问题
Basically, the question is - are the following essentially the same?
NSString *value1 = ...;
NSString *value2 = [[NSString alloc] initWithString:value1];
and
NSString *value1 = ...;
NSString *value2 = [value1 copy];
回答1:
Conceptually, yes. However, there is one difference: alloc always creates a new string, whereas copy may return the same string.
In particular, immutable objects, such as immutable strings, are likely respond to copy by returning themselves rather than creating and returning a copy. (After all, if you can't change anything about the original, why would you really need a copy?) Mutable strings will respond to it by creating and returning a copy, as you'd expect.
initWithString: is in the middle: It may release the receiver and return the string you gave it, similar to how copy may return the receiver. However, if that happens, it means you wasted the creation of the string you created with alloc. With copy, you may not need to create any additional objects at all.
About the only reason to use alloc and initWithString: is if you have your own subclass of NSString and want to make an instance of it from an existing string. copy won't use your desired subclass. Since subclassing NSString is practically never warranted in Cocoa, the same is true of using initWithString: (or stringWithString:).
So the bottom line is, just use copy (or mutableCopy). It's shorter, clearer about your intent, and can be faster.
回答2:
Non-mutable strings are treated a bit special, compared to ordinary objects, so in this case, yes, the two operations are the same.
To wit:
NSString *str1 = @"string";
NSString *str2 = [str1 copy];
NSString *str3 = [[NSString alloc] initWithString: str1];
NSLog(@"str1: %p, str2: %p, str3: %p", str1, str2, str3);
Which gives me the following output:
str1: 0x108a960b0, str2: 0x108a960b0, str3: 0x108a960b0
Since the pointer addresses are the same, we are talking about the same object.
来源：https://stackoverflow.com/questions/8386084/is-allocinitwithstring-same-as-copy