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[英]Warning when trying to get NSString type argument from NSInvocation
[英]How can I get an NSString result from an NSInvocation?
以下代码按预期工作:
NSLog(@"%@", [NSString stringWithString:@"test"]; // Logs "test"
但是当我用NSInvocation
替换它时,我得到一个完全不同的结果:
Class class = [NSString class];
SEL selector = @selector(stringWithString:);
NSInvocation *invocation = [NSInvocation invocationWithMethodSignature:
[class methodSignatureForSelector:selector]];
[invocation setTarget:class];
[invocation setSelector:selector];
[invocation setArgument:@"test" atIndex:2];
[invocation invoke];
id returnValue = nil;
[invocation getReturnValue:&returnValue];
NSLog(@"%@", returnValue); // Logs "NSCFString"
我搜索过高低,但无法弄清楚这一点。 有帮助吗? 谢谢!
从NSInvocation类引用:
当参数值是对象时,将指针传递给应从中复制对象的变量(或内存):
NSArray *anArray;
[invocation setArgument:&anArray atIndex:3];
因为@“test”实际上构建了NSString的一个实例,所以你应该使用
NSString *testString = @"test";
[invocation setArgument:&testString atIndex:2];
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