Output of C++ programs | Set 42

Prerequisite : Pointers and References

Q.1 What Is The Output Of this program?

 `#include ` `using` `namespace` `std; ` `void` `fun(``int``& a, ``int` `b) ` `{ ` `    ``a += 2; ` `    ``b += 1; ` `} ` `int` `main() ` `{ ` `    ``int` `x = 10, y = 2; ` `    ``fun(x, y); ` `    ``cout << x << ``" "` `<< y << ``" "``; ` `    ``fun(x, y); ` `    ``cout << x << ``" "` `<< y; ` `    ``return` `0; ` `} `

Option
a) 10 2 10 2
b) 12 2 14 2
c) 12 3 14 3
d) 12 2 14 3

`Answer : b`

Explanation : In this program, in main() we pass two values x or y in fun() and fun() received the reference of x so increment it’s value but y have incremented but not a reference so y is same in main block.

Q.2 What Is The Output Of this program?

 `#include ` `using` `namespace` `std; ` `void` `f2(``int` `p = 30) ` `{ ` `    ``for` `(``int` `i = 20; i <= p; i += 5) ` `        ``cout << i << ``" "``; ` `} ` `void` `f1(``int``& m) ` `{ ` `    ``m += 10; ` `    ``f2(m); ` `} ` `int` `main() ` `{ ` `    ``int` `n = 20; ` `    ``f1(n); ` `    ``cout << n << ``" "``; ` `    ``return` `0; ` `} `

Option
a) 25 30 35 20
b) 20 25 30 20
c) 25 30 25 30
d) 20 25 30 30

`Answer : d`

Explanation : In this program, main() call the f1() and pass the value of n in f1() and f1() is received the reference of n and increments 10 of it’s value now n is 30. Again call f2() and pass the value of m in f2(), f2() receive the value of m and check the condition and print the value.

Q.3 What Is The Output Of this program?

 `#include ` `using` `namespace` `std; ` `void` `fun(``char` `s[], ``int` `n) ` `{ ` `    ``for` `(``int` `i = 0; s[i] != ``'\0'``; i++) ` `        ``if` `(i % 2 == 0) ` `            ``s[i] = s[i] - n; ` `        ``else` `            ``s[i] = s[i] + n; ` `} ` `int` `main() ` `{ ` `    ``char` `str[] = ``"Hello_World"``; ` `    ``fun(str, 2); ` `    ``cout << str << endl; ` `    ``return` `0; ` `} `

Option
a) EgjnmaTqpnb
b) FgjnmaUqpnb
c) Fgjnm_Uqpnb
d) EgjnmaTqpnb

`Answer : b`

Explanation : In main(), call a fun() and pass string and one integer is 2. In fun(), loop i is iterated and check the condition if i is Even then decrease index value by 2 and if i is odd then increase by 2.

Q.4 What Is The Output Of this program?

 `#include ` `using` `namespace` `std; ` `int` `main() ` `{ ` `    ``int` `x[] = { 12, 25, 30, 55, 110 }; ` `    ``int``* p = x; ` `    ``while` `(*p < 110) { ` `        ``if` `(*p % 3 != 0) ` `            ``*p = *p + 1; ` `        ``else` `            ``*p = *p + 2; ` `        ``p++; ` `    ``} ` `    ``for` `(``int` `i = 4; i >= 1; i--) { ` `        ``cout << x[i] << ``" "``; ` `    ``} ` `    ``return` `0; ` `} `

Option
a) 110 56 32 26
b) 110 57 30 27
c) 110 56 32 25
d) 100 55 30 25

`Answer : a`

Explanation : In this program, a pointer p contains a base address of array x so *p contain a value of x[0] and p++ is pointing address of next element.

Q.5 What Is The Output Of this program?

 `#include ` `using` `namespace` `std; ` `int` `main() ` `{ ` `    ``char``* str = ``"GEEKSFORGEEK"``; ` `    ``int``* p, arr[] = { 10, 15, 70, 19 }; ` `    ``p = arr; ` `    ``str++; ` `    ``p++; ` `    ``cout << *p << ``" "` `<< str << endl; ` `    ``return` `0; ` `} `

Option
a) 10 EEKSFORGEEK
b) 15 GEEKSFORGEEK
c) 15 EEKSFORGEEK
d) 10 GEEKSFORGEEK

`Answer : C`

Explanation : In this, the str contain a base address of string and pointer p contain a base address of array. If string and pointer is incremented by 1 – it’s pointing the next address value of string and array.

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