Output of C++ programs | Set 21

07 May, 2018
What will be the output of this program?
#include<iostream>using namespace std;int x; // Global x  int main(){    int x = 10; // Local x    cout << "Value of global x is " << ::x << endl;    cout << "Value of local x is " << x;     return 0;}
Output:
Value of global x is 0
Value of local x is 10

In C++, global variable (if we have a local variable with same name), can be accessed using scope resolution operator (::).
 What will be the output of this program?
#include <iostream>using namespace std;int a = 90;  int fun(int x, int *y = &a){    *y = x + *y;    return x + *y;}  int main(){    int a = 5, b = 10;      a = fun(a);    cout << a << " " << b << endl;      b = fun(::a,&a);    cout << a << " " << b << endl;      return 0;}
100   10
195   290
There are two variables with name ‘a’, one is global and other is local. When we call a = fun(a);, it calls int fun(int x, int *y=&a), here pointer to global variable (which is a = 90) is assigned to y. Therefore.*y = x + *y; // becomes 5 + 90return x + *y; // becomes 5 + 95
 What will be the output of this program?
#include <iostream>using namespace std;int a = 2;  int fun(int *a){    ::a *= *a;    cout << ::a << endl;    return *a;}  int main(){    int a = 9;    int &x = ::a;      ::a += fun(&x);    cout << x;}
Output:
4
8

The global variable is being accessed by the function fun every time using ::a. The local variable value does’nt impact on the value of a.
 What will be the output of this program?
#include <iostream>using namespace std;  int main(){    char *A[] = { "abcx", "dbba", "cccc"};    char var = *(A+1) - *A+1;    cout << (*A + var);}
Output:
 bba

Here the array representation is A[0] = “abcx”, A[1] = “dbba”, A[2] = “cccc”. Precedence of (pointer)* >(binary) +, and order of execution of ‘*’ is right to left . If ‘A’ address is ‘x’ then address of ‘*(A+1)’ is ‘x+6’ and address of ‘*A+1’ is ‘x+1’. So integer value of var = 6 (total no of character between two points (x+6)-(x+1)+1). During printing the operator ‘+’ is overloaded now the pointer points to ‘x+7’ . For this reason the output of the program. What will be the output of this program?
#include <iostream>using namespace std;  int main(){    char a = 'a', b = 'x';    char c = (b ^ a >> 1 * 2) +(b && a >> 1 * 2 );    cout << " c = " << c;}
Output:
 c = 97

Integer value of a = 97 (01100001), integer value of b = 120 (01111000), precedence of ‘*’ > ‘>>’ > ‘^’ > ‘&&’.So expression is ‘ ((b ^ (a >> (1 * 2))) – (b && (a >> (1 * 2 ))))’. Integer Value of expressiona >> 1 * 2 = 24 b ^ a >> 1 * 2 = 96 b && a >> 1 * 2 = 1 which is 97.
 
What will be the output of this program?
#include <iostream>using namespace std;  int main(){    int i = 5, j = 3;    switch(j)    {    case 1:        if (i < 10) cout << "\ncase 1";        else if (i > 10)           case 2:               cout << "case 2";        else if (i==10)           case 3:    default:        cout << "case 3";        cout << "hello";    }}
Output:
case 3hello
Since j=3 satisfies the condition, it goes in case 3. There is nothing in case 3, also there is no break. So default is executed. Please refer switch statement in C for details.
This article is contributed by Punit, Smriti Satyanarayana, Akash Tiwari, Sagar Tripathy . If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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