Among the operations that don’t change the value of an object is initialization— when we use an object to initialize another object, it doesn’t matter whether either or both of the objects are consts

belongs to book: C++ Primer|Stanley B.Lippman, Josee Lajoie, Barbara E.Moo|5th Edition| Chapter number:2| Question number:e.g2.4.4

KasimKamal

Save to bookmark

2022-07-03

365

Very Easy

Among the operations that don’t change the value of an object is initialization— when we use an object to initialize another object, it doesn’t matter whether either or both of the objects are consts.

C++ programming, textbook solutions, C++ programming textbook-solutions,

All Answers

KasimKamal

2022-07-03

need an explanation for this answer? contact us directly to get an explanation for this answer

int i = 42;
const int ci = i; // ok: the value in i is copied into ci
int j = ci; // ok: the value in ci is copied into j

need an explanation for this answer? contact us directly to get an explanation for this answer

total answers (1)

Among the operations that don’t change the value of an object is initialization— when we use an object to initialize another object, it doesn’t matter whether either or both of the objects are consts

All Answers

Similar questions

need a help?

find thousands of online teachers now