在Objective-C中有没有(stringByAppendingString:)字符串连接的快捷方式，或者一般使用NSString的快捷方式?
例如，我想做:
NSString *myString = @"This";
NSString *test = [myString stringByAppendingString:@" is just a test"];
更像是:
string myString = "This";
string test = myString + " is just a test";
当前回答
当处理字符串时，我经常发现使源文件objc++更容易，然后我可以使用问题中显示的第二个方法连接std::字符串。
std::string stdstr = [nsstr UTF8String];
//easier to read and more portable string manipulation goes here...
NSString* nsstr = [NSString stringWithUTF8String:stdstr.c_str()];
其他回答
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];
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];
使c = [a stringByAppendingString: b]更短的唯一方法是在st点附近使用自动补全。+运算符是C的一部分，它不知道Objective-C对象。
NSString *result=[NSString stringWithFormat:@"%@ %@", @"Hello", @"World"];
