Math.toIntExact inside lambda expression?

I'm learning about lambda expressions.
Given a list of names, I want to count the numbers of names that start with N.

I did that:

final static List<String> friends = Arrays.asList("Brian", "Nate", "Neal", "Raju", "Sara", "Scott");



public static int countFriendsStartWithN() {

    return  Math.toIntExact(friends

            .stream()

            .filter(name -> name.startsWith("N"))

            .count());

}

The call to the count method returns a primitive long but I want an int.

I used Math.toIntExact to get the long value as int.

Is it possible to get the int value directly inside the lambda expression?

edited Nov 16 '18 at 8:38

asked Nov 16 '18 at 8:09

Jesus Zavarce

1,069819

What do you mean by lambda experssion? That looks alright to me.

– SamzSakerz
Nov 16 '18 at 8:10

3

If you know the long value is not too large, you can simply cast it to int.

– Eran
Nov 16 '18 at 8:11

2

@Eran and you know it's not too large, because it is necessarily no greater than the number of items in a collection, which is an int.

– Andy Turner
Nov 16 '18 at 8:14

I just want to know if is possible to do the int conversion inside the stream process.

– Jesus Zavarce
Nov 16 '18 at 8:16

add a comment |

I'm learning about lambda expressions.
Given a list of names, I want to count the numbers of names that start with N.

I did that:

final static List<String> friends = Arrays.asList("Brian", "Nate", "Neal", "Raju", "Sara", "Scott");



public static int countFriendsStartWithN() {

    return  Math.toIntExact(friends

            .stream()

            .filter(name -> name.startsWith("N"))

            .count());

}

The call to the count method returns a primitive long but I want an int.

I used Math.toIntExact to get the long value as int.

Is it possible to get the int value directly inside the lambda expression?

edited Nov 16 '18 at 8:38

asked Nov 16 '18 at 8:09

Jesus Zavarce

1,069819

What do you mean by lambda experssion? That looks alright to me.

– SamzSakerz
Nov 16 '18 at 8:10

3

If you know the long value is not too large, you can simply cast it to int.

– Eran
Nov 16 '18 at 8:11

2

@Eran and you know it's not too large, because it is necessarily no greater than the number of items in a collection, which is an int.

– Andy Turner
Nov 16 '18 at 8:14

I just want to know if is possible to do the int conversion inside the stream process.

– Jesus Zavarce
Nov 16 '18 at 8:16

add a comment |

I'm learning about lambda expressions.
Given a list of names, I want to count the numbers of names that start with N.

I did that:

final static List<String> friends = Arrays.asList("Brian", "Nate", "Neal", "Raju", "Sara", "Scott");



public static int countFriendsStartWithN() {

    return  Math.toIntExact(friends

            .stream()

            .filter(name -> name.startsWith("N"))

            .count());

}

The call to the count method returns a primitive long but I want an int.

I used Math.toIntExact to get the long value as int.

Is it possible to get the int value directly inside the lambda expression?

edited Nov 16 '18 at 8:38

asked Nov 16 '18 at 8:09

Jesus Zavarce

1,069819

I'm learning about lambda expressions.
Given a list of names, I want to count the numbers of names that start with N.

I did that:

final static List<String> friends = Arrays.asList("Brian", "Nate", "Neal", "Raju", "Sara", "Scott");



public static int countFriendsStartWithN() {

    return  Math.toIntExact(friends

            .stream()

            .filter(name -> name.startsWith("N"))

            .count());

}

The call to the count method returns a primitive long but I want an int.

I used Math.toIntExact to get the long value as int.

Is it possible to get the int value directly inside the lambda expression?

java lambda

edited Nov 16 '18 at 8:38

asked Nov 16 '18 at 8:09

Jesus Zavarce

1,069819

edited Nov 16 '18 at 8:38

asked Nov 16 '18 at 8:09

Jesus Zavarce

1,069819

edited Nov 16 '18 at 8:38

asked Nov 16 '18 at 8:09

Jesus Zavarce

1,069819

asked Nov 16 '18 at 8:09

Jesus Zavarce

1,069819

asked Nov 16 '18 at 8:09

Jesus Zavarce

1,069819

What do you mean by lambda experssion? That looks alright to me.

– SamzSakerz
Nov 16 '18 at 8:10

3

If you know the long value is not too large, you can simply cast it to int.

– Eran
Nov 16 '18 at 8:11

2

@Eran and you know it's not too large, because it is necessarily no greater than the number of items in a collection, which is an int.

– Andy Turner
Nov 16 '18 at 8:14

I just want to know if is possible to do the int conversion inside the stream process.

– Jesus Zavarce
Nov 16 '18 at 8:16

add a comment |

What do you mean by lambda experssion? That looks alright to me.

– SamzSakerz
Nov 16 '18 at 8:10

3

If you know the long value is not too large, you can simply cast it to int.

– Eran
Nov 16 '18 at 8:11

2

@Eran and you know it's not too large, because it is necessarily no greater than the number of items in a collection, which is an int.

– Andy Turner
Nov 16 '18 at 8:14

I just want to know if is possible to do the int conversion inside the stream process.

– Jesus Zavarce
Nov 16 '18 at 8:16

What do you mean by lambda experssion? That looks alright to me.

– SamzSakerz
Nov 16 '18 at 8:10

If you know the long value is not too large, you can simply cast it to int.

– Eran
Nov 16 '18 at 8:11

@Eran and you know it's not too large, because it is necessarily no greater than the number of items in a collection, which is an int.

– Andy Turner
Nov 16 '18 at 8:14

I just want to know if is possible to do the int conversion inside the stream process.

– Jesus Zavarce
Nov 16 '18 at 8:16

add a comment |

5 Answers
5

active

oldest

votes

No, it is not possible to fit your call to toIntExact into your chain of method calls, your stream pipeline. This is because count is a terminal operation and returns a primitive long on which no method call is possible. A terminal operation is an operation that ends the stream pipeline and produces a result (or a side effect).

So I believe the best thing you can do is to live with the code you already have. IMHO it’s fine.

answered Nov 16 '18 at 8:14

Ole V.V.

31.8k74257

add a comment |

Well, here's a somewhat silly way of calculating the count as an int without casting:

public static int countFriendsStartWithN() {

    return friends.stream()

                  .filter(name -> name.startsWith("N"))

                  .mapToInt (s -> 1)

                  .sum();

}

answered Nov 16 '18 at 8:18

Eran

290k37478563

add a comment |

You can't do anything inside the lambda expression you currently have, since that's a Predicate: it returns a boolean. Math.toIntExact returns an int.

You can do it without the Math.toIntExact (or a simple cast) like so:

return /* create stream, filter */

    .mapToInt(a -> 1).sum();

But this is likely to be slower than doing what you are doing at the moment.

answered Nov 16 '18 at 8:18

Andy Turner

84k983143

add a comment |

Yet another option that is not really better - it is possible to use a collector that applies a finisher:

public static int countFriendsStartWithN() {

    return friends.stream()

                .filter(name -> name.startsWith("N"))

                .collect(Collectors.collectingAndThen(Collectors.counting(), Math::toIntExact));

}

This may have an advantage if you need it frequenty - you could build a utility method returning this Collector to make it reusable.

edited Nov 16 '18 at 9:01

answered Nov 16 '18 at 8:56

Hulk

3,42812142

add a comment |

Here's a way to do this with reduce

public static int countFriendsStartWithN2() {

    return friends

            .stream()

            .filter(name -> name.startsWith("N"))

            .map(s -> 1)

            .reduce(0, Integer::sum);

}

answered Nov 16 '18 at 8:25

SamzSakerz

1,6691627

add a comment |

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5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

So I believe the best thing you can do is to live with the code you already have. IMHO it’s fine.

answered Nov 16 '18 at 8:14

Ole V.V.

31.8k74257

add a comment |

So I believe the best thing you can do is to live with the code you already have. IMHO it’s fine.

answered Nov 16 '18 at 8:14

Ole V.V.

31.8k74257

add a comment |

So I believe the best thing you can do is to live with the code you already have. IMHO it’s fine.

answered Nov 16 '18 at 8:14

Ole V.V.

31.8k74257

So I believe the best thing you can do is to live with the code you already have. IMHO it’s fine.

answered Nov 16 '18 at 8:14

Ole V.V.

31.8k74257

answered Nov 16 '18 at 8:14

Ole V.V.

31.8k74257

answered Nov 16 '18 at 8:14

Ole V.V.

31.8k74257

answered Nov 16 '18 at 8:14

Ole V.V.

31.8k74257

add a comment |

Well, here's a somewhat silly way of calculating the count as an int without casting:

public static int countFriendsStartWithN() {

    return friends.stream()

                  .filter(name -> name.startsWith("N"))

                  .mapToInt (s -> 1)

                  .sum();

}

answered Nov 16 '18 at 8:18

Eran

290k37478563

add a comment |

Well, here's a somewhat silly way of calculating the count as an int without casting:

public static int countFriendsStartWithN() {

    return friends.stream()

                  .filter(name -> name.startsWith("N"))

                  .mapToInt (s -> 1)

                  .sum();

}

answered Nov 16 '18 at 8:18

Eran

290k37478563

add a comment |

Well, here's a somewhat silly way of calculating the count as an int without casting:

public static int countFriendsStartWithN() {

    return friends.stream()

                  .filter(name -> name.startsWith("N"))

                  .mapToInt (s -> 1)

                  .sum();

}

answered Nov 16 '18 at 8:18

Eran

290k37478563

Well, here's a somewhat silly way of calculating the count as an int without casting:

public static int countFriendsStartWithN() {

    return friends.stream()

                  .filter(name -> name.startsWith("N"))

                  .mapToInt (s -> 1)

                  .sum();

}

answered Nov 16 '18 at 8:18

Eran

290k37478563

answered Nov 16 '18 at 8:18

Eran

290k37478563

answered Nov 16 '18 at 8:18

Eran

290k37478563

answered Nov 16 '18 at 8:18

Eran

290k37478563

add a comment |

You can't do anything inside the lambda expression you currently have, since that's a Predicate: it returns a boolean. Math.toIntExact returns an int.

You can do it without the Math.toIntExact (or a simple cast) like so:

return /* create stream, filter */

    .mapToInt(a -> 1).sum();

But this is likely to be slower than doing what you are doing at the moment.

answered Nov 16 '18 at 8:18

Andy Turner

84k983143

add a comment |

You can't do anything inside the lambda expression you currently have, since that's a Predicate: it returns a boolean. Math.toIntExact returns an int.

You can do it without the Math.toIntExact (or a simple cast) like so:

return /* create stream, filter */

    .mapToInt(a -> 1).sum();

But this is likely to be slower than doing what you are doing at the moment.

answered Nov 16 '18 at 8:18

Andy Turner

84k983143

add a comment |

You can't do anything inside the lambda expression you currently have, since that's a Predicate: it returns a boolean. Math.toIntExact returns an int.

You can do it without the Math.toIntExact (or a simple cast) like so:

return /* create stream, filter */

    .mapToInt(a -> 1).sum();

But this is likely to be slower than doing what you are doing at the moment.

answered Nov 16 '18 at 8:18

Andy Turner

84k983143

You can't do anything inside the lambda expression you currently have, since that's a Predicate: it returns a boolean. Math.toIntExact returns an int.

You can do it without the Math.toIntExact (or a simple cast) like so:

return /* create stream, filter */

    .mapToInt(a -> 1).sum();

But this is likely to be slower than doing what you are doing at the moment.

answered Nov 16 '18 at 8:18

Andy Turner

84k983143

answered Nov 16 '18 at 8:18

Andy Turner

84k983143

answered Nov 16 '18 at 8:18

Andy Turner

84k983143

answered Nov 16 '18 at 8:18

Andy Turner

84k983143

add a comment |

Yet another option that is not really better - it is possible to use a collector that applies a finisher:

public static int countFriendsStartWithN() {

    return friends.stream()

                .filter(name -> name.startsWith("N"))

                .collect(Collectors.collectingAndThen(Collectors.counting(), Math::toIntExact));

}

This may have an advantage if you need it frequenty - you could build a utility method returning this Collector to make it reusable.

edited Nov 16 '18 at 9:01

answered Nov 16 '18 at 8:56

Hulk

3,42812142

add a comment |

Yet another option that is not really better - it is possible to use a collector that applies a finisher:

public static int countFriendsStartWithN() {

    return friends.stream()

                .filter(name -> name.startsWith("N"))

                .collect(Collectors.collectingAndThen(Collectors.counting(), Math::toIntExact));

}

This may have an advantage if you need it frequenty - you could build a utility method returning this Collector to make it reusable.

edited Nov 16 '18 at 9:01

answered Nov 16 '18 at 8:56

Hulk

3,42812142

add a comment |

Yet another option that is not really better - it is possible to use a collector that applies a finisher:

public static int countFriendsStartWithN() {

    return friends.stream()

                .filter(name -> name.startsWith("N"))

                .collect(Collectors.collectingAndThen(Collectors.counting(), Math::toIntExact));

}

This may have an advantage if you need it frequenty - you could build a utility method returning this Collector to make it reusable.

edited Nov 16 '18 at 9:01

answered Nov 16 '18 at 8:56

Hulk

3,42812142

Yet another option that is not really better - it is possible to use a collector that applies a finisher:

public static int countFriendsStartWithN() {

    return friends.stream()

                .filter(name -> name.startsWith("N"))

                .collect(Collectors.collectingAndThen(Collectors.counting(), Math::toIntExact));

}

This may have an advantage if you need it frequenty - you could build a utility method returning this Collector to make it reusable.

edited Nov 16 '18 at 9:01

answered Nov 16 '18 at 8:56

Hulk

3,42812142

edited Nov 16 '18 at 9:01

answered Nov 16 '18 at 8:56

Hulk

3,42812142

answered Nov 16 '18 at 8:56

Hulk

3,42812142

answered Nov 16 '18 at 8:56

Hulk

3,42812142

add a comment |

Here's a way to do this with reduce

public static int countFriendsStartWithN2() {

    return friends

            .stream()

            .filter(name -> name.startsWith("N"))

            .map(s -> 1)

            .reduce(0, Integer::sum);

}

answered Nov 16 '18 at 8:25

SamzSakerz

1,6691627

add a comment |

Here's a way to do this with reduce

public static int countFriendsStartWithN2() {

    return friends

            .stream()

            .filter(name -> name.startsWith("N"))

            .map(s -> 1)

            .reduce(0, Integer::sum);

}

answered Nov 16 '18 at 8:25

SamzSakerz

1,6691627

add a comment |

Here's a way to do this with reduce

public static int countFriendsStartWithN2() {

    return friends

            .stream()

            .filter(name -> name.startsWith("N"))

            .map(s -> 1)

            .reduce(0, Integer::sum);

}

answered Nov 16 '18 at 8:25

SamzSakerz

1,6691627

Here's a way to do this with reduce

public static int countFriendsStartWithN2() {

    return friends

            .stream()

            .filter(name -> name.startsWith("N"))

            .map(s -> 1)

            .reduce(0, Integer::sum);

}

answered Nov 16 '18 at 8:25

SamzSakerz

1,6691627

answered Nov 16 '18 at 8:25

SamzSakerz

1,6691627

answered Nov 16 '18 at 8:25

SamzSakerz

1,6691627

answered Nov 16 '18 at 8:25

SamzSakerz

1,6691627

add a comment |

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