Step 1: Execute the following program and record the results. #include <stdlib.h

Question

Step 1: Execute the following program and record the results.

#include <stdlib.h>
#include <iostream>
using namespace std;

void fun3();
void fun4();

int main()
{
char x;

x = 'Z';
cout << "x = " << x << endl;
fun3();
cout << "x = " << x << endl;
fun4();
cout << "x = " << x << endl;

return 0;
}

void fun3() {
char x;
x = 'H';
cout << x << "ey Fargo is in North Dakota not Texas! ";
}

void fun4()
{
char x;
x = 'Y';
cout << x << "ou bet it's cold! ";
}

Step 2: Explain the results obtained-in Step 1 by tracing the program's execution while describing the creation, destruction, and assignment of variables. At each point in the execution at which a cout statement is executed, identify the local variables that have been created but not destroyed. What is the scope of each? What is the value of each?

Explanation / Answer

Output is x = Z Hey Fargo is in North Dakota not Texas! x = Z You bet it's cold! x = Z At each cout in main(), the only variable in scope and still extant is main::x = 'Z'. In fun3() main::x still exists and has the value 'Z'. fun3::x also exists with the value 'H' In fun4() main::x still exists and has the value 'Z'. fun4::x also exists with the value 'Y', fun3::x has been destroyed fun3::x is created on entrance to fun3() and destroyed on exist. similarly for fun4::x. main::x exists right up to the point when main() exits.

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