宏:

// stringConcat(...)
//     A shortcut for concatenating strings (or objects' string representations).
//     Input: Any number of non-nil NSObjects.
//     Output: All arguments concatenated together into a single NSString.

#define stringConcat(...) \
    [@[__VA_ARGS__] componentsJoinedByString:@""]

测试用例:

- (void)testStringConcat {
    NSString *actual;

    actual = stringConcat(); //might not make sense, but it's still a valid expression.
    STAssertEqualObjects(@"", actual, @"stringConcat");

    actual = stringConcat(@"A");
    STAssertEqualObjects(@"A", actual, @"stringConcat");

    actual = stringConcat(@"A", @"B");
    STAssertEqualObjects(@"AB", actual, @"stringConcat");

    actual = stringConcat(@"A", @"B", @"C");
    STAssertEqualObjects(@"ABC", actual, @"stringConcat");

    // works on all NSObjects (not just strings):
    actual = stringConcat(@1, @" ", @2, @" ", @3);
    STAssertEqualObjects(@"1 2 3", actual, @"stringConcat");
}

备用宏:(如果你想强制一个最小数量的参数)

// stringConcat(...)
//     A shortcut for concatenating strings (or objects' string representations).
//     Input: Two or more non-nil NSObjects.
//     Output: All arguments concatenated together into a single NSString.

#define stringConcat(str1, str2, ...) \
    [@[ str1, str2, ##__VA_ARGS__] componentsJoinedByString:@""];

我能想到两个答案……这两种方法都没有使用连接运算符那么令人愉快。

首先，使用一个NSMutableString，它有一个appendString方法，消除了对额外临时字符串的一些需求。

其次，使用NSArray通过componentsJoinedByString方法进行连接。

假设你不知道这里有多少根弦。

NSMutableArray *arrForStrings = [[NSMutableArray alloc] init];
for (int i=0; i<[allMyStrings count]; i++) {
    NSString *str = [allMyStrings objectAtIndex:i];
    [arrForStrings addObject:str];
}
NSString *readyString = [[arrForStrings mutableCopy] componentsJoinedByString:@", "];

通过创建AppendString宏的快捷方式…

#define AS(A,B)    [(A) stringByAppendingString:(B)]

NSString *myString = @"This";

NSString *test = AS(myString,@" is just a test");

注意:

如果使用宏，当然只需使用可变参数，请参阅ethb的答案。

尝试stringWithFormat:

NSString *myString = [NSString stringWithFormat:@"%@ %@ %@ %d", "The", "Answer", "Is", 42];

如何缩短stringByAppendingString和使用#define:

#define and stringByAppendingString

因此你可以使用:

NSString* myString = [@"Hello " and @"world"];

问题是它只适用于两个字符串，你需要包装额外的括号更多的追加:

NSString* myString = [[@"Hello" and: @" world"] and: @" again"];