NSString *result=[NSString stringWithFormat:@"%@ %@", @"Hello", @"World"];

正在尝试在lldb窗格中执行以下操作

[NSString stringWithFormat:@"%@/%@/%@", three, two, one];

这错误。

而是使用alloc和initWithFormat方法:

[[NSString alloc] initWithFormat:@"%@/%@/%@", @"three", @"two", @"one"];

NSString *result=[NSString stringWithFormat:@"%@ %@", @"Hello", @"World"];

NSString *label1 = @"Process Name: ";
NSString *label2 = @"Process Id: ";
NSString *processName = [[NSProcessInfo processInfo] processName];
NSString *processID = [NSString stringWithFormat:@"%d", [[NSProcessInfo processInfo] processIdentifier]];
NSString *testConcat = [NSString stringWithFormat:@"%@ %@ %@ %@", label1, processName, label2, processID];

一个选项:

[NSString stringWithFormat:@"%@/%@/%@", one, two, three];

另一个选择:

我猜你不满意多个追加(a+b+c+d)，在这种情况下，你可以这样做:

NSLog(@"%@", [Util append:one, @" ", two, nil]); // "one two"
NSLog(@"%@", [Util append:three, @"/", two, @"/", one, nil]); // three/two/one

使用类似于

+ (NSString *) append:(id) first, ...
{
    NSString * result = @"";
    id eachArg;
    va_list alist;
    if(first)
    {
        result = [result stringByAppendingString:first];
        va_start(alist, first);
        while (eachArg = va_arg(alist, id)) 
        result = [result stringByAppendingString:eachArg];
        va_end(alist);
    }
    return result;
}

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。