Chapter 6, Problem 6E

Explanation of Solution

Verification:

• The single-precision value of “0.1” is “0x3dcccccd”.
• Here, for the “x”, the value is obtained by negating that and then flipping the least significant bit of the mantissa “0x8dcccccc”.

Verifying the given statement using C program:

• The following program depicts the checking the given the single-precision code.

#include <stdio.h>

union

{

    volatile float f;

    volatile int i;

} a, b, c, d, e;

volatile float s;

volatile float t;

void test()

{

    a.f = 0.1;

    b.i = 0xbdcccccc;

    printf("a = %x %e\n", a.i, a.f);

    printf("b = %x %e\n", b.i, b.f);

    c.f = (float) (a.f + b.f);

    printf("a + b = %x %e\n", c.i, c.f);   

    d.f = (float) (a.f * (float) 10.0);

    e.f = (float) (b.f * (float) 10.0);

    printf("a*10 = %x %e\n", d.i, d.f);

    printf("b*10 = %x %e\n", e.i, e.f);

    t = a.f + b.f;

    d...