Computer Science, asked by madgaokar57, 1 month ago

Determine the output of the following program.

#include<iostream>

using namespace std;

int g=20;

void func(int &x, int y) {

x=x-y;

y=x*5;

cout<<x<<”,”<<y<<endl;

}

int main()

{

int g=10;

func(g,::g);

cout<<g<<”,”<<::g<<endl;

return 0;

}​

Answers

Answered by prithikashree235
0

Answer:

Question 1:

#include <cstdlib>

#include <iostream>

using namespace std;

int main()

{

int ran = rand();

cout << ran << endl;

return 0;

}

Output:

1804289383

Explanation: As the declared number is an integer, It will produce the random number from 0 to RAND_MAX. The value of RAND_MAX is library-dependent but is guaranteed to be at least 32767 on any standard library implementation.

Question 2:

#include <cstdlib>

#include <iostream>

using namespace std;

int main()

{

cout << RAND_MAX << endl;

return 0;

}

Output:

2147483647

Explanation: The output is Compiler Dependent. RAND_MAX is a function used by the compiler to create a maximum random number.

Question 3:

#include <iostream>

using namespace std;

int main()

{

void a = 10, b = 10;

int c;

c = a + b;

cout << c;

return 0;

}

Output:

Compile time error

Explanation: void will not accept any values to its type.

Question 4:

#include <iostream>

using namespace std;

int array1[] = { 1200, 200, 2300, 1230, 1543 };

int array2[] = { 12, 14, 16, 18, 20 };

int temp, result = 0;

int main()

{

for (temp = 0; temp < 5; temp++) {

result += array1[temp];

}

for (temp = 0; temp < 5; temp++) {

result += array2[temp];

}

cout << result;

return 0;

}

Output:

6553

Explanation: In this program we are adding the every element of two arrays. All the elements of array1[] and array2[] will be added and the sum will be stored in result and hence output is 6553.

Question 5:

#include <iostream>

using namespace std;

int main()

{

int a = 5, b = 10, c = 15;

int arr[3] = { &a, &b, &c };

cout << *arr[*arr[1] - 8];

return 0;

}

Output:

Compile time error!

Explanation: The conversion is invalid in this array. The array arr[] is declared to hold integer type value but we are trying to assign references(addresses) to the array so it will arise error. The following compilation error will be raised:

cannot convert from ‘int *’ to ‘int’

Question 6:

#include <iostream>

using namespace std;

int main()

{

int array[] = { 10, 20, 30 };

cout << -2 [array];

return 0;

}

Output:

-30

Explanation: -2[array]: this statement is equivalent to -(array[2]). At the zero index 30 is stored hence negation of 30 will be printed due to unary operator (-).

Question 7:

#include <iostream>

using namespace std;

int main()

{

int const p = 5;

cout << ++p;

return 0;

}

Output:

Compile time Error!

Explanation: Constant variables are those whose value can’t be changed throughout the execution. ++p statement try to change the value hence compiler will raise an error.

Question 8:

#include <iostream>

using namespace std;

int main()

{

char arr[20];

int i;

for (i = 0; i < 10; i++)

*(arr + i) = 65 + i;

*(arr + i) = '\0';

cout << arr;

return (0);

}

Output:

ABCDEFGHIJ

Explanation: Each time we are assigning 65 + i. In first iteration i = 0 and 65 is assigned. So it will print from A to J.

Question 9:

#include <iostream>

using namespace std;

int Add(int X, int Y, int Z)

{

return X + Y;

}

double Add(double X, double Y, double Z)

{

return X + Y;

}

int main()

{

cout << Add(5, 6);

cout << Add(5.5, 6.6);

return 0;

}

Output:

Compile time error!

Explanation: Here we want to add two element but in the given functions we take 3 arguments.So compiler doesn’t get the required function(function with 2 arguments)

Question 10:

#include <iostream>

using namespace std;

#define PR(id) cout << "The value of " #id " is " << id

int main()

{

int i = 10;

PR(i);

return 0;

}

Output:

The value of i is 10

Explanation: In this program, we are just printing the declared value through a macro. Carefully observe that in macro there is no semicolon(;) used as a termination statement.

HOPE IT IS HELPFULL.

PLS MARK AS BRAINLIST

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