在Java中等于與Arrays相等在Java中比較數組時，以下2條語句之間有什么區別嗎？array1.equals(array2);Arrays.equals(array1, array2);如果是的話，他們是什么？
3 回答
不負相思意
TA貢獻1777條經驗 獲得超10個贊
array1.equals(array2)array1 == array2
Arrays.equals(array1, array2)
array.toString()Arrays.toString(array).
心有法竹
TA貢獻1866條經驗 獲得超5個贊
.equals()
equalsObjectequals
Arrays.equals(array1, array2)array1.equals(array2)Object.equals==null).
Arrays.equals(array1, array2)equals
equalsArrays.equals(array1, array2)
Qyouu
TA貢獻1786條經驗 獲得超11個贊
深入了解這兩種方法的實現情況：
array1.equals(array2);
/**
* Indicates whether some other object is "equal to" this one.
* <p>
* The {@code equals} method implements an equivalence relation
* on non-null object references:
* <ul>
* <li>It is <i>reflexive</i>: for any non-null reference value
* {@code x}, {@code x.equals(x)} should return
* {@code true}.
* <li>It is <i>symmetric</i>: for any non-null reference values
* {@code x} and {@code y}, {@code x.equals(y)}
* should return {@code true} if and only if
* {@code y.equals(x)} returns {@code true}.
* <li>It is <i>transitive</i>: for any non-null reference values
* {@code x}, {@code y}, and {@code z}, if
* {@code x.equals(y)} returns {@code true} and
* {@code y.equals(z)} returns {@code true}, then
* {@code x.equals(z)} should return {@code true}.
* <li>It is <i>consistent</i>: for any non-null reference values
* {@code x} and {@code y}, multiple invocations of
* {@code x.equals(y)} consistently return {@code true}
* or consistently return {@code false}, provided no
* information used in {@code equals} comparisons on the
* objects is modified.
* <li>For any non-null reference value {@code x},
* {@code x.equals(null)} should return {@code false}.
* </ul>
* <p>
* The {@code equals} method for class {@code Object} implements
* the most discriminating possible equivalence relation on objects;
* that is, for any non-null reference values {@code x} and
* {@code y}, this method returns {@code true} if and only
* if {@code x} and {@code y} refer to the same object
* ({@code x == y} has the value {@code true}).
* <p>
* Note that it is generally necessary to override the {@code hashCode}
* method whenever this method is overridden, so as to maintain the
* general contract for the {@code hashCode} method, which states
* that equal objects must have equal hash codes.
*
* @param obj the reference object with which to compare.
* @return {@code true} if this object is the same as the obj
* argument; {@code false} otherwise.
* @see #hashCode()
* @see java.util.HashMap
*/
public boolean equals(Object obj) {
return (this == obj);
}
同時：
Arrays.equals(array1, array2);
/**
* Returns <tt>true</tt> if the two specified arrays of Objects are
* <i>equal</i> to one another. The two arrays are considered equal if
* both arrays contain the same number of elements, and all corresponding
* pairs of elements in the two arrays are equal. Two objects <tt>e1</tt>
* and <tt>e2</tt> are considered <i>equal</i> if <tt>(e1==null ? e2==null
* : e1.equals(e2))</tt>. In other words, the two arrays are equal if
* they contain the same elements in the same order. Also, two array
* references are considered equal if both are <tt>null</tt>.<p>
*
* @param a one array to be tested for equality
* @param a2 the other array to be tested for equality
* @return <tt>true</tt> if the two arrays are equal
*/
public static boolean equals(Object[] a, Object[] a2) {
if (a==a2)
return true;
if (a==null || a2==null)
return false;
int length = a.length;
if (a2.length != length)
return false;
for (int i=0; i<length; i++) {
Object o1 = a[i];
Object o2 = a2[i];
if (!(o1==null ? o2==null : o1.equals(o2)))
return false;
}
return true;
}
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