英文:
Is there any difference between these two codes?
问题
@Autowired
private StringRedisTemplate stringRedisTemplate;
public void test(String key) {
// IDEA提示错误
Map<String, String> entries1 = stringRedisTemplate.opsForHash().entries(key);
// 这是正确的。
HashOperations<String, String, String> opsForHash = stringRedisTemplate.opsForHash();
Map<String, String> entries = opsForHash.entries(key);
}
英文:
@Autowired
private StringRedisTemplate stringRedisTemplate;
public void test(String key) {
// IDEA prompts an error
Map<String, String> entries1 = stringRedisTemplate.opsForHash().entries(key);
// This is OK.
HashOperations<String, String, String> opsForHash = stringRedisTemplate.opsForHash();
Map<String, String> entries = opsForHash.entries(key);
}
答案1
得分: 4
问题是方法opsForHash()使用了两种泛型，签名如下：
public <HK, HV> HashOperations<K, HK, HV> opsForHash()
如果你想要在一行中使用，你需要设置这些泛型，就像这样：
Map<String, String> entries1 = stringRedisTemplate.<String, String>opsForHash().entries(key);
在你的代码中，第二种方法有效是因为编译器从=操作符左边定义的变量中推断出了泛型。
英文:
The problem Is that the method opsForHash() uses 2 generics, This is the signature:
public <HK, HV> HashOperations<K, HK, HV> opsForHash()
If you want to use a single line, you need to set the generics, just like:
Map<String, String> entries1 = stringRedisTemplate.<String, String>opsForHash().entries(key);
In your code, the second approach works because the compiler finds out the generics from the defined variable on the left side of the = operator.
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