/*
 
 The IEEE 754 standard regarding floating point representations does
 not provide a specification for 8-bit floating point numbers. Imagine,
 however, that such numbers adhere to the following format:
 
    x     x x x     x x x x
 (sign) (exponent) (mantissa)
 
 
 That is, 1 sign bit, 3 exponent bits, and 4 bits for the mantissa.
 
 Write a program that prompts a user to enter 8 bits and outputs the
 decimal equivalent. Make sure the result is printed with three digit precision
 (i.e., "%.3f" in printf).
 
 
 (Hint: The exponent bias for this 8-bit representation is 3)
 
 ============= START OF SAMPLE RUN =======================
 Enter 8-bit floating point number:
 0 0 0 0 0 0 0 0
 The decimal value is 0.125
 =============  END OF SAMPLE RUN  =======================
 
 ============= START OF SAMPLE RUN =======================
 Enter 8-bit floating point number:
 1 1 0 1 0 1 1 0
 The decimal value is -5.500
 =============  END OF SAMPLE RUN  =======================
 
 */

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

#define BIAS 3

int main()
{
	int d1, d2, d3, d4, d5, d6, d7, d8;
	double sign, exponent, mantissa, result;
	
	printf("Enter 8-bit floating point number:\n");
	scanf("%d %d %d %d %d %d %d %d", &d1, &d2, &d3, &d4, &d5, &d6, &d7, &d8);
	
	sign = pow(-1, d1);
	
	exponent = 4*d2 + 2*d3 + d4 - BIAS;
	
	mantissa = 1 + 0.5*d5 + 0.25*d6 + 0.125*d7 + 0.0675*d8;
	
	result = sign * pow(2, exponent) * mantissa;
	
	printf("The decimal value is %.3f\n", result);
	
	system("pause");
	
	return 0;
}