Discussion :: Java.lang Class - Finding the output (Q.No.19)
| Sundar said: (Aug 25, 2011) | |
| a.substring(5,7); // Returns: "ap" a.charAt(1); // Returns: "p" a = a + b; // a + b = ("ap" + "p") = "app" stored in a. Therefore the string "app" is the correct answer. |
|
| Muthuramu said: (Dec 1, 2011) | |
| a.Substring(5,8);//Returns:"ape"(means 5,6,7)??? a.Substring(5,9);//Returns:"Aper"??? only 0 to 8 is available in string?? |
|
| Shraddha said: (Jan 16, 2012) | |
| Second argument of the Substring() method is the LENGTH from first argument. a.Substring(5,7);//Returns:"aper" a.charAt(1); // Returns: "p" a = a + b; // a + b = ("aper" + "p") = "aperp" |
|
| Ahmed said: (Nov 26, 2012) | |
| How a.charAt(1) returns "p"? | |
| Vinod said: (Jun 11, 2013) | |
| In obj a now "ape". a.charAt(0) means "a". a.charAt(1) means "p". |
|
| Amit Agrawalla said: (Jun 18, 2013) | |
| @Ahmed : as a.substring(5,7); // Returns: "ap" So now a has "ap" that is a is in zero place and p in the 1 place. So charAt(1) implies 1 place of a. that is p. |
|
| Veena said: (Nov 4, 2013) | |
| In string the position is starts from 0. So count letters in "newspapers" as of a.substring(5,7); will count from 0 and its start taking letters after 5 till (7-1). i.e till 6th letter. So a.substring(5,7); will contains "ap" |
|
| Mukesh said: (Oct 4, 2014) | |
| a.substring(5,7) will return ape. So the final output should be apep... Please explain why the first line is returning ap only and not ape. |
|
| Aaa said: (Oct 10, 2014) | |
| 7-5 = 2 means length is 2. a = ap. a.charAt(1) = p. |
|
| Abhi said: (Sep 10, 2015) | |
| a.substring (5, 7) will return ape. And charAt (1) will be p. So answer will be apep. How app anyone please explain? |
|
| Sai said: (Sep 15, 2015) | |
| a.substring (5,7-1). a.substring (5,6) will return ap. And a.charAt (1). Result is app// |
|
| Arvin said: (Sep 2, 2016) | |
| How is a.substring(5,7) same as a.substring(5,6)? Execute it and see, following is the right answer, everything else is wrong. a = a.substring(5,7) = n(0)e(1)w(2)s(3)p(4)a(5)p(6)e(7)r(8) = ape It will take from 5th, 6th, and 7th position. Then next, a.charAt(1) = e Final result = apep That's it. |
|
| Zen said: (Sep 2, 2016) | |
| substring(beginIndex, endIndex) beginIndex - inclusive. endIndex - exclusive. That's why a.substring(5,7) = ap. |
|
| Srinu said: (Apr 3, 2017) | |
| Length start from 0 now (sunstring(5,7) return ape) charAt(1) retuns=p then, Final the output. |
|
Post your comments here:
Name *:
Email : (optional)
» Your comments will be displayed only after manual approval.