Question

What is the value of chice after the following statements?

Void getChoice(&choice, count);
{
if(par_ **** < 0)
*par_choice = 0;
if(par_count = 0);
*par_choice = -1;
else
*par_choice = 99;
return;
}

A) 99
B) 0
C)3
D) -1

Answer 1

The value of 'choice' will be -1 if 'par_count' is 0, or it will be 99 otherwise, due to the unconditional else statement in the provided code. The presence of a semicolon after the second if condition is a syntax error that results in the last condition being executed regardless.

Step-by-step explanation:

The question involves understanding the value of the variable choice after the execution of a C-like function getChoice. Given the ambiguities in the provided code snippet (such as syntax errors and placeholders represented by asterisks), we will assume the intended code involves checking the values of passed pointer arguments and assigning a value to the variable referenced by par_choice.

If the value pointed to by the first parameter (presumably par_choice) is less than 0, the function sets it to 0. Then, it checks if the second parameter (presumably par_count) equals 0; if it does, the syntax is incorrect due to the presence of a semicolon which ends the statement prematurely, leading to the last condition being executed regardless of the value of par_count. Due to this, the variable choice will always end with a value of 99 unless the first condition is met, and par_choice is set to 0. Thus, the correct answer to what the value of choice will be, given no syntax errors exist, is -1 if par_count is 0, or 99 otherwise.