Chapter 4

Expressing computations

Use	For
size_t	sizes, cardinalities, or ordinal numbers
unsigned	small quantities that can’t be negative
signed	small quantities that bear a sign
ptrdiff_t	large differences that bear a sign
double	floating-point calculations
double complex	complex calculations

Exs 13. Show that if the minimal and maximal values for signed long long have similar properties, the smallest integer value for the platform can’t be written as a combination of one literal with a minus sign.

maximum value is $2^p-1$
⇒ constant $2^p$ doesn’t fit into signed long long

⇒ minimal signed value is $-2^p$ cannot be written as a literal of the same type with a minus sign

Exs 14. Show that if the maximum unsigned is $2^{16} -1$ , then -0x8000 has value 32768, too.

0x8000 has value 32768

⇒ -0x8000 is UINT_MAX $+ 1 - 32768 = 2^{16} - 2^{15} = 2^{15} = 32768$

For unsigned type:
UINT_MAX + 1 ≡ V + 1 ≡ 0 (mod UINT_MAX + 1)

Ex 17. Under the assumption that the maximum value for unsigned int is 0xFFFFFFFF ( $2^{32}-1$ ), prove that -0x80000000 == 0x80000000 ( $2^{31}$ ).

Similar argument applies here.

Binary representation

$\sum_{i = 0}^{p-1}b_i2^i$

bit

precision

max: $2^p -1$

Exs 22. Show that A \ B can be computed by A - (A&B).

$RHS = \sum_{i = 0}^{p-1}a_i2^i - \sum_{i = 0}^{p-1}\bar{a}_i2^i= \sum_{i = 0}^{p-1}\hat{a}_i2^i$

where $a_i \in A, \bar{a}_i \in A \cap B \implies \hat{a}_i \in A \setminus (A \cap B) = A \setminus B$

Exs 23. Show that V + 1 is 0.

$V + 1 = 2^p - 1 + 1 = 2^p \equiv 0 \mod 2^p$

Exs 24. Show that A^B is equivalent to (A - (A&B)) + (B - (A&B)) and A + B - 2*(A&B).

Exs 25. Show that A|B is equivalent to A + B - (A&B).

$x \in A \Delta B \Longleftrightarrow x \in A\setminus B \cap B\setminus A$

$\implies A \wedge B = (A - (A\&B)) + (B - (B\&A)) = A + B - 2*(A\&B)$

$A|B = A\&B + A\wedge B$

$= A\&B +A + B - 2*(A\&B) = A + B - A\&B$

Shift operators

A left-shift operation << corresponds to the multiplication
by a power of two.

A right-shift operation >> corresponds to an integer division
by a power of two.

A = 240; $A << 9 \equiv 240 \cdot 2^9 = 122880 = 57344 \mod 2^{16}$

Exs 28. Show that the bits that are “lost” in an operation x>>n correspond to the remainder x % (1ULL << n).

$x = \sum_{i = 0}^{p-1}b_i2^i = \sum_{i = 0}^{n-1}b_i2^i + \sum_{i = n}^{p-1}b_i2^i=2^n \cdot \left(\frac{1}{2^n}\sum_{i = 0}^{n-1}b_i2^{i} + \sum_{i = 0}^{p-n-1}b_i2^i\right)$

bits lost
$<2^n$
remainder

$x$ % (1ULL << $n$ ) = $x \mod 2^n$

number of times $2^n$ fit in $x$

Floating-point data

s \cdot2^e \cdot\sum_{k=1}^{p}f_k2^{-k}

s \cdot2^e \cdot\sum_{k=1}^{p}f_k2^{-k}

p, emin, emax: type dependent

Never compare floating-point values for equality.

Exs 35. Show that all representable floating-point values with $e > p$ are multiples of $2^{e−p}$ .

s \cdot2^e \cdot\sum_{k=1}^{p}f_k2^{-k}

s \cdot2^e \cdot\sum_{k=1}^{p}f_k2^{-k}

= s \cdot\sum_{k=1}^{p}f_k2^{e-k}

= s \cdot\sum_{k=1}^{p}f_k2^{e-k}

= s \cdot\sum_{k=1}^{p}f_k2^{e-p}2^{p-k}

= s \cdot\sum_{k=1}^{p}f_k2^{e-p}2^{p-k}

= s \cdot2^{e-p} \cdot\sum_{i=0}^{p-1}f_{p-i}2^{i}

= s \cdot2^{e-p} \cdot\sum_{i=0}^{p-1}f_{p-i}2^{i}

	#include <stdio.h>
	#include <stdlib.h>

	size_t clock_24(size_t a, size_t t) { return (a + t) % 24; }

	int main(void) {
	size_t x[][2] = {
	{10, 3}, // 3 hours after 10:00
	{20, 8}, // 8 hours after 20:00
	{5, 12}, // 12 hours after 5:00
	{23, 2}, // 2 hours after 23:00
	};
	size_t nx = sizeof(x) / sizeof(x[0]);

	for (size_t i = 0; i < nx; i++) {
	printf("%zu hours after %02zu:00 is %02zu:00\n", x[i][1], x[i][0],
	clock_24(x[i][0], x[i][1]));
	}

	return EXIT_SUCCESS;
	}

	#include <stdio.h>

	int main() {
	double s1 = 1.0E-13 + 1.0E-13;
	double s2 = (1.0E-13 + (1.0E-13 + 1.0)) - 1.0;
	printf("s1 = %.15lf\ns2 = %.15lf\n", s1, s2);

	return 0;
	}

	#include <stdio.h>

	int main() {
	double s1 = 1.0E-13 + 1.0E-13;
	double s2 = (1.0E-13 + (1.0E-13 + 1.0E13)) - 1.0E13;
	printf("s1 = %.15lf\ns2 = %.15lf\n", s1, s2);

	return 0;
	}

Modern C

Last time

Chapter 4

`a + b`

boolean

conditional operator

short-circuit evaluation

Chapter 5

abstract state machine

Example

string literals

enumerations

Binary representation

Shift operators

Floating-point data

Modern C Chapter 4—5 Summary

Modern C Chapter 4—5 Summary

Trang Le

Modern C

Modern C Chapter 4—5 Summary

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