecimal[j++] = temp + 48;
       i=i-4;
   }

   if(i ==1)
      hexaDecimal[j] = binaryNumber[i-1] *2 + binaryNumber[i] + 48 ;
   else if(i==0)
      hexaDecimal[j] =  binaryNumber[i] + 48 ;
    else
      --j;

   printf("Equivalent hexadecimal value: ");
   while(j>=0){
      printf("%c",hexaDecimal[j--]);
   }

   return 0;
}
Sample output:
Enter any number any binary number: 1010011011100011110
001001111011110001000100011101110111011110
Equivalent hexadecimal value: 14DC789EF111DDDE

Question:C code for sum of two binary numbers 

#include<stdio.h>

int main(){

    long int binary1,binary2;
    int i=0,remainder = 0,sum[20];

    printf("Enter any first binary number: ");
    scanf("%ld",&binary1);
    printf("Enter any second binary number: ");
    scanf("%ld",&binary2);

    while(binary1!=0||binary2!=0){
         sum[i++] =  (binary1 %10 + binary2 %10 + remainder ) % 2;
         remainder = (binary1 %10 + binary2 %10 + remainder ) / 2;
         binary1 = binary1/10;
         binary2 = binary2/10;
    }

    if(remainder!=0)
         sum[i++] = remainder;

    --i;
    printf("Sum of two binary numbers: ");
    while(i>=0)
         printf("%d",sum[i--]);

   return 0;
}
Sample output:
Enter any first binary number: 1100011
Enter any second binary number: 1101
Sum of two binary numbers: 1110000

Question:C program for multiplication of two binary numbers 

#include<stdio.h>

int binaryAddition(int,int);

int main(){

    long int binary1,binary2,multiply=0;
    int digit,factor=1;

    printf("Enter any first binary number: ");
    scanf("%ld",&binary1);
    printf("Enter any second binary number: ");
    scanf("%ld",&binary2);

    while(binary2!=0){
         digit =  binary2 %10;

         if(digit ==1){
                 binary1=binary1*factor;
                 multiply = binaryAddition(binary1,multiply);
         }
         else
             binary1=binary1*factor;
   
         binary2 = binary2/10;
         factor = 10;
    }

    printf("Product of two binary numbers: %ld",multiply);
   
   return 0;
}

int binaryAddition(int binary1,int binary2){

    int i=0,remainder = 0,sum[20];
    int binarySum=0;

    while(binary1!=0||binary2!=0){
         sum[i++] =  (binary1 %10 + binary2 %10 + remainder ) % 2;
         remainder = (binary1 %10 + binary2 %10 + remainder ) / 2;
         binary1 = binary1/10;
         binary2 = binary2/10;
    }

    if(remainder!=0)
         sum[i++] = remainder;
    --i;
    while(i>=0)
         binarySum = binarySum*10 + sum[i--];

    return binarySum;
}
Sample output:
Enter any first binary number: 1101
Enter any second binary number: 11
Product of two binary numbers: 100111

Question:C code for fractional binary to decimal converter 

#include<stdio.h>
#define MAX 1000

int main(){

    long double fraDecimal=0.0,dFractional=0.0 ,fraFactor=0.5;
    long int dIntegral = 0,bIntegral=0,bFractional[MAX];
    long int intFactor=1,remainder,i=0,k=0,flag=0;
    char fraBinary[MAX];

    printf("Enter any fractional binary number: ");
    scanf("%s",&fraBinary);
   
    while(fraBinary[i]){
        
         if(fraBinary[i] == '.')
             flag = 1;
         else if(flag==0)
             bIntegral = bIntegral * 10 + (fraBinary[i] -48);
         else
              bFractional[k++] = fraBinary[i] -48;
         i++;
    }
   
    while(bIntegral!=0){
        remainder=bIntegral%10;
        dIntegral= dIntegral+remainder*intFactor;
        intFactor=intFactor*2;
        bIntegral=bIntegral/10;
    }
   
    for(i=0;i<k;i++){
         dFractional  = dFractional  + bFractional[i] * fraFactor;
         fraFactor = fraFactor / 2;
    }

    fraDecimal = dIntegral + dFractional ;

    printf("Equivalent decimal value: %Lf",fraDecimal);
   
    return 0;
}
Sample output:
Enter any fractional binary number: 11.11
Equivalent decimal value: 3.750000

Question:C code for fractional decimal to binary converter 

#include<stdio.h>

int main(){
   
    long double fraDecimal,fraBinary,bFractional = 0.0,dFractional,fraFactor=0.1;
    long int dIntegral,bIntegral=0;
    long int intFactor=1,remainder,temp,i;

    printf("Enter any fractional decimal number: ");
    scanf("%Lf",&fraDecimal);
   
    dIntegral = fraDecimal;
    dFractional =  fraDecimal - dIntegral;

    while(dIntegral!=0){
         remainder=dIntegral%2;
         bIntegral=bIntegral+remainder*intFactor;
         dIntegral=dIntegral/2;
         intFactor=intFactor*10;
    }

   for(i=1;i<=6;i++){
      
       dFractional = dFractional * 2;
       temp =  dFractional;
        
       bFractional = bFractional + fraFactor* temp;
       if(temp ==1)
             dFractional = dFractional - temp;

       fraFactor=fraFactor/10;
   }
  
   fraBinary =  bIntegral +  bFractional;
   printf("Equivalent binary value: %lf",fraBinary);
   
   return 0;
}
Sample output:
Enter any fractional decimal number: 5.7
Equivalent binary value: 101.101100

Question:Convert numbers to roman numerals in c 

#include<stdio.h>

void predigits(char c1,char c2);
void postdigits(char c,int n);

char roman_Number[1000];
int i=0;

int main(){

    int j;
    long int number;
   
    printf("Enter any natural number: ");
    scanf("%d",&number);
   
    if(number <= 0){
         printf("Invalid number");
         return 0;
    }

    while(number != 0){

         if(number >= 1000){
             postdigits('M',number/1000);
             number = number - (number/1000) * 1000;
         }
         else if(number >=500){
             if(number < (500 + 4 * 100)){
                 postdigits('D',number/500);
                 number = number - (number/500) * 500;
             }
             else{
                 predigits('C','M');
                 number = number - (1000-100);
             }
         }
         else if(number >=100){
             if(number < (100 + 3 * 100)){
                 postdigits('C',number/100);
                 number = number - (number/100) * 100;
             }
             else{
                 predigits('L','D');
                 number = number - (500-100);
             }
         }
         else if(number >=50){
             if(number < (50 + 4 * 10)){
                 postdigits('L',number/50);
                 number = number - (number/50) * 50;
             }
             else{
                 predigits('X','C');
                 number = number - (100-10);
             }
         }
         else if(number >=10){
             if(number < (10 + 3 * 10)){
                 postdigits('X',number/10);
                 number = number - (number/10) * 10;
             }
             else{
                 predigits('X','L');
                 number = number - (50-10);
             }
         }
         else if(number >=5){
             if(number < (5 + 4 * 1)){
                 postdigits('V',number/5);
                 number = number - (number/5) * 5;
             }
             else{
                 predigits('I','X');
                 number = number - (10-1);
             }
         }
         else if(number >=1){
             if(number < 4){
                 postdigits('I',number/1);
                 number = number - (number/1) * 1;
             }
             else{
                 predigits('I','V');
                 number = number - (5-1);
             }
         }
    }

    printf("Roman number will be: ");
    for(j=0;j<i;j++)
         printf("%c",roman_Number[j]);

    return 0;

}

void predigits(char c1,char c2){
    roman_Number[i++] = c1;
    roman_Number[i++] = c2;
}

void postdigits(char c,int n){
    int j;
    for(j=0;j<n;j++)
         roman_Number[i++] = c;
   
}
Sample output:
Enter any natural number: 87
Roman number will be: LXXXVII

Question:C code for roman numbers to English numbers 

#include<stdio.h>
#include<string.h>

int digitValue(char);

int main(){

    char roman_Number[1000];
    int i=0;
    long int number =0;
   
    printf("Enter any roman number (Valid digits are I, V, X, L, C, D, M):  \n");
    scanf("%s",roman_Number);
   
    while(roman_Number[i]){
        
         if(digitValue(roman_Number[i]) < 0){
             printf("Invalid roman digit : %c",roman_Number[i]);
             return 0;
         }
            
         if((strlen(roman_Number) -i) > 2){
             if(digitValue(roman_Number[i]) < digitValue(roman_Number[i+2])){
                 printf("Invalid roman number");
                 return 0;
             }
         }

         if(digitValue(roman_Number[i]) >= digitValue(roman_Number[i+1]))
             number = number + digitValue(roman_Number[i]);
         else{
             number = number + (digitValue(roman_Number[i+1]) - digitValue(roman_Number[i]));
             i++;
         }
         i++;
    }
        
    printf("Its decimal value is : %ld",number);
   
    return 0;

}

int digitValue(char c){

    int value=0;

    switch(c){
         case 'I': value = 1; break;
         case 'V': value = 5; break;
         case 'X': value = 10; break;
         case 'L': value = 50; break;
         case 'C': value = 100; break;
         case 'D': value = 500; break;
         case 'M': value = 1000; break;
         case '\0': value = 0; break;
         default: value = -1; 
    }

    return value;
}
Sample output:
Enter any roman number (Valid digits are I, V, X, L, C, D, M):
XVII
Its decimal value is: 17

Question:C code to covert each digits of a number in English word 

#include<stdio.h>

int main(){

    int number,i=0,j,digit;
    char * word[1000];

    printf("Enter any integer: ");
    scanf("%d",&number);

    while(number){

    digit = number %10;
    number = number /10;

         switch(digit){
             case 0: word[i++] = "zero"; break;
             case 1: word[i++] = "one"; break;
             case 2: word[i++] = "two"; break;
             case 3: word[i++] = "three"; break;
             case 4: word[i++] = "four"; break;
             case 5: word[i++] = "five"; break;
             case 6: word[i++] = "six"; break;
             case 7: word[i++] = "seven"; break;
             case 8: word[i++] = "eight"; break;
             case 9: word[i++] = "nine"; break;

         }
    }
   
    for(j=i-1;j>=0;j--){
         printf("%s ",word[j]);
    }

    return 0;

}
Sample output:
Enter any integer: 23451208
two three four five one two zero eight

Question:C code to convert any number to English word 

#include<stdio.h>
#include<string.h>

void toWord(int,int);
char * getPositionValue(int);
char * digitToWord(int);

char  word[100][30];
int i =0;

int main(){

    int j,k,subnumer;
    unsigned long int number;

    printf("Enter any postive number: ");
    scanf("%lu",&number);
   
    if(number ==0){
         printf("Zero");
         return 0;
    }

    while(number){

         if(i==1){
             toWord(number %10,i);
             number = number/10;
         }else{
             toWord(number %100,i);
             number = number/100;
         }

         i++;
        
    }

    printf("Number in word: ");
    *word[i-1] = *word[i-1] - 32;
    for(j=i-1;j>=0;j--){
         printf("%s",word[j]);
    }

    return 0;

}

void toWord(int number,int position){

    char  numberToword[100]={" "};

    if(number ==0){
    }else if (number < 20 ||number %10==0){
         strcpy(numberToword,digitToWord(number));
    }else{
         strcpy(numberToword,digitToWord((number/10)*10));
         strcat(numberToword,digitToWord(number%10));
    }
   
    strcat(numberToword,getPositionValue(position));
    strcpy(word[i],numberToword);
}

char * getPositionValue(int postion){

    static char positionValue[10]=" ";
   
    switch(postion){

         case 1: strcpy(positionValue,"hundreds "); break;
         case 2: strcpy(positionValue,"thousand "); break;
         case 3: strcpy(positionValue,"lakh "); break;
         case 4: strcpy(positionValue,"crore "); break;
         case 5: strcpy(positionValue,"arab "); break;
         case 6: strcpy(positionValue,"kharab "); break;
         case 7: strcpy(positionValue,"neel "); break;
         case 8: strcpy(positionValue,"padam "); break;
    }
    
    return positionValue;
}

char * digitToWord(int digit){

     static char digitInWord[10]=" ";

    switch(digit){
         case 1: strcpy(digitInWord , "one "); break;
         case 2: strcpy(digitInWord , "two "); break;
         case 3: strcpy(digitInWord , "three "); break;
         case 4: strcpy(digitInWord , "four "); break;
         case 5: strcpy(digitInWord , "five "); break;
         case 6: strcpy(digitInWord , "six "); break;
         case 7: strcpy(digitInWord , "seven "); break;
         case 8: strcpy(digitInWord , "eight "); break;
         case 9: strcpy(digitInWord , "nine ");break;
         case 10: strcpy(digitInWord , "ten "); break;
         case 11: strcpy(digitInWord , "eleven "); break;
         case 12: strcpy(digitInWord , "twelve "); break;
         case 13: strcpy(digitInWord , "thirteen "); break;
         case 14: strcpy(digitInWord , "fourteen "); break;
         case 15: strcpy(digitInWord , "fifteen "); break;
         case 16: strcpy(digitInWord , "sixteen "); break;
         case 17: strcpy(digitInWord , "seventeen "); break;
         case 18: strcpy(digitInWord , "eighteen "); break;
         case 19: strcpy(digitInWord , "nineteen "); break;
         case 20: strcpy(digitInWord , "twenty "); break;
         case 30: strcpy(digitInWord , "thirty "); break;
         case 40: strcpy(digitInWord , "fourty "); break;
         case 50: strcpy(digitInWord , "fifty "); break;
         case 60: strcpy(digitInWord , "sixty "); break;
         case 70: strcpy(digitInWord , "seventy "); break;
         case 80: strcpy(digitInWord , "eighty "); break;
         case 90: strcpy(digitInWord,"ninety "); break;
    }
   
    return digitInWord;
}

Question:C code to convert one unit to other unit 

#include<stdio.h>
#include<math.h>

int main(){
   
    char fromUnit,toUnit;
    char *fUnit,*tUnit;
    long double fromValue,meterValue,toValue;
    int power =0;

    printf("To convert the 12 Inch to Foot\n");
    printf("enter the the unit in the format : dc12\n");

    printf("Ell: a\n");
    printf("Femi: b\n");
    printf("Foot: c\n");
    printf("Inch: d\n");
    printf("Light year: e\n");
    printf("Metre: f\n");
    printf("Mile: g\n");
    printf("Nano meter: h\n");
    printf("Pace: i\n");
    printf("Point: j\n");
    printf("Yard: k\n");
    printf("Mili meter: l\n");
    printf("Centi meter: m\n");
    printf("Deci meter: n\n");
    printf("Deca meter: o\n");
    printf("Hecto meter: p\n");
    printf("Kilo meter: q\n");

    scanf("%c%c%Lf",&fromUnit,&toUnit,&fromValue); 

    switch(fromUnit){
         case 'a': meterValue = fromValue * 1.143; fUnit="ell"; break;
         case 'b': meterValue = fromValue ; power = -15; fUnit="fm"; break;
         case 'c': meterValue = fromValue * 0.3048; fUnit="ft"; break;
         case 'd': meterValue = fromValue * 0.0254; fUnit="in"; break;
         case 'e': meterValue = fromValue * 9.4607304725808; power =15; fUnit="ly"; break;
         case 'f': meterValue = fromValue; fUnit="m"; break;
         case 'g': meterValue = fromValue * 1609.344; fUnit="mi"; break;
         case 'h': meterValue = fromValue; fUnit="nm"; power = -9; break;
         case 'i': meterValue = fromValue * 0.762 ; fUnit="pace"; break;
         case 'j': meterValue = fromValue * 0.000351450; fUnit="pt"; break;
         case 'k': meterValue = fromValue * 0.9144; fUnit="yd"; break;
         case 'l': meterValue = fromValue * 0.001; fUnit="mm"; break;
         case 'm': meterValue = fromValue * 0.01; fUnit="cm"; break;
         case 'n': meterValue = fromValue * 0.1; fUnit="deci meter"; break;
         case 'o': meterValue = fromValue * 10; fUnit="deca meter"; break;
         case 'p': meterValue = fromValue * 100; fUnit="hm"; break;
         case 'q': meterValue = fromValue * 1000; fUnit="km"; break;
         default:
             printf("Invalid input"); exit(0);
    }

    switch(toUnit){
         case 'a': toValue = meterValue/1.143; tUnit="ell"; break;
         case 'b': toValue = meterValue; tUnit="fm"; break;
         case 'c': toValue = meterValue/0.3048; tUnit="ft"; break;
         case 'd': toValue = meterValue/0.0254; tUnit="in"; break;
         case 'e': toValue = meterValue/9.4607304725808; tUnit="ly"; break;
         case 'f': toValue = meterValue; tUnit="m";break;
         case 'g': toValue = meterValue/1609.344; tUnit="mi"; break;
         case 'h': toValue = meterValue; tUnit="nm"; break;
         case 'i': toValue = meterValue/0.762; tUnit="pace"; break;
         case 'j': toValue = meterValue/0.000351450; tUnit="pt"; break;
         case 'k': toValue = meterValue/0.9144; tUnit="yd"; break;
         case 'l': toValue = meterValue/0.001; tUnit="mm"; break;
         case 'm': toValue = meterValue/0.01; tUnit="cm"; break;
         case 'n': toValue = meterValue/0.1; tUnit="deci meter"; break;
         case 'o': toValue = meterValue/10; tUnit="deca meter"; break;
         case 'p': toValue = meterValue/100; tUnit="hm"; break;
         case 'q': toValue = meterValue/1000; tUnit="km"; break;
         default:
             printf("Invalid input"); exit(0);
    }
   
    if(power==0)
         printf("%.4Lf %s = %.4Lf %s",fromValue,fUnit,toValue,tUnit);
    else{
         while(tovalue > 10 
         printf("%.4Lf %s = %.4Lf*10^%d %s",fromValue,fUnit,toValue,power,tUnit);
    }


    return 0;
}