xcode：如何在xcode中调试？

Question

I have a Person class which has only a property: name. I want to list the property value when debug, but xcode just display "isa", how can I do it like in eclipse?

Xcode :

eclipse:

Answer 1

Under the hood, properties are accessed using methods. A property named name can be read using the name method, and it can be set using the setName: method. You can use the debugger's po command to print a description of an object. Try typing this at the debugger console:

po [p name]

The po command works by sending the debugDescription message to the object you're printing, and by default, debugDescription just sends the description message. So you could add this method to your Person class:

- (NSString *)description {
    return [NSString stringWithFormat:@"<%@: %p name=%@>", self.class, self, self.name];
}

Then you can use a debugger command like this:

po p

and get output like this:

<Person: 0x10013fd60 name=Jack>

Answer 2

如果rob的帖子不起作用，那么我将尝试在控制台中输入bt（用于回溯）