据我所知,如果 lambda 实现的抽象方法的签名中没有 throws
,则无法处理 lambda 中抛出的异常。
我遇到了以下代码,它有效。为什么 openStream()
不需要处理 IOException
?我可以在 tryWithResources
中看到 try-catch
但我不明白它背后的机制。
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.net.URL;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collection;
import java.util.Map;
import java.util.concurrent.Callable;
import java.util.concurrent.ConcurrentSkipListMap;
import java.util.function.Function;
import java.util.function.Supplier;
public class Main {
public static <AUTOCLOSEABLE extends AutoCloseable, OUTPUT> Supplier<OUTPUT> tryWithResources(
Callable<AUTOCLOSEABLE> callable, Function<AUTOCLOSEABLE, Supplier<OUTPUT>> function,
Supplier<OUTPUT> defaultSupplier) {
return () -> {
try (AUTOCLOSEABLE autoCloseable = callable.call()) {
return function.apply(autoCloseable).get();
} catch (Throwable throwable) {
return defaultSupplier.get();
}
};
}
public static <INPUT, OUTPUT> Function<INPUT, OUTPUT> function(Supplier<OUTPUT> supplier) {
return i -> supplier.get();
}
public static void main(String... args) {
Map<String, Collection<String>> anagrams = new ConcurrentSkipListMap<>();
int count = tryWithResources(
() -> new BufferedReader(new InputStreamReader(
new URL("http://www.puzzlers.org/pub/wordlists/unixdict.txt").openStream())),
reader -> () -> reader.lines().parallel().mapToInt(word -> {
char[] chars = word.toCharArray();
Arrays.parallelSort(chars);
String key = Arrays.toString(chars);
Collection<String> collection = anagrams.computeIfAbsent(key, function(ArrayList::new));
collection.add(word);
return collection.size();
}).max().orElse(0), () -> 0).get();
anagrams.values().stream().filter(ana -> ana.size() >= count).forEach((list) -> {
for (String s : list)
System.out.print(s + " ");
System.out.println();
});
}
}
最佳答案
我已将您的示例简化为核心部分:
public static void main(String[] args) {
withCallable(() -> new URL("url").openStream()); // compiles
withSupplier(() -> new URL("url").openStream()); // does not compile
}
public static <T> void withCallable(Callable<T> callable) { }
public static <T> void withSupplier(Supplier<T> callable) { }
如果您尝试这样做,您会发现 withCallable
可以正常编译,但 withSupplier
无法编译;即使 lambda 表达式与两个功能接口(interface)的签名兼容。
这背后的原因是 Callable
的功能方法接口(interface),即call()
, 在其签名中声明 throws Exception
。 Supplier.get()
没有。
引用 JLS section 11.2.3 :
It is a compile-time error if a lambda body can throw some exception class E when E is a checked exception class and E is not a subclass of some class declared in the throws clause of the function type targeted by the lambda expression.
关于java - 在没有 try-catch 的情况下处理 lambda 中的异常,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/33776451/