String[] a = c.toArray(new String[0]) - 未名空间MITBBS历史存档

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String[] a = c.toArray(new String[0])

String[] a = c.toArray(new String[0])# Java - 爪哇娇娃

y*o2006-11-25 08:11

1 楼

At the very bottom of the following page:
http://java.sun.com/docs/books/tutorial/collections/interfaces/collection.ht
ml

Suppose that c is known to contain only strings (perhaps because c is of typ
e Collection). The following snippet dumps the contents of c into a
newly allocated array of String whose length is identical to the number of e
lements in c.
String[] a = c.toArray(new String[0]);

The API doc for public T[] toArray(T[] a) says:

g*g2006-11-25 08:11

2 楼

don't make much difference, new String[5] is an array too.

typ
a
e

【在 y********o 的大作中提到】

: At the very bottom of the following page:
: http://java.sun.com/docs/books/tutorial/collections/interfaces/collection.ht
: ml
:
: Suppose that c is known to contain only strings (perhaps because c is of typ
: e Collection). The following snippet dumps the contents of c into a
: newly allocated array of String whose length is identical to the number of e
: lements in c.
: String[] a = c.toArray(new String[0]);
:

y*o2006-11-25 08:11

3 楼

Thx, I was just curious about the capacity of the array at initialization.
It makes no difference if a.length <= c.size(), otherwise, you get null for
a[i] where i>c.size().

【在 g*****g 的大作中提到】

: don't make much difference, new String[5] is an array too.
:
: typ
: a
: e

m*t2006-11-25 08:11

4 楼

String[] a = c.toArray(new String[c.size()]);
is what I usually do.

for

【在 y********o 的大作中提到】

: Thx, I was just curious about the capacity of the array at initialization.
: It makes no difference if a.length <= c.size(), otherwise, you get null for
: a[i] where i>c.size().