Question

An "unsigned char" typed variable in C++ has 8 bits of storage. When a character is assigned to it, the variable has the ASCII code of the character.

char a = 'F';
unsigned char b;
unsigned char c;
unsigned char d = 'T';

a = a + 18;
b = 123 - 'D';
c = 'e' - ':';
d = d + ''';

Answer 1

The question involves the manipulation of unsigned char variables in C++, which store 8-bit values corresponding to ASCII characters. Mathematical operations are performed on the ASCII values, and the result of these operations are stored in the unsigned char variables.

Step-by-step explanation:

The subject of this question is related to the C++ programming language, specifically regarding the usage and manipulation of unsigned char variables. An unsigned char can store values in the range of 0 to 255, as it is an 8-bit variable. Each character in the ASCII table is represented by a number from 0 to 127, and when a character is assigned to an unsigned char variable, the ASCII code of the character is stored.

Let's analyze the code provided, step by step:

• char a = 'F'; - This initializes variable 'a' with the ASCII code of 'F', which is 70.

• a = a + 18; - The ASCII code of 'F' (70) is incremented by 18, so 'a' now stores the value 88, which corresponds to the character 'X' in the ASCII table.

• unsigned char b;

  b = 123 - 'D'; - 123 corresponds to the curly brace '{', while 'D' has an ASCII code of 68. Thus, b stores the result of 123 - 68, which is 55.

• unsigned char c;

  c = 'e' - ':'; - The ASCII code for 'e' is 101, and for ':' it is 58. Thus, c stores the result of 101 - 58, which is 43.

• unsigned char d = 'T'; - Initializes 'd' with the ASCII code of 'T', which is 84.

• d = d + '''; - This line seems to contain a typo as it's not clear what is being added to 'd'. The correct expression would need another character or number after the '+' symbol.