HashMap put() 与compute() 的返回类型

Question

HashMap put 选择返回与该键关联的先前值

    /**
     * Associates the specified value with the specified key in this map.
     * If the map previously contained a mapping for the key, the old
     * value is replaced.
     *
     * @param key key with which the specified value is to be associated
     * @param value value to be associated with the specified key
     * @return the previous value associated with <tt>key</tt>, or
     *         <tt>null</tt> if there was no mapping for <tt>key</tt>.
     *         (A <tt>null</tt> return can also indicate that the map
     *         previously associated <tt>null</tt> with <tt>key</tt>.)
     */
    public V put(K key, V value) {
        return putVal(hash(key), key, value, false, true);
    }

而compute选择返回新值

    /**
     * Attempts to compute a mapping for the specified key and its current
     * mapped value (or {@code null} if there is no current mapping). For
     * example, to either create or append a {@code String} msg to a value
     * mapping:
     *
     * <pre> {@code
     * map.compute(key, (k, v) -> (v == null) ? msg : v.concat(msg))}</pre>
     * (Method {@link #merge merge()} is often simpler to use for such purposes.)
     *
     * <p>If the function returns {@code null}, the mapping is removed (or
     * remains absent if initially absent).  If the function itself throws an
     * (unchecked) exception, the exception is rethrown, and the current mapping
     * is left unchanged.
     *
     * @implSpec
     * The default implementation is equivalent to performing the following
     * steps for this {@code map}, then returning the current value or
     * {@code null} if absent:
     *
     * <pre> {@code
     * V oldValue = map.get(key);
     * V newValue = remappingFunction.apply(key, oldValue);
     * if (oldValue != null ) {
     *    if (newValue != null)
     *       map.put(key, newValue);
     *    else
     *       map.remove(key);
     * } else {
     *    if (newValue != null)
     *       map.put(key, newValue);
     *    else
     *       return null;
     * }
     * }</pre>
     *
     * <p>The default implementation makes no guarantees about synchronization
     * or atomicity properties of this method. Any implementation providing
     * atomicity guarantees must override this method and document its
     * concurrency properties. In particular, all implementations of
     * subinterface {@link java.util.concurrent.ConcurrentMap} must document
     * whether the function is applied once atomically only if the value is not
     * present.
     *
     * @param key key with which the specified value is to be associated
     * @param remappingFunction the function to compute a value
     * @return the new value associated with the specified key, or null if none
     * @throws NullPointerException if the specified key is null and
     *         this map does not support null keys, or the
     *         remappingFunction is null
     * @throws UnsupportedOperationException if the {@code put} operation
     *         is not supported by this map
     *         (<a href="{@docRoot}/java/util/Collection.html#optional-restrictions">optional</a>)
     * @throws ClassCastException if the class of the specified key or value
     *         prevents it from being stored in this map
     *         (<a href="{@docRoot}/java/util/Collection.html#optional-restrictions">optional</a>)
     * @since 1.8
     */
    default V compute(K key,
            BiFunction<? super K, ? super V, ? extends V> remappingFunction) {

我很想知道return为 put 和compute 选择不同类型值的原因是什么。我知道这是一个主观问题，但想听听不同的意见 - 这样可以帮助理解它的逻辑意义。

Answer 1

只是因为当您调用 时put，您知道新值 - 您将其作为参数传递，因此返回它是毫无意义的。

当您调用 时compute，您不知道新值，因此返回它很有用。