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ex16a.c
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/*
* 16. Modify Program 7.14 so that the user can convert any number of integers.
* Make provision for the program to terminate when a zero is typed in as the
* value of number to be converted. (alternate version)
* By Faisal Saadatmand
*/
#include <stdio.h>
int convertedNumber[32];
long long int numberToConvert;
int base, digit, numberOfBits;
void getNumberAndBase(void)
{
printf("Number to be converted? ");
scanf("%lli", &numberToConvert);
printf("Base? ");
scanf("%i", &base);
if (base < 2 || base > 16) { /* issue: a value that is
not a number in this range will cause an infinit loop */
do {
printf("Invalid base - must be between 2 and 16\n");
printf("Base? ");
scanf("%i", &base);
} while (base < 2 || base > 16);
}
if (numberToConvert < 0 && base != 10) {
do {
printf("Representation? (4, 8, 16, or 32-bit) ");
scanf("%i", &numberOfBits);
switch (numberOfBits) {
case 4:
printf("4-bit represention\n");
break;
case 8:
printf("8-bit representation\n");
break;
case 16:
printf("16-bit representation\n");
break;
case 32:
printf("32-bit representation\n");
break;
default:
printf("Invalid input\n");
numberOfBits = 0;
break;
}
} while (numberOfBits == 0);
}
}
void convertNumber(int base)
{
digit = 0; /* it's a good idea to reset the value
locally since the functions is called
multiple of times */
do {
convertedNumber[digit] = numberToConvert % base;
++digit;
numberToConvert /= base;
} while (numberToConvert != 0);
}
void twosComplement(void)
{
int i, remainder = 1;
for (i = numberOfBits - 1; i >= 0; --i) /* invert binary number */
if (convertedNumber[i] == 0)
convertedNumber[i] = 1;
else
convertedNumber[i] = 0;
for (i = 0; i <= digit; ++i) /* add 1 to binary number */
if (convertedNumber[i] == 0 && remainder == 1) {
convertedNumber[i] = 1;
remainder = 0;
break; /* optimises algorithm */
} else if (convertedNumber[i] == 1 && remainder == 0) {
convertedNumber[i] = 1;
remainder = 0;
} else if (convertedNumber[i] == 1 && remainder == 1) {
convertedNumber[i] = 0;
remainder = 1;
} else
convertedNumber[i] = 0;
}
long long int x_to_the_n(int x, int n)
{
long long int result = 1LL;
while (n > 0) {
result *= x;
--n;
}
return result;
}
long long int arraySum(long long int a[], int n)
{
int i;
long long int sum = 0;
for (i = 0; i < n; ++i)
sum += a[i];
return sum;
}
void binaryToBase10(void)
{
int i;
long long int array[numberOfBits];
for (i = 0; i <= numberOfBits - 1; ++i)
array[i] = convertedNumber[i] * x_to_the_n(2, i);
numberToConvert = arraySum(array, numberOfBits);
}
void displayConvertedNumber(void)
{
int nextDigit;
const char baseDigits[16] =
{ '0', '1', '2', '3', '4', '5', '6', '7',
'8', '9', 'A', 'B', 'C', 'D', 'E', 'F' };
printf("Converted number = ");
for (--digit; digit >= 0; --digit) {
nextDigit = convertedNumber[digit];
printf("%c", baseDigits[nextDigit]);
}
printf("\n");
}
int main(void)
{
void getNumberAndBase(void), convertNumber(int base), twosComplement(void),
binaryToBase10(void), displayConvertedNumber(void);
long long int x_to_the_n(int x, int n), arraySum(long long int a[], int n);
getNumberAndBase();
if (numberToConvert == 0) /* terminate programme when 0 is entered for the number to convert */
return 1;
else if (base == 10) /* exclude base 10 from the conversion algorithm */
printf("Converted number = %lli\n", numberToConvert);
else if (numberToConvert < 0) { /* handles negative integer input */
numberToConvert = -numberToConvert;
convertNumber(2);
twosComplement();
binaryToBase10(); /* issue: could be optimise to exclude base 2 from this function */
convertNumber(base);
displayConvertedNumber();
} else { /* handles positive integer input */
convertNumber(base);
displayConvertedNumber();
}
return 0;
}