Printing the letters from A to Z using a Java stream
I have this code, but it gives me an error:
Type mismatch: cannot convert from int to Character
Stream.iterate('a', i -> i + 1).limit(26).forEach(System.out::println);
Although it is fine to write int i = 'a';
I know I can write it like this, but that seems like too much code for a simple task.
Stream.iterate('a', i -> (char)(i + 1)).limit(26).forEach(System.out::println);
Why is the Java type inference failing?
java java-8 char java-stream
edited Dec 28 '18 at 10:55
Peter Mortensen
13.5k1983111
asked Dec 27 '18 at 21:41
fastcodejava
24k19109162
add a comment |
I have this code, but it gives me an error:
Type mismatch: cannot convert from int to Character
Stream.iterate('a', i -> i + 1).limit(26).forEach(System.out::println);
Although it is fine to write int i = 'a';
I know I can write it like this, but that seems like too much code for a simple task.
Stream.iterate('a', i -> (char)(i + 1)).limit(26).forEach(System.out::println);
Why is the Java type inference failing?
java java-8 char java-stream
edited Dec 28 '18 at 10:55
Peter Mortensen
13.5k1983111
asked Dec 27 '18 at 21:41
fastcodejava
24k19109162
5
Related stackoverflow.com/a/32424763/1746118
– nullpointer
Dec 28 '18 at 1:21
add a comment |
I have this code, but it gives me an error:
Type mismatch: cannot convert from int to Character
Stream.iterate('a', i -> i + 1).limit(26).forEach(System.out::println);
Although it is fine to write int i = 'a';
I know I can write it like this, but that seems like too much code for a simple task.
Stream.iterate('a', i -> (char)(i + 1)).limit(26).forEach(System.out::println);
Why is the Java type inference failing?
java java-8 char java-stream
edited Dec 28 '18 at 10:55
Peter Mortensen
13.5k1983111
asked Dec 27 '18 at 21:41
fastcodejava
24k19109162
I have this code, but it gives me an error:
Type mismatch: cannot convert from int to Character
Stream.iterate('a', i -> i + 1).limit(26).forEach(System.out::println);
Although it is fine to write int i = 'a';
I know I can write it like this, but that seems like too much code for a simple task.
Stream.iterate('a', i -> (char)(i + 1)).limit(26).forEach(System.out::println);
Why is the Java type inference failing?
java java-8 char java-stream
java java-8 char java-stream
edited Dec 28 '18 at 10:55
Peter Mortensen
13.5k1983111
asked Dec 27 '18 at 21:41
fastcodejava
24k19109162
edited Dec 28 '18 at 10:55
Peter Mortensen
13.5k1983111
asked Dec 27 '18 at 21:41
fastcodejava
24k19109162
edited Dec 28 '18 at 10:55
Peter Mortensen
13.5k1983111
edited Dec 28 '18 at 10:55
Peter Mortensen
13.5k1983111
edited Dec 28 '18 at 10:55
Peter Mortensen
13.5k1983111
13.5k1983111
asked Dec 27 '18 at 21:41
fastcodejava
24k19109162
asked Dec 27 '18 at 21:41
fastcodejava
24k19109162
asked Dec 27 '18 at 21:41
fastcodejava
24k19109162
24k19109162
5
Related stackoverflow.com/a/32424763/1746118
– nullpointer
Dec 28 '18 at 1:21
add a comment |
5
Related stackoverflow.com/a/32424763/1746118
– nullpointer
Dec 28 '18 at 1:21
5
5
Related stackoverflow.com/a/32424763/1746118
– nullpointer
Dec 28 '18 at 1:21
Related stackoverflow.com/a/32424763/1746118
– nullpointer
Dec 28 '18 at 1:21
add a comment |
2 Answers
2
active
oldest
votes
The reason why i -> i + 1 does not compile is because you're attempting to implicitly convert an int to a Character which the compiler cannot do itself alone.
In other words, you can think of Stream.iterate('a', i -> i + 1) as:
Stream.iterate('a', (Character i) -> {
int i1 = i + 1;
return i1; // not possible
});
As you have noted, explicitly casting to char solves it:
Stream.iterate('a', i -> (char)(i + 1))...
Btw this is better done as:
IntStream.rangeClosed('a', 'z').forEach(c -> System.out.println((char)c));
This is better because:
- No boxing overhead thus more efficient
- if you were to stop at say letter
hwith the use ofiterateyou'd have to do more brain processing than just enteringhas the upper bound withrangeClosedbecause you'd need to find the number to truncate the infinite stream upon. - Along with the boxing
iterategenerates an infinite stream which in this specific case has more overhead than the finite one withrangeClosed. Further, it's far easier to runIntStream.rangeClosedin parallel, not that you want to in this specific case but it's something to keep in mind. here is some discussion on Generators as sources by Brian Goetz.
etc...
answered Dec 27 '18 at 21:42
Aomine
40.6k73870
"iterate generates an infinite stream which again has more overhead than a finite one." why's that?
– Alexander
Dec 28 '18 at 6:11
@Alexander Firstly, I will improve my wording as it may not be the best to remove ambiguity. Thanks. I have also included a reference to a blog which might be of interest.
– Aomine
Dec 28 '18 at 7:53
2
"an infinite stream which in this specific case has more overhead than the finite one with" that's better :) Good callout on the split performance of this.
– Alexander
Dec 29 '18 at 17:39
add a comment |
How about just:
Stream.iterate('a', i -> ++i).limit(26).forEach(System.out::println);
i -> i + 1 does not work because i is a Character and i + 1 causes an implicit narrowing conversion (JLS 5.1.3), which is not allowed. You can explicitly cast it as was shown. However ++i works because (From JLS 15.15.1):
Before the addition, binary numeric promotion (§5.6.2) is performed on the value 1 and the value of the variable. If necessary, the sum is narrowed by a narrowing primitive conversion (§5.1.3) and/or subjected to boxing conversion (§5.1.7) to the type of the variable before it is stored.
The ++ operator takes care of the narrowing conversion without us having to explicitly cast it
answered Dec 27 '18 at 21:47
GBlodgett
9,24541633
1
Good answer, would have been even better if you explain why++iworks andi + idoesn't.
– fastcodejava
Dec 28 '18 at 0:45
1
@fastcodejava I have edited my answer to try to explain.
– GBlodgett
Dec 28 '18 at 1:43
3
1 👏🏻 for awesome explanation @GBlodgett
– Deadpool
Dec 28 '18 at 1:50
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
The reason why i -> i + 1 does not compile is because you're attempting to implicitly convert an int to a Character which the compiler cannot do itself alone.
In other words, you can think of Stream.iterate('a', i -> i + 1) as:
Stream.iterate('a', (Character i) -> {
int i1 = i + 1;
return i1; // not possible
});
As you have noted, explicitly casting to char solves it:
Stream.iterate('a', i -> (char)(i + 1))...
Btw this is better done as:
IntStream.rangeClosed('a', 'z').forEach(c -> System.out.println((char)c));
This is better because:
- No boxing overhead thus more efficient
- if you were to stop at say letter
hwith the use ofiterateyou'd have to do more brain processing than just enteringhas the upper bound withrangeClosedbecause you'd need to find the number to truncate the infinite stream upon. - Along with the boxing
iterategenerates an infinite stream which in this specific case has more overhead than the finite one withrangeClosed. Further, it's far easier to runIntStream.rangeClosedin parallel, not that you want to in this specific case but it's something to keep in mind. here is some discussion on Generators as sources by Brian Goetz.
etc...
answered Dec 27 '18 at 21:42
Aomine
40.6k73870
"iterate generates an infinite stream which again has more overhead than a finite one." why's that?
– Alexander
Dec 28 '18 at 6:11
@Alexander Firstly, I will improve my wording as it may not be the best to remove ambiguity. Thanks. I have also included a reference to a blog which might be of interest.
– Aomine
Dec 28 '18 at 7:53
2
"an infinite stream which in this specific case has more overhead than the finite one with" that's better :) Good callout on the split performance of this.
– Alexander
Dec 29 '18 at 17:39
add a comment |
The reason why i -> i + 1 does not compile is because you're attempting to implicitly convert an int to a Character which the compiler cannot do itself alone.
In other words, you can think of Stream.iterate('a', i -> i + 1) as:
Stream.iterate('a', (Character i) -> {
int i1 = i + 1;
return i1; // not possible
});
As you have noted, explicitly casting to char solves it:
Stream.iterate('a', i -> (char)(i + 1))...
Btw this is better done as:
IntStream.rangeClosed('a', 'z').forEach(c -> System.out.println((char)c));
This is better because:
- No boxing overhead thus more efficient
- if you were to stop at say letter
hwith the use ofiterateyou'd have to do more brain processing than just enteringhas the upper bound withrangeClosedbecause you'd need to find the number to truncate the infinite stream upon. - Along with the boxing
iterategenerates an infinite stream which in this specific case has more overhead than the finite one withrangeClosed. Further, it's far easier to runIntStream.rangeClosedin parallel, not that you want to in this specific case but it's something to keep in mind. here is some discussion on Generators as sources by Brian Goetz.
etc...
answered Dec 27 '18 at 21:42
Aomine
40.6k73870
"iterate generates an infinite stream which again has more overhead than a finite one." why's that?
– Alexander
Dec 28 '18 at 6:11
@Alexander Firstly, I will improve my wording as it may not be the best to remove ambiguity. Thanks. I have also included a reference to a blog which might be of interest.
– Aomine
Dec 28 '18 at 7:53
2
"an infinite stream which in this specific case has more overhead than the finite one with" that's better :) Good callout on the split performance of this.
– Alexander
Dec 29 '18 at 17:39
add a comment |
The reason why i -> i + 1 does not compile is because you're attempting to implicitly convert an int to a Character which the compiler cannot do itself alone.
In other words, you can think of Stream.iterate('a', i -> i + 1) as:
Stream.iterate('a', (Character i) -> {
int i1 = i + 1;
return i1; // not possible
});
As you have noted, explicitly casting to char solves it:
Stream.iterate('a', i -> (char)(i + 1))...
Btw this is better done as:
IntStream.rangeClosed('a', 'z').forEach(c -> System.out.println((char)c));
This is better because:
- No boxing overhead thus more efficient
- if you were to stop at say letter
hwith the use ofiterateyou'd have to do more brain processing than just enteringhas the upper bound withrangeClosedbecause you'd need to find the number to truncate the infinite stream upon. - Along with the boxing
iterategenerates an infinite stream which in this specific case has more overhead than the finite one withrangeClosed. Further, it's far easier to runIntStream.rangeClosedin parallel, not that you want to in this specific case but it's something to keep in mind. here is some discussion on Generators as sources by Brian Goetz.
etc...
answered Dec 27 '18 at 21:42
Aomine
40.6k73870
The reason why i -> i + 1 does not compile is because you're attempting to implicitly convert an int to a Character which the compiler cannot do itself alone.
In other words, you can think of Stream.iterate('a', i -> i + 1) as:
Stream.iterate('a', (Character i) -> {
int i1 = i + 1;
return i1; // not possible
});
As you have noted, explicitly casting to char solves it:
Stream.iterate('a', i -> (char)(i + 1))...
Btw this is better done as:
IntStream.rangeClosed('a', 'z').forEach(c -> System.out.println((char)c));
This is better because:
- No boxing overhead thus more efficient
- if you were to stop at say letter
hwith the use ofiterateyou'd have to do more brain processing than just enteringhas the upper bound withrangeClosedbecause you'd need to find the number to truncate the infinite stream upon. - Along with the boxing
iterategenerates an infinite stream which in this specific case has more overhead than the finite one withrangeClosed. Further, it's far easier to runIntStream.rangeClosedin parallel, not that you want to in this specific case but it's something to keep in mind. here is some discussion on Generators as sources by Brian Goetz.
etc...
answered Dec 27 '18 at 21:42
Aomine
40.6k73870
edited Dec 28 '18 at 7:53
answered Dec 27 '18 at 21:42
Aomine
40.6k73870
answered Dec 27 '18 at 21:42
Aomine
40.6k73870
answered Dec 27 '18 at 21:42
Aomine
40.6k73870
40.6k73870
"iterate generates an infinite stream which again has more overhead than a finite one." why's that?
– Alexander
Dec 28 '18 at 6:11
@Alexander Firstly, I will improve my wording as it may not be the best to remove ambiguity. Thanks. I have also included a reference to a blog which might be of interest.
– Aomine
Dec 28 '18 at 7:53
2
"an infinite stream which in this specific case has more overhead than the finite one with" that's better :) Good callout on the split performance of this.
– Alexander
Dec 29 '18 at 17:39
add a comment |
"iterate generates an infinite stream which again has more overhead than a finite one." why's that?
– Alexander
Dec 28 '18 at 6:11
@Alexander Firstly, I will improve my wording as it may not be the best to remove ambiguity. Thanks. I have also included a reference to a blog which might be of interest.
– Aomine
Dec 28 '18 at 7:53
2
"an infinite stream which in this specific case has more overhead than the finite one with" that's better :) Good callout on the split performance of this.
– Alexander
Dec 29 '18 at 17:39
"iterate generates an infinite stream which again has more overhead than a finite one." why's that?
– Alexander
Dec 28 '18 at 6:11
"iterate generates an infinite stream which again has more overhead than a finite one." why's that?
– Alexander
Dec 28 '18 at 6:11
@Alexander Firstly, I will improve my wording as it may not be the best to remove ambiguity. Thanks. I have also included a reference to a blog which might be of interest.
– Aomine
Dec 28 '18 at 7:53
@Alexander Firstly, I will improve my wording as it may not be the best to remove ambiguity. Thanks. I have also included a reference to a blog which might be of interest.
– Aomine
Dec 28 '18 at 7:53
2
2
"an infinite stream which in this specific case has more overhead than the finite one with" that's better :) Good callout on the split performance of this.
– Alexander
Dec 29 '18 at 17:39
"an infinite stream which in this specific case has more overhead than the finite one with" that's better :) Good callout on the split performance of this.
– Alexander
Dec 29 '18 at 17:39
add a comment |
How about just:
Stream.iterate('a', i -> ++i).limit(26).forEach(System.out::println);
i -> i + 1 does not work because i is a Character and i + 1 causes an implicit narrowing conversion (JLS 5.1.3), which is not allowed. You can explicitly cast it as was shown. However ++i works because (From JLS 15.15.1):
Before the addition, binary numeric promotion (§5.6.2) is performed on the value 1 and the value of the variable. If necessary, the sum is narrowed by a narrowing primitive conversion (§5.1.3) and/or subjected to boxing conversion (§5.1.7) to the type of the variable before it is stored.
The ++ operator takes care of the narrowing conversion without us having to explicitly cast it
answered Dec 27 '18 at 21:47
GBlodgett
9,24541633
1
Good answer, would have been even better if you explain why++iworks andi + idoesn't.
– fastcodejava
Dec 28 '18 at 0:45
1
@fastcodejava I have edited my answer to try to explain.
– GBlodgett
Dec 28 '18 at 1:43
3
1 👏🏻 for awesome explanation @GBlodgett
– Deadpool
Dec 28 '18 at 1:50
add a comment |
How about just:
Stream.iterate('a', i -> ++i).limit(26).forEach(System.out::println);
i -> i + 1 does not work because i is a Character and i + 1 causes an implicit narrowing conversion (JLS 5.1.3), which is not allowed. You can explicitly cast it as was shown. However ++i works because (From JLS 15.15.1):
Before the addition, binary numeric promotion (§5.6.2) is performed on the value 1 and the value of the variable. If necessary, the sum is narrowed by a narrowing primitive conversion (§5.1.3) and/or subjected to boxing conversion (§5.1.7) to the type of the variable before it is stored.
The ++ operator takes care of the narrowing conversion without us having to explicitly cast it
answered Dec 27 '18 at 21:47
GBlodgett
9,24541633
1
Good answer, would have been even better if you explain why++iworks andi + idoesn't.
– fastcodejava
Dec 28 '18 at 0:45
1
@fastcodejava I have edited my answer to try to explain.
– GBlodgett
Dec 28 '18 at 1:43
3
1 👏🏻 for awesome explanation @GBlodgett
– Deadpool
Dec 28 '18 at 1:50
add a comment |
How about just:
Stream.iterate('a', i -> ++i).limit(26).forEach(System.out::println);
i -> i + 1 does not work because i is a Character and i + 1 causes an implicit narrowing conversion (JLS 5.1.3), which is not allowed. You can explicitly cast it as was shown. However ++i works because (From JLS 15.15.1):
Before the addition, binary numeric promotion (§5.6.2) is performed on the value 1 and the value of the variable. If necessary, the sum is narrowed by a narrowing primitive conversion (§5.1.3) and/or subjected to boxing conversion (§5.1.7) to the type of the variable before it is stored.
The ++ operator takes care of the narrowing conversion without us having to explicitly cast it
answered Dec 27 '18 at 21:47
GBlodgett
9,24541633
How about just:
Stream.iterate('a', i -> ++i).limit(26).forEach(System.out::println);
i -> i + 1 does not work because i is a Character and i + 1 causes an implicit narrowing conversion (JLS 5.1.3), which is not allowed. You can explicitly cast it as was shown. However ++i works because (From JLS 15.15.1):
Before the addition, binary numeric promotion (§5.6.2) is performed on the value 1 and the value of the variable. If necessary, the sum is narrowed by a narrowing primitive conversion (§5.1.3) and/or subjected to boxing conversion (§5.1.7) to the type of the variable before it is stored.
The ++ operator takes care of the narrowing conversion without us having to explicitly cast it
answered Dec 27 '18 at 21:47
GBlodgett
9,24541633
edited Dec 28 '18 at 1:41
answered Dec 27 '18 at 21:47
GBlodgett
9,24541633
answered Dec 27 '18 at 21:47
GBlodgett
9,24541633
answered Dec 27 '18 at 21:47
GBlodgett
9,24541633
9,24541633
1
Good answer, would have been even better if you explain why++iworks andi + idoesn't.
– fastcodejava
Dec 28 '18 at 0:45
1
@fastcodejava I have edited my answer to try to explain.
– GBlodgett
Dec 28 '18 at 1:43
3
1 👏🏻 for awesome explanation @GBlodgett
– Deadpool
Dec 28 '18 at 1:50
add a comment |
1
Good answer, would have been even better if you explain why++iworks andi + idoesn't.
– fastcodejava
Dec 28 '18 at 0:45
1
@fastcodejava I have edited my answer to try to explain.
– GBlodgett
Dec 28 '18 at 1:43
3
1 👏🏻 for awesome explanation @GBlodgett
– Deadpool
Dec 28 '18 at 1:50
1
1
Good answer, would have been even better if you explain why
++i works and i + i doesn't.– fastcodejava
Dec 28 '18 at 0:45
Good answer, would have been even better if you explain why
++i works and i + i doesn't.– fastcodejava
Dec 28 '18 at 0:45
1
1
@fastcodejava I have edited my answer to try to explain.
– GBlodgett
Dec 28 '18 at 1:43
@fastcodejava I have edited my answer to try to explain.
– GBlodgett
Dec 28 '18 at 1:43
3
3
1 👏🏻 for awesome explanation @GBlodgett
– Deadpool
Dec 28 '18 at 1:50
1 👏🏻 for awesome explanation @GBlodgett
– Deadpool
Dec 28 '18 at 1:50
add a comment |
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5
Related stackoverflow.com/a/32424763/1746118
– nullpointer
Dec 28 '18 at 1:21