Question

Hello, I have a question about a C program problem below.

The question is " Write ONE C Program to accomplish the following task: Convert a double number to its hexadecimal form"

But, there are constraints below.

Constraints:
(1) The double number x is located between 0 and 1000 (0 <= x < =1000), e.g. 678.345.

(2) The double number with the hexadecimal form contains 6 significant digits. e.g. “5D.32FA45”.

(3) The double number with the hexadecimal form is represented by a string (or a character array), e.g. “5D.32FA45”.

Thanks for helping me!

Answer 1

=========================

C PROGRAM

=========================

/******************************************************************************
File: DoubleToHexadecimal.c

The program converts double number >=0 and <=1000

to its hex form (upto 6 significant digits)

*******************************************************************************/

#include <stdio.h>
/*prototype for function intToHex*/
char intToHex(int);

/*main program starts*/
int main()
{
double dblValue;
long intPart;
float fractionPart;
char hex[10]; /*here the actual hex digits will be stored*/
char tempIntToHex[3]; /*temp storage for int part*/
int remain;
int i ,counter ,j;
  
printf("Enter double value: ");
scanf("%lf",&dblValue);/*read double value from user*/
  
if(dblValue >=0 && dblValue <=1000) /*check if the input number is in range [0,1000]*/
{
intPart = (int)dblValue; /*get the integer part*/
fractionPart = dblValue - intPart; /*get the fraction part*/
counter = 0; /* this will maintain total characters in hex array*/
  
/*conversion of int part to its equivalent hex form*/
while(intPart > 0)
{
remain = intPart%16;
intPart = intPart/16;
tempIntToHex[counter] = intToHex(remain);
counter++;
}
  
/*store contents of tempIntToHex into hex in reverse order*/
for(j = 0; j < counter; j++)
{
hex[j] = tempIntToHex[counter -j-1];
}
  
hex[counter]='.'; /*store the point*/
  
counter++;
  
/*now convert the fractinPart to its equivalent hex form
this will be done for max 6 significant digits only*/
for(i = 1; i <=6 ;i++)
{

fractionPart = fractionPart * 16;
/*get int part and convert it to hex digit*/
intPart = (int)fractionPart;
hex[counter] = intToHex(intPart);
fractionPart = fractionPart - intPart;/*get next fraction part*/
counter++;
/*if at any point fraction part becomes zero
before 6th iteration , break out of the loop*/
if(fractionPart == 0){
break;
}
}
  
}else{
/*the number is out of range*/
printf("Input number is outside the range [0,1000].");
return 0;
}
  
/*print the contents of hex array, which holds the hex form of input number*/
printf("\nThe hexadecimal number is: ");
for(i = 0; i < counter ; i++)
{
printf("%c",hex[i]);
}
  
return 0;
}

/*function to convert num in 0 to 15 to its equivalent hex digit*/
char intToHex(int num)
{
char hexDigit;
  
if(num >=0 && num <=15){
if(num >=0 && num<=9){
hexDigit = num+'0';/*hexDigit will be the num itself*/
}else{
hexDigit = 'A' + (num -10); /*10 will be 'A',11 will be 'B' ,....15 will be 'F'*/
}
}
/*printf("\n%d = %c",num,hexDigit);*/
return hexDigit;
}

============================

OUTPUT

============================

RUN1

Enter double value: 456.2569

The hexadecimal number is: 1C8.41C433

RUN2

Enter double value: 999.987

The hexadecimal number is: 3E7.FCAC08

RUN3

Enter double value: 143.25

The hexadecimal number is: 8F.4

RUN4

Enter double value: 1011
Input number is outside the range [0,1000].