What is wrong in my approach to create a List of Integers using IntStream and forEach?

up vote
6
down vote

favorite

The naive code I have is:

class $
 public static void main(String _) 
 final List<Integer> ints = new ArrayList<>();
 IntStream.iterate(0, i -> i++).limit(5).forEach(val -> ints.add(val));
 System.out.println(ints);

where my expectation was to see the following in the console:

[0, 1, 2, 3, 4]

but the actual is:

[0, 0, 0, 0, 0]

Probably something very simple but what am I missing?

asked 4 hours ago

Koray Tugay

8,04526108210

add a commentÂ |Â

up vote
6
down vote

favorite

The naive code I have is:

class $
 public static void main(String _) 
 final List<Integer> ints = new ArrayList<>();
 IntStream.iterate(0, i -> i++).limit(5).forEach(val -> ints.add(val));
 System.out.println(ints);

where my expectation was to see the following in the console:

[0, 1, 2, 3, 4]

but the actual is:

[0, 0, 0, 0, 0]

Probably something very simple but what am I missing?

asked 4 hours ago

Koray Tugay

8,04526108210

add a commentÂ |Â

up vote
6
down vote

favorite

The naive code I have is:

class $
 public static void main(String _) 
 final List<Integer> ints = new ArrayList<>();
 IntStream.iterate(0, i -> i++).limit(5).forEach(val -> ints.add(val));
 System.out.println(ints);

where my expectation was to see the following in the console:

[0, 1, 2, 3, 4]

but the actual is:

[0, 0, 0, 0, 0]

Probably something very simple but what am I missing?

asked 4 hours ago

Koray Tugay

8,04526108210

The naive code I have is:

class $
 public static void main(String _) 
 final List<Integer> ints = new ArrayList<>();
 IntStream.iterate(0, i -> i++).limit(5).forEach(val -> ints.add(val));
 System.out.println(ints);

where my expectation was to see the following in the console:

[0, 1, 2, 3, 4]

but the actual is:

[0, 0, 0, 0, 0]

Probably something very simple but what am I missing?

java java-8 java-stream

asked 4 hours ago

Koray Tugay

8,04526108210

asked 4 hours ago

Koray Tugay

8,04526108210

asked 4 hours ago

Koray Tugay

8,04526108210

asked 4 hours ago

Koray Tugay

8,04526108210

asked 4 hours ago

Koray Tugay

8,04526108210

add a commentÂ |Â

4 Answers
4

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up vote
4
down vote

accepted

You need to return the incremented value. You postfix incremented a local variable and returned the non-incremented value. Use ++i not i++

final List<Integer> ints = new ArrayList<>();
IntStream.iterate(0, i -> ++i).limit(5).forEach(val -> ints.add(val));
System.out.println(ints);

answered 4 hours ago

flakes

5,80611848

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up vote
8
down vote

i++ has the value of i before it's incremented. You need to use the prefix operator.

IntStream.iterate(0, i -> ++i).limit(5).forEach(val -> ints.add(val));

Actually, don't do that. There's no reason to mutate i, it's simply thrown away. Use the side effect free version. That's the whole idea behind functional programming, after all: avoiding side effects.

IntStream.iterate(0, i -> i + 1).limit(5).forEach(val -> ints.add(val));

It would also be better to collect the stream into a list rather than adding values with forEach. It directly expresses the intention to create a list and again avoids side effects.

final List<Integer> ints = IntStream.iterate(0, i -> i + 1).limit(5)
 .boxed().collect(Collectors.toList());

edited 4 hours ago

answered 4 hours ago

John Kugelman

236k51394447

add a commentÂ |Â

up vote
2
down vote

You are using a postfix i++, instead of a prefix ++i while passing on the value to forEach. Changing to the following shall provide you the expected output :

IntStream.iterate(0, i -> ++i).limit(5).forEach(ints::add);

Aside, an alternate way of iterating and combining limit with Java9+ is using IntStream.iterate with an IntPredicate as :

IntStream.iterate(0, i -> i < 5, i -> i + 1).forEach(ints::add);

edited 4 hours ago

answered 4 hours ago

nullpointer

33.7k1068136

add a commentÂ |Â

up vote
1
down vote

You could use a print to see what was happend:

final List<Integer> ints = new ArrayList<>();
 IntStream.iterate(0, i -> 
 System.out.println(i);
 return i++;
 ).limit(5)
 .forEach(val -> ints.add(val));
 System.out.println(ints);

In that case, the value of i always will be 0, because the increment occurs after the value is returned, the correct way is

final List<Integer> ints = new ArrayList<>();
 IntStream.iterate(0, i -> 
 System.out.println(i);
 return ++i;
 ).limit(5)
 .forEach(val -> ints.add(val));
 System.out.println(ints);

answered 4 hours ago

Damian Lattenero

8,86821642

add a commentÂ |Â

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

up vote
4
down vote

accepted

You need to return the incremented value. You postfix incremented a local variable and returned the non-incremented value. Use ++i not i++

final List<Integer> ints = new ArrayList<>();
IntStream.iterate(0, i -> ++i).limit(5).forEach(val -> ints.add(val));
System.out.println(ints);

answered 4 hours ago

flakes

5,80611848

add a commentÂ |Â

up vote
4
down vote

accepted

You need to return the incremented value. You postfix incremented a local variable and returned the non-incremented value. Use ++i not i++

final List<Integer> ints = new ArrayList<>();
IntStream.iterate(0, i -> ++i).limit(5).forEach(val -> ints.add(val));
System.out.println(ints);

answered 4 hours ago

flakes

5,80611848

add a commentÂ |Â

up vote
4
down vote

accepted

You need to return the incremented value. You postfix incremented a local variable and returned the non-incremented value. Use ++i not i++

final List<Integer> ints = new ArrayList<>();
IntStream.iterate(0, i -> ++i).limit(5).forEach(val -> ints.add(val));
System.out.println(ints);

answered 4 hours ago

flakes

5,80611848

You need to return the incremented value. You postfix incremented a local variable and returned the non-incremented value. Use ++i not i++

final List<Integer> ints = new ArrayList<>();
IntStream.iterate(0, i -> ++i).limit(5).forEach(val -> ints.add(val));
System.out.println(ints);

answered 4 hours ago

flakes

5,80611848

answered 4 hours ago

flakes

5,80611848

answered 4 hours ago

flakes

5,80611848

answered 4 hours ago

flakes

5,80611848

add a commentÂ |Â

up vote
8
down vote

i++ has the value of i before it's incremented. You need to use the prefix operator.

IntStream.iterate(0, i -> ++i).limit(5).forEach(val -> ints.add(val));

IntStream.iterate(0, i -> i + 1).limit(5).forEach(val -> ints.add(val));

It would also be better to collect the stream into a list rather than adding values with forEach. It directly expresses the intention to create a list and again avoids side effects.

final List<Integer> ints = IntStream.iterate(0, i -> i + 1).limit(5)
 .boxed().collect(Collectors.toList());

edited 4 hours ago

answered 4 hours ago

John Kugelman

236k51394447

add a commentÂ |Â

up vote
8
down vote

i++ has the value of i before it's incremented. You need to use the prefix operator.

IntStream.iterate(0, i -> ++i).limit(5).forEach(val -> ints.add(val));

IntStream.iterate(0, i -> i + 1).limit(5).forEach(val -> ints.add(val));

It would also be better to collect the stream into a list rather than adding values with forEach. It directly expresses the intention to create a list and again avoids side effects.

final List<Integer> ints = IntStream.iterate(0, i -> i + 1).limit(5)
 .boxed().collect(Collectors.toList());

edited 4 hours ago

answered 4 hours ago

John Kugelman

236k51394447

add a commentÂ |Â

up vote
8
down vote

i++ has the value of i before it's incremented. You need to use the prefix operator.

IntStream.iterate(0, i -> ++i).limit(5).forEach(val -> ints.add(val));

IntStream.iterate(0, i -> i + 1).limit(5).forEach(val -> ints.add(val));

It would also be better to collect the stream into a list rather than adding values with forEach. It directly expresses the intention to create a list and again avoids side effects.

final List<Integer> ints = IntStream.iterate(0, i -> i + 1).limit(5)
 .boxed().collect(Collectors.toList());

edited 4 hours ago

answered 4 hours ago

John Kugelman

236k51394447

i++ has the value of i before it's incremented. You need to use the prefix operator.

IntStream.iterate(0, i -> ++i).limit(5).forEach(val -> ints.add(val));

IntStream.iterate(0, i -> i + 1).limit(5).forEach(val -> ints.add(val));

It would also be better to collect the stream into a list rather than adding values with forEach. It directly expresses the intention to create a list and again avoids side effects.

final List<Integer> ints = IntStream.iterate(0, i -> i + 1).limit(5)
 .boxed().collect(Collectors.toList());

edited 4 hours ago

answered 4 hours ago

John Kugelman

236k51394447

edited 4 hours ago

answered 4 hours ago

John Kugelman

236k51394447

answered 4 hours ago

John Kugelman

236k51394447

answered 4 hours ago

John Kugelman

236k51394447

add a commentÂ |Â

up vote
2
down vote

You are using a postfix i++, instead of a prefix ++i while passing on the value to forEach. Changing to the following shall provide you the expected output :

IntStream.iterate(0, i -> ++i).limit(5).forEach(ints::add);

Aside, an alternate way of iterating and combining limit with Java9+ is using IntStream.iterate with an IntPredicate as :

IntStream.iterate(0, i -> i < 5, i -> i + 1).forEach(ints::add);

edited 4 hours ago

answered 4 hours ago

nullpointer

33.7k1068136

add a commentÂ |Â

up vote
2
down vote

You are using a postfix i++, instead of a prefix ++i while passing on the value to forEach. Changing to the following shall provide you the expected output :

IntStream.iterate(0, i -> ++i).limit(5).forEach(ints::add);

Aside, an alternate way of iterating and combining limit with Java9+ is using IntStream.iterate with an IntPredicate as :

IntStream.iterate(0, i -> i < 5, i -> i + 1).forEach(ints::add);

edited 4 hours ago

answered 4 hours ago

nullpointer

33.7k1068136

add a commentÂ |Â

up vote
2
down vote

You are using a postfix i++, instead of a prefix ++i while passing on the value to forEach. Changing to the following shall provide you the expected output :

IntStream.iterate(0, i -> ++i).limit(5).forEach(ints::add);

Aside, an alternate way of iterating and combining limit with Java9+ is using IntStream.iterate with an IntPredicate as :

IntStream.iterate(0, i -> i < 5, i -> i + 1).forEach(ints::add);

edited 4 hours ago

answered 4 hours ago

nullpointer

33.7k1068136

You are using a postfix i++, instead of a prefix ++i while passing on the value to forEach. Changing to the following shall provide you the expected output :

IntStream.iterate(0, i -> ++i).limit(5).forEach(ints::add);

Aside, an alternate way of iterating and combining limit with Java9+ is using IntStream.iterate with an IntPredicate as :

IntStream.iterate(0, i -> i < 5, i -> i + 1).forEach(ints::add);

edited 4 hours ago

answered 4 hours ago

nullpointer

33.7k1068136

edited 4 hours ago

answered 4 hours ago

nullpointer

33.7k1068136

answered 4 hours ago

nullpointer

33.7k1068136

answered 4 hours ago

nullpointer

33.7k1068136

add a commentÂ |Â

up vote
1
down vote

You could use a print to see what was happend:

final List<Integer> ints = new ArrayList<>();
 IntStream.iterate(0, i -> 
 System.out.println(i);
 return i++;
 ).limit(5)
 .forEach(val -> ints.add(val));
 System.out.println(ints);

In that case, the value of i always will be 0, because the increment occurs after the value is returned, the correct way is

final List<Integer> ints = new ArrayList<>();
 IntStream.iterate(0, i -> 
 System.out.println(i);
 return ++i;
 ).limit(5)
 .forEach(val -> ints.add(val));
 System.out.println(ints);

answered 4 hours ago

Damian Lattenero

8,86821642

add a commentÂ |Â

up vote
1
down vote

You could use a print to see what was happend:

final List<Integer> ints = new ArrayList<>();
 IntStream.iterate(0, i -> 
 System.out.println(i);
 return i++;
 ).limit(5)
 .forEach(val -> ints.add(val));
 System.out.println(ints);

In that case, the value of i always will be 0, because the increment occurs after the value is returned, the correct way is

final List<Integer> ints = new ArrayList<>();
 IntStream.iterate(0, i -> 
 System.out.println(i);
 return ++i;
 ).limit(5)
 .forEach(val -> ints.add(val));
 System.out.println(ints);

answered 4 hours ago

Damian Lattenero

8,86821642

add a commentÂ |Â

up vote
1
down vote

You could use a print to see what was happend:

final List<Integer> ints = new ArrayList<>();
 IntStream.iterate(0, i -> 
 System.out.println(i);
 return i++;
 ).limit(5)
 .forEach(val -> ints.add(val));
 System.out.println(ints);

In that case, the value of i always will be 0, because the increment occurs after the value is returned, the correct way is

final List<Integer> ints = new ArrayList<>();
 IntStream.iterate(0, i -> 
 System.out.println(i);
 return ++i;
 ).limit(5)
 .forEach(val -> ints.add(val));
 System.out.println(ints);

answered 4 hours ago

Damian Lattenero

8,86821642

You could use a print to see what was happend:

final List<Integer> ints = new ArrayList<>();
 IntStream.iterate(0, i -> 
 System.out.println(i);
 return i++;
 ).limit(5)
 .forEach(val -> ints.add(val));
 System.out.println(ints);

In that case, the value of i always will be 0, because the increment occurs after the value is returned, the correct way is

final List<Integer> ints = new ArrayList<>();
 IntStream.iterate(0, i -> 
 System.out.println(i);
 return ++i;
 ).limit(5)
 .forEach(val -> ints.add(val));
 System.out.println(ints);

answered 4 hours ago

Damian Lattenero

8,86821642

answered 4 hours ago

Damian Lattenero

8,86821642

answered 4 hours ago

Damian Lattenero

8,86821642

answered 4 hours ago

Damian Lattenero

8,86821642

add a commentÂ |Â

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