递增属性int(self.int++)

integer++ integer 直接将新值赋给 整数 而不是发送消息和使用访问器。假设 整数 声明为 NSInteger 属性以下语句对整数值的影响相同,但直接访问不符合kvo。

[self setInteger:0];
self.integer = self.integer + 1; // use generated accessors
NSLog(@"Integer is :%d",[self integer]); // Integer is: 1
integer++;
NSLog(@"Integer is :%d",[self integer]); // Integer is: 2

I believe Obj-C was updated in the last year or two so that this kind of code works. 我写了一个快速测试,发现以下代码都是有效的:

我的H:

#import <Cocoa/Cocoa.h>

@interface TheAppDelegate : NSObject <NSApplicationDelegate> {
    NSUInteger value;
}

@property NSUInteger otherValue;

- (NSUInteger) value;
- (void) setValue:(NSUInteger)value;

@end

我的M:

#import "TheAppDelegate.h"

@implementation TheAppDelegate

- (void)applicationDidFinishLaunching:(NSNotification *)aNotification
{
    value = self.otherValue = 1;
    NSLog(@"%lu %lu", (unsigned long)value, (unsigned long)self.value);
    NSLog(@"%lu %lu", (unsigned long)_otherValue, (unsigned long)self.otherValue);
    self.value++;
    self.otherValue++;
    NSLog(@"%lu %lu", (unsigned long)value, (unsigned long)self.value);
    NSLog(@"%lu %lu", (unsigned long)_otherValue, (unsigned long)self.otherValue);
}

- (NSUInteger) value
{
    return value;
}

- (void) setValue:(NSUInteger)_value
{
    value = _value;
}

@end

我的输出:

1 1
1 1
2 2
2 2

我相信在我读到的某个地方有一个技术文档解释了这一点,但我记不起在哪里找到了它。我相信它说的是:

x++ 将改为 x+=1

那 x+=y x=x+y

x=y.a=z y.a=z,x=y.a