Chapter 2.2, Problem 2.21PP

Program Plan Intro

Signed versus unsigned in C:

• Both signed and unsigned arithmetic is used in “C”.
  • It is applied to all of integer data type in “C”.
• Generally, most numbers are signed by default.
  • Consider an example, when defining a constant “12345” or “0x1A2B”, the given value is assumed as “signed”.
  • Adding character “U” or “u” as a suffix generates an unsigned constant that is “67812U” or “0x321Du”.
• In “C”, the expression is defined by “signed” or “unsigned”. Assume if the expression contains two operands.
  • When any one operand in a comparison is unsigned, then the other operand is indirectly converted to unsigned.
    • That is considering the given numbers are nonnegative.
• The evaluation for given “C” expression is identified by “0” or “1”.
  • If the given expression is “true”, then the result of evaluation is “1”.
  • If the given expression is “false”, then the result of evaluation is “0”.

Example:

The example for applying signed and unsigned type in “C” is shown below:

For expression “0 == 0U”:

• Type: Unsigned.
  • From the given expression, both operands are unsigned value. So, the type will be unsigned.
• Evaluation: 1
  • The relational operator “==” is used to check both operands are equal.
  • From the given expression, the condition “0 == 0” is satisfied. So, the evaluation will be “1”.

For expression “-1 < 0”:

• Type: Signed.
  • From the given expression, one operand contains negative value. So, the type will be signed.
• Evaluation: 1
  • From the given expression, the condition “-1 < 0” is satisfied. So, the evaluation will be “1”.

For expression “-1 < 0U”:

• Type: Unsigned.
  • From the given expression, one operand contain unsigned.
  • When any one operand in a comparison is unsigned, then the other operand is indirectly converted to unsigned.
    • Hence, it is unsigned.
• Evaluation: 0
  • From the given expression, the right operand is unsigned, so the left side operand is implicitly converts to unsigned.
  • For converting “-1” to unsigned value, first need to subtract “1” from ${\text{2}}^{\text{N}}$.
    • Here “N” is the number of bits used to denote the “int” data type.
    • So, value of “N” is “32”.
    • Then,

$\begin{array}{l}{\text{2}}^{\text{N}}\text{-}1\text{=}{\text{2}}^{32}\text{-}1\\ =4294967296\text{-}\text{1}\\ =4294967295\end{array}$

• Therefore, after converting “-1U” is “4294967295U”

$\begin{array}{l}\text{-1U}\text{<}\text{0U}\\ \text{4294967295U}\text{<}\text{0U}\end{array}$

• So, the expression “4294967295U < OU” implies “false”.
• Therefore, the evaluation will be “0”.