Able to insert more values than its length in an array in JAVA
0
I have created a Java program.I have defined array length to be 10 but when I am giving input 3,5,18,1,3,7,16,14,15,11,13,19 which contains more than 10 values the program is running fine. I am new to Java. Please help me understand how it is happening. My code is :
package ds;
import java.util.Scanner;
public class Trade {
public static void maxprofit(Stringprice)
{
int prices= new int[price.length];
for(int i=0;i<price.length;i++)
{
prices[i]=Integer.parseInt(price[i]);
}
int min=prices[0];
int max=prices[prices.length-1];
for(int i=0;i<prices.length;i++)
{
if(i<prices.length-1-i)
{
if(prices[i]<min)
{
min=prices[i];
}
if(prices[prices.length-1-i]>max)
{
max=prices[prices.length-1-i];
}
}
else
break;
}
System.out.println(max-min);
}
public static void main(String args) {
// TODO Auto-generated method stub
Scanner sc = new Scanner(System.in);
String prices=new String[10];
String input=new String();
input=sc.next();
prices=input.split(",");
maxprofit(prices);
}
}
java arrays
asked Nov 18 '18 at 6:46
user3797489user3797489
346
add a comment |
0
I have created a Java program.I have defined array length to be 10 but when I am giving input 3,5,18,1,3,7,16,14,15,11,13,19 which contains more than 10 values the program is running fine. I am new to Java. Please help me understand how it is happening. My code is :
package ds;
import java.util.Scanner;
public class Trade {
public static void maxprofit(Stringprice)
{
int prices= new int[price.length];
for(int i=0;i<price.length;i++)
{
prices[i]=Integer.parseInt(price[i]);
}
int min=prices[0];
int max=prices[prices.length-1];
for(int i=0;i<prices.length;i++)
{
if(i<prices.length-1-i)
{
if(prices[i]<min)
{
min=prices[i];
}
if(prices[prices.length-1-i]>max)
{
max=prices[prices.length-1-i];
}
}
else
break;
}
System.out.println(max-min);
}
public static void main(String args) {
// TODO Auto-generated method stub
Scanner sc = new Scanner(System.in);
String prices=new String[10];
String input=new String();
input=sc.next();
prices=input.split(",");
maxprofit(prices);
}
}
java arrays
asked Nov 18 '18 at 6:46
user3797489user3797489
346
add a comment |
0
0
0
I have created a Java program.I have defined array length to be 10 but when I am giving input 3,5,18,1,3,7,16,14,15,11,13,19 which contains more than 10 values the program is running fine. I am new to Java. Please help me understand how it is happening. My code is :
package ds;
import java.util.Scanner;
public class Trade {
public static void maxprofit(Stringprice)
{
int prices= new int[price.length];
for(int i=0;i<price.length;i++)
{
prices[i]=Integer.parseInt(price[i]);
}
int min=prices[0];
int max=prices[prices.length-1];
for(int i=0;i<prices.length;i++)
{
if(i<prices.length-1-i)
{
if(prices[i]<min)
{
min=prices[i];
}
if(prices[prices.length-1-i]>max)
{
max=prices[prices.length-1-i];
}
}
else
break;
}
System.out.println(max-min);
}
public static void main(String args) {
// TODO Auto-generated method stub
Scanner sc = new Scanner(System.in);
String prices=new String[10];
String input=new String();
input=sc.next();
prices=input.split(",");
maxprofit(prices);
}
}
java arrays
asked Nov 18 '18 at 6:46
user3797489user3797489
346
I have created a Java program.I have defined array length to be 10 but when I am giving input 3,5,18,1,3,7,16,14,15,11,13,19 which contains more than 10 values the program is running fine. I am new to Java. Please help me understand how it is happening. My code is :
package ds;
import java.util.Scanner;
public class Trade {
public static void maxprofit(Stringprice)
{
int prices= new int[price.length];
for(int i=0;i<price.length;i++)
{
prices[i]=Integer.parseInt(price[i]);
}
int min=prices[0];
int max=prices[prices.length-1];
for(int i=0;i<prices.length;i++)
{
if(i<prices.length-1-i)
{
if(prices[i]<min)
{
min=prices[i];
}
if(prices[prices.length-1-i]>max)
{
max=prices[prices.length-1-i];
}
}
else
break;
}
System.out.println(max-min);
}
public static void main(String args) {
// TODO Auto-generated method stub
Scanner sc = new Scanner(System.in);
String prices=new String[10];
String input=new String();
input=sc.next();
prices=input.split(",");
maxprofit(prices);
}
}
java arrays
java arrays
asked Nov 18 '18 at 6:46
user3797489user3797489
346
asked Nov 18 '18 at 6:46
user3797489user3797489
346
asked Nov 18 '18 at 6:46
user3797489user3797489
346
asked Nov 18 '18 at 6:46
user3797489user3797489
346
asked Nov 18 '18 at 6:46
user3797489user3797489
346
346
add a comment |
add a comment |
1 Answer
1
active
oldest
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2
input.split(",") returns a new array. So after you assign that array to your input variable, input references the new (larger) array instead of the original 10 elements array.
Both of these initializations are pointless:
String prices=new String[10];
String input=new String();
Instead, you can write:
String input = sc.next();
String prices = input.split(",");
answered Nov 18 '18 at 6:48
EranEran
283k37458543
are u saying that my original array size is increased? as I am using the same name "prices" to store the result of split function so how it is possible that it has created a new array with same name.
– user3797489
Nov 18 '18 at 6:52
1
@user3797489 no. Your original array size is not increased. You are not using the original array.pricesis a reference variable that references an array object. Before theprices = input.split(",")assignment, it references your original array (or length 10) and after the assignment it references a different array, which is larger. The original array is no longer referenced by any variable, which means it can be garbage collected.
– Eran
Nov 18 '18 at 6:56
Thanks for the explaination. I understood your point but I have a doubt that when try to initialize an array with a same name again in Java program then it gives error. So in that case why the reference is not changed. Suppose we define String prices=new String[10]; String prices=new String[11]; together in a java program then it throws error So in this case why the reference is not changing
– user3797489
Nov 18 '18 at 6:59
@user3797489 that's because you declare the same variable twice in the same scope, which is not allowed. You can declare the variable once and assign to it multiple times. For example, you can writeString prices=new String[10]; prices=new String[11];instead.
– Eran
Nov 18 '18 at 7:04
@user3797489 Java doesn't allow to declare or initialize variables with the same names within the same scope.
– Maseed
Nov 18 '18 at 7:05
add a comment |
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1 Answer
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2
input.split(",") returns a new array. So after you assign that array to your input variable, input references the new (larger) array instead of the original 10 elements array.
Both of these initializations are pointless:
String prices=new String[10];
String input=new String();
Instead, you can write:
String input = sc.next();
String prices = input.split(",");
answered Nov 18 '18 at 6:48
EranEran
283k37458543
are u saying that my original array size is increased? as I am using the same name "prices" to store the result of split function so how it is possible that it has created a new array with same name.
– user3797489
Nov 18 '18 at 6:52
1
@user3797489 no. Your original array size is not increased. You are not using the original array.pricesis a reference variable that references an array object. Before theprices = input.split(",")assignment, it references your original array (or length 10) and after the assignment it references a different array, which is larger. The original array is no longer referenced by any variable, which means it can be garbage collected.
– Eran
Nov 18 '18 at 6:56
Thanks for the explaination. I understood your point but I have a doubt that when try to initialize an array with a same name again in Java program then it gives error. So in that case why the reference is not changed. Suppose we define String prices=new String[10]; String prices=new String[11]; together in a java program then it throws error So in this case why the reference is not changing
– user3797489
Nov 18 '18 at 6:59
@user3797489 that's because you declare the same variable twice in the same scope, which is not allowed. You can declare the variable once and assign to it multiple times. For example, you can writeString prices=new String[10]; prices=new String[11];instead.
– Eran
Nov 18 '18 at 7:04
@user3797489 Java doesn't allow to declare or initialize variables with the same names within the same scope.
– Maseed
Nov 18 '18 at 7:05
add a comment |
2
input.split(",") returns a new array. So after you assign that array to your input variable, input references the new (larger) array instead of the original 10 elements array.
Both of these initializations are pointless:
String prices=new String[10];
String input=new String();
Instead, you can write:
String input = sc.next();
String prices = input.split(",");
answered Nov 18 '18 at 6:48
EranEran
283k37458543
are u saying that my original array size is increased? as I am using the same name "prices" to store the result of split function so how it is possible that it has created a new array with same name.
– user3797489
Nov 18 '18 at 6:52
1
@user3797489 no. Your original array size is not increased. You are not using the original array.pricesis a reference variable that references an array object. Before theprices = input.split(",")assignment, it references your original array (or length 10) and after the assignment it references a different array, which is larger. The original array is no longer referenced by any variable, which means it can be garbage collected.
– Eran
Nov 18 '18 at 6:56
Thanks for the explaination. I understood your point but I have a doubt that when try to initialize an array with a same name again in Java program then it gives error. So in that case why the reference is not changed. Suppose we define String prices=new String[10]; String prices=new String[11]; together in a java program then it throws error So in this case why the reference is not changing
– user3797489
Nov 18 '18 at 6:59
@user3797489 that's because you declare the same variable twice in the same scope, which is not allowed. You can declare the variable once and assign to it multiple times. For example, you can writeString prices=new String[10]; prices=new String[11];instead.
– Eran
Nov 18 '18 at 7:04
@user3797489 Java doesn't allow to declare or initialize variables with the same names within the same scope.
– Maseed
Nov 18 '18 at 7:05
add a comment |
2
2
2
input.split(",") returns a new array. So after you assign that array to your input variable, input references the new (larger) array instead of the original 10 elements array.
Both of these initializations are pointless:
String prices=new String[10];
String input=new String();
Instead, you can write:
String input = sc.next();
String prices = input.split(",");
answered Nov 18 '18 at 6:48
EranEran
283k37458543
input.split(",") returns a new array. So after you assign that array to your input variable, input references the new (larger) array instead of the original 10 elements array.
Both of these initializations are pointless:
String prices=new String[10];
String input=new String();
Instead, you can write:
String input = sc.next();
String prices = input.split(",");
answered Nov 18 '18 at 6:48
EranEran
283k37458543
answered Nov 18 '18 at 6:48
EranEran
283k37458543
answered Nov 18 '18 at 6:48
EranEran
283k37458543
answered Nov 18 '18 at 6:48
EranEran
283k37458543
283k37458543
are u saying that my original array size is increased? as I am using the same name "prices" to store the result of split function so how it is possible that it has created a new array with same name.
– user3797489
Nov 18 '18 at 6:52
1
@user3797489 no. Your original array size is not increased. You are not using the original array.pricesis a reference variable that references an array object. Before theprices = input.split(",")assignment, it references your original array (or length 10) and after the assignment it references a different array, which is larger. The original array is no longer referenced by any variable, which means it can be garbage collected.
– Eran
Nov 18 '18 at 6:56
Thanks for the explaination. I understood your point but I have a doubt that when try to initialize an array with a same name again in Java program then it gives error. So in that case why the reference is not changed. Suppose we define String prices=new String[10]; String prices=new String[11]; together in a java program then it throws error So in this case why the reference is not changing
– user3797489
Nov 18 '18 at 6:59
@user3797489 that's because you declare the same variable twice in the same scope, which is not allowed. You can declare the variable once and assign to it multiple times. For example, you can writeString prices=new String[10]; prices=new String[11];instead.
– Eran
Nov 18 '18 at 7:04
@user3797489 Java doesn't allow to declare or initialize variables with the same names within the same scope.
– Maseed
Nov 18 '18 at 7:05
add a comment |
are u saying that my original array size is increased? as I am using the same name "prices" to store the result of split function so how it is possible that it has created a new array with same name.
– user3797489
Nov 18 '18 at 6:52
1
@user3797489 no. Your original array size is not increased. You are not using the original array.pricesis a reference variable that references an array object. Before theprices = input.split(",")assignment, it references your original array (or length 10) and after the assignment it references a different array, which is larger. The original array is no longer referenced by any variable, which means it can be garbage collected.
– Eran
Nov 18 '18 at 6:56
Thanks for the explaination. I understood your point but I have a doubt that when try to initialize an array with a same name again in Java program then it gives error. So in that case why the reference is not changed. Suppose we define String prices=new String[10]; String prices=new String[11]; together in a java program then it throws error So in this case why the reference is not changing
– user3797489
Nov 18 '18 at 6:59
@user3797489 that's because you declare the same variable twice in the same scope, which is not allowed. You can declare the variable once and assign to it multiple times. For example, you can writeString prices=new String[10]; prices=new String[11];instead.
– Eran
Nov 18 '18 at 7:04
@user3797489 Java doesn't allow to declare or initialize variables with the same names within the same scope.
– Maseed
Nov 18 '18 at 7:05
are u saying that my original array size is increased? as I am using the same name "prices" to store the result of split function so how it is possible that it has created a new array with same name.
– user3797489
Nov 18 '18 at 6:52
are u saying that my original array size is increased? as I am using the same name "prices" to store the result of split function so how it is possible that it has created a new array with same name.
– user3797489
Nov 18 '18 at 6:52
1
1
@user3797489 no. Your original array size is not increased. You are not using the original array.
prices is a reference variable that references an array object. Before the prices = input.split(",") assignment, it references your original array (or length 10) and after the assignment it references a different array, which is larger. The original array is no longer referenced by any variable, which means it can be garbage collected.– Eran
Nov 18 '18 at 6:56
@user3797489 no. Your original array size is not increased. You are not using the original array.
prices is a reference variable that references an array object. Before the prices = input.split(",") assignment, it references your original array (or length 10) and after the assignment it references a different array, which is larger. The original array is no longer referenced by any variable, which means it can be garbage collected.– Eran
Nov 18 '18 at 6:56
Thanks for the explaination. I understood your point but I have a doubt that when try to initialize an array with a same name again in Java program then it gives error. So in that case why the reference is not changed. Suppose we define String prices=new String[10]; String prices=new String[11]; together in a java program then it throws error So in this case why the reference is not changing
– user3797489
Nov 18 '18 at 6:59
Thanks for the explaination. I understood your point but I have a doubt that when try to initialize an array with a same name again in Java program then it gives error. So in that case why the reference is not changed. Suppose we define String prices=new String[10]; String prices=new String[11]; together in a java program then it throws error So in this case why the reference is not changing
– user3797489
Nov 18 '18 at 6:59
@user3797489 that's because you declare the same variable twice in the same scope, which is not allowed. You can declare the variable once and assign to it multiple times. For example, you can write
String prices=new String[10]; prices=new String[11]; instead.– Eran
Nov 18 '18 at 7:04
@user3797489 that's because you declare the same variable twice in the same scope, which is not allowed. You can declare the variable once and assign to it multiple times. For example, you can write
String prices=new String[10]; prices=new String[11]; instead.– Eran
Nov 18 '18 at 7:04
@user3797489 Java doesn't allow to declare or initialize variables with the same names within the same scope.
– Maseed
Nov 18 '18 at 7:05
@user3797489 Java doesn't allow to declare or initialize variables with the same names within the same scope.
– Maseed
Nov 18 '18 at 7:05
add a comment |
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