Sam*_*Sam 29
这是测试两个Bundles是否相同的一种方法:
one
可以是null
,请确保这两个值都是null
,并且键实际存在two
码:
public boolean equalBundles(Bundle one, Bundle two) {
if(one.size() != two.size())
return false;
Set<String> setOne = new HashSet<>(one.keySet());
setOne.addAll(two.keySet());
Object valueOne;
Object valueTwo;
for(String key : setOne) {
if (!one.containsKey(key) || !two.containsKey(key))
return false;
valueOne = one.get(key);
valueTwo = two.get(key);
if(valueOne instanceof Bundle && valueTwo instanceof Bundle &&
!equalBundles((Bundle) valueOne, (Bundle) valueTwo)) {
return false;
}
else if(valueOne == null) {
if(valueTwo != null)
return false;
}
else if(!valueOne.equals(valueTwo))
return false;
}
return true;
}
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我已经测试了Sam的答案,它包含一个缺陷。另外我现在很喜欢 Kotlin,所以这是我的版本。
Bundle
递归测试。代码:
fun equalBundles(one: Bundle, two: Bundle): Boolean {
if (one.size() != two.size())
return false
if (!one.keySet().containsAll(two.keySet()))
return false
for (key in one.keySet()) {
val valueOne = one.get(key)
val valueTwo = two.get(key)
if (valueOne is Bundle && valueTwo is Bundle) {
if (!equalBundles(valueOne , valueTwo)) return false
} else if (valueOne != valueTwo) return false
}
return true
}
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private static boolean equalsBundles(Bundle a, Bundle b) {
Set<String> aks = a.keySet();
Set<String> bks = b.keySet();
if (!aks.containsAll(bks)) {
return false;
}
for (String key : aks) {
if (!a.get(key).equals(b.get(key))) {
return false;
}
}
return true;
}
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