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Chapter 10: Problem 2

Consider the following type definition: struct ShoeType \\{ char style double price \\}; Given this structure type definition, what will be the output produced by the following code? ShoeType shoe1, shoe2; shoe1.style \(=^{\prime} A^{\prime}\) shoe1.price \(=9.99\) cout \( < < \) shoe \(1 .\) style \( < < " \$^{\prime \prime} < < \) shoe1.price \( < < \) endl shoe \(2=\) shoe 1 shoe2.price \(=\) shoe \(2 .\) price \(/ 9\) cout \( < < \) shoe \(2 .\) style \( < < " \$ " < < \) shoe \(2 .\) price \( < < \) end 1;

Short Answer

Expert verified
Answer: The printed output values are "A $9.99" for shoe1 and "A $1.11" for shoe2.

Step by step solution

01

Analyze the structure definition

struct ShoeType { char style; double price; }; The struct ShoeType has two members - style (a character) and price (a double).
02

Initialize shoe1 and shoe2 objects

ShoeType shoe1, shoe2; Here, two objects of type ShoeType, shoe1 and shoe2, are created.
03

Assign values to shoe1

shoe1.style = 'A'; shoe1.price = 9.99; The character 'A' is assigned to the style member of shoe1, and the value 9.99 is assigned to the price member of shoe1.
04

Print shoe1 values

cout << shoe1.style << " $" << shoe1.price << endl; The style and price of shoe1 are printed, separated by a " \(". The output will be: A \)9.99
05

Assign shoe1 values to shoe2

shoe2 = shoe1; Both the style and price members of shoe2 are assigned the corresponding values of shoe1, which means shoe2.style equals 'A' and shoe2.price equals 9.99.
06

Update shoe2 price

shoe2.price = shoe2.price / 9; The price of shoe2 is divided by 9 and assigned back to shoe2.price. So, shoe2.price = 9.99 / 9 \(\approx 1.11\).
07

Print shoe2 values

cout << shoe2.style << " $" << shoe2.price << endl; The style and updated price of shoe2 are printed, separated by a " \(". The output will be: A \)1.11 To summarize, the output of the given code is: A $9.99 A $1.11

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