Consider the following C++ code:int n =1; // globalprintl (int x) {cout << x + n;}plus_n () {n= n + 2;printl (n);}main () {int n;n = 20;printl (5);n= 40;plus_n ();cout << n;}What is the output of the program under?b. Static scoping rulesc. Dynamic scoping rules

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int n = 1; // global printl (int x) { cout << x + n; } plus_n () { - n = n + 2; printl (n); } main () { int n; n = 20; print1 (5); n = 40; plus_n (); cout << n; } What is the output of the program under? b. Static scoping rules c. Dynamic scoping rules

int n = 1; // global printl (int x) { cout << x + n; } plus_n () { - n = n + 2; printl (n); } main () { int n; n = 20; print1 (5); n = 40; plus_n (); cout << n; } What is the output of the program under? b. Static scoping rules c. Dynamic scoping rules

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Question

Consider the following C++ code:
int n =
1; // global
printl (int x) {
cout << x + n;
}
plus_n () {
n
= n + 2;
printl (n);
}
main () {
int n;
n = 20;
printl (5);
n
= 40;
plus_n ();
cout << n;
}
What is the output of the program under?
b. Static scoping rules
c. Dynamic scoping rules

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