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2. Problem 1: Floating Point (floating)
In this problem, we will investigate how floating-point numbers are represented in
memory. Recall that a float is a 32-bit value with a single sign bit, eight exponent bits,
and 23 mantissa bits. Specifically, a floating point number x with sign bit ‘sign’, exponent
e, and mantissa bits m0, m1,. . . , m22 can be
written1
x = (−1)sign · (1.m22m21m20 . . . m0) ·2e-bias
where the mantissa is, of course, in base two. You will be given a list of N floating point
values x1, x2, .. . , xN . For each xi , your program should write its binary representation to
the output file as indicated below.
Suggested approach: You’ll need to use bitwise operations, but you cannot do so on a
floating-point number directly. Instead, you will need a way of considering a variable as
either a float or an unsigned int. We will use a union, which is valid in this case because
we assume the size of the two data types is the same.
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C Programming Assignment and Solution

3. union float bits {
float f;
unsigned int bits;
};
II print hex( 5.Of ) outputs "The float looks like Ox4OaOOOOO in hex."
void print hex( float f) {
union float bits t;
t.f = f;
printf( "The float looks like Ox%x in hex.n", t.bits );
}
Input Format
Line 1: One integer N
Lines 2 . . .N + 1: Line i + 1 contains floating point number xi
Sample Input (file floating.in)
3
1.5
O.15625
-7.333
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4. 1Except for the case where x is a denormal floating point number, as discussed in class,
in which case the (unbiased) exponent is -126 and mantissa is written 0.m22m21m20 . . .
m0.
Output Format
Lines 1 . . .N : Line i contains a representation of the floating-point number xi ,
formatted as shown in the sample output.
Sample Output (file floating.out)
1.1OOOOOOOOOOOOOOOOOOOOOO * 2^O
1.O1OOOOOOOOOOOOOOOOOOOOO * 2^-3
-1.11O1O1O1O1OO1111111OOOO * 2^2
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5. /*
PROG: floating
LANG: C
*/
#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>
#include <math.h>
#define ABSOLUTE_WIDTH 31
#define MANTISSA_WIDTH 23
#define EXPONENT_WIDTH 8
#define EXPONENT_MASK 0xffu
#define MANTISSA_MASK 0x007fffffu
#define EXPONENT_BIAS 127
union float_bits {
float f;
uint32_t bits;
};
void print_float( FILE *output, float f ) {
union float_bits t; t.f = f;
uint32_t sign_bit = ( t.bits » ABSOLUTE_WIDTH );
uint32_t exponent = ( t.bits » MANTISSA_WIDTH ) & EXPONENT_MASK;
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6. uint32_t mantissa = ( t.bits & MANTISSA_MASK );
if( sign_bit != 0 ) {
fprintf( output, “-” );
}
if( exponent > 2 * EXPONENT_BIAS ) {
fprintf( output, “Infn” ); /* Infinity */
return;
} else if( exponent == 0 ) {
fprintf( output, “0.” ); /* Zero or Denormal */
exponent = ( mantissa != 0 ) ? exponent + 1 : exponent;
} else {
fprintf( output, “1.” ); /* Usual */
}
for( int k = MANTISSA_WIDTH - 1; k >= 0; –k ) {
fprintf( output, “%d”, ( mantissa » k ) & 1 );
}
if( exponent != 0 || mantissa != 0 ) {
fprintf( output, " * 2^%dn", (int) ( exponent - EXPONENT_BIAS ) );
}
}
int main() {
FILE *input = fopen( “floating.in”, “r” ),
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7. *output = fopen( “floating.out”, “w” );
size_t N; float f;
fscanf( input, “%zu”, &N );
for( size_t i = 0; i < N; ++i ) {
fscanf( input, “%f”, &f );
print_float( output, f );
}
fclose( input );
fclose( output );
return 0;
}
Below is the output using the test data:
floating:
1: OK [0.004 seconds] OK!
2: OK [0.004 seconds] OK!
3: OK [0.004 seconds] OK!
4: OK [0.004 seconds] OK!
5: OK [0.005 seconds] OK!
6: OK [0.004 seconds] OK!
7: OK [0.004 seconds] OK!
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