Differences between built-in type default-value-initialization

Asked Oct 10 '19 at 20:16

Active Oct 10 '19 at 20:29

Viewed 22 times

Is there any difference/advantage between using any of this ways of initialization of a variable?

float A = 0;
float B = float();
float C(0);

cout << A << endl
     << B << endl
     << C << endl

Of course, the result will be the same in the 3 cases:

0
0
0

edited Oct 10 '19 at 20:25

asked Oct 10 '19 at 20:16

Ali Rojas

I've changed the dupe target since you've updated the question. It should answer all of your questions. – NathanOliver Oct 10 '19 at 20:30
FYI, that last print you deleted is no garbage. It's required to be 1. – StoryTeller - Unslander Monica Oct 10 '19 at 20:31

0 Answers0