Why does HashMap.put both compare hashes and test equality?

I analyse the HashMap source code in Java and get a question about the put method.

Below is the put method in JDK1.6:

public V put(K key, V value) {
    if (key == null)
        return putForNullKey(value);
    int hash = hash(key.hashCode());
    int i = indexFor(hash, table.length);
    for (Entry<K,V> e = table[i]; e != null; e = e.next) {
        Object k;
        if (e.hash == hash && ((k = e.key) == key || key.equals(k))) {
            V oldValue = e.value;
            e.value = value;
            e.recordAccess(this);
            return oldValue;
        }
    }

    modCount++;
    addEntry(hash, key, value, i);
    return null;
}

I get confused about the if (e.hash == hash && ((k = e.key) == key || key.equals(k)))

Why is the condition like this?

Because in Java super class Object, there is a contract of hashCode and equals:

If two objects are equal according to the equals(Object) method, then calling the hashCode method on each of the two objects must produce the same integer result.

So the key.equals(k) implies key.hashCode() == k.hashCode().

The hash() is below:

 static int hash(int h) {
    // This function ensures that hashCodes that differ only by
    // constant multiples at each bit position have a bounded
    // number of collisions (approximately 8 at default load factor).
    h ^= (h >>> 20) ^ (h >>> 12);
    return h ^ (h >>> 7) ^ (h >>> 4);
}

thus key.hashCode() == k.hashCode() implies e.hash == hash.

So why isn't the condition like if ((k = e.key) == key || key.equals(k))?