This must be quite basic, but I was wondering how to add an integer to an array?
这一定是非常基础的,但我想知道如何在数组中添加一个整数?
I know I can add strings like this:
我知道我可以添加这样的字符串:
NSMutableArray *trArray = [[NSMutableArray alloc] init];
[trArray addObject:@"0"];
[trArray addObject:@"1"];
[trArray addObject:@"2"];
[trArray addObject:@"3"];
But I guess I can't simply add integers like so:
但我想我不能简单地添加这样的整数:
NSMutableArray *trArray = [[NSMutableArray alloc] init];
[trArray addObject:0];
[trArray addObject:1];
[trArray addObject:2];
[trArray addObject:3];
At least the compiler isn't happy with that and tells me that I'm doing a cast without having told it so.
至少编译器对此并不满意,并且告诉我,我正在进行演员而没有告诉它。
Any explanations would be very much appreciated.
任何解释都将非常感激。
3 个解决方案
#1
27
Yes that's right. The compiler won't accept your code like this. The difference is the following:
恩,那就对了。编译器不会接受这样的代码。区别在于:
If you write @"a String"
, it's the same as if you created a string and autoreleased it. So you create an object by using @"a String"
.
如果你写@“一个字符串”,它就像你创建一个字符串并自动释放它一样。因此,您使用@“a String”创建对象。
But an array can only store objects (more precise: pointers to object). So you have to create objects which store your integer.
但是数组只能存储对象(更精确:指向对象的指针)。所以你必须创建存储整数的对象。
NSNumber *anumber = [NSNumber numberWithInteger:4];
[yourArray addObject:anumber];
To retrive the integer again, do it like this
要再次检索整数,请执行此操作
NSNumber anumber = [yourArray objectAtIndex:6];
int yourInteger = [anumber intValue];
I hope my answer helps you to understand why it doesn't work. You can't cast an integer to a pointer. And that is the warning you get from Xcode.
我希望我的回答可以帮助你理解为什么它不起作用。您不能将整数强制转换为指针。这是你从Xcode得到的警告。
EDIT:
编辑:
It is now also possible to write the following
现在也可以写下面的内容
[yourArray addObject:@3];
which is a shortcut to create a NSNumber. The same syntax is available for arrays
这是创建NSNumber的快捷方式。数组的语法相同
@[@1, @2];
will give you an NSArray containing 2 NSNumber objects with the values 1 and 2.
将为您提供一个NSArray,其中包含2个值为1和2的NSNumber对象。
#2
7
You have to use NSNumbers I think, try adding these objects to your array: [NSNumber numberWithInteger:myInt];
您必须使用NSNumbers,尝试将这些对象添加到您的数组中:[NSNumber numberWithInteger:myInt];
NSMutableArray *trArray = [[NSMutableArray alloc] init];
NSNumber *yourNumber = [[NSNumber alloc] numberWithInt:5];
[trArray addObject: yourNumber];
#3
4
You can also use this if you want to use strings:
如果要使用字符串,也可以使用它:
NSMutableArray *array = [[NSMutableArray alloc] init];
[array addObject:[NSString stringWithFormat:@"%d",1]];
[[array objectAtIndex:0] intValue];
#1
27
Yes that's right. The compiler won't accept your code like this. The difference is the following:
恩,那就对了。编译器不会接受这样的代码。区别在于:
If you write @"a String"
, it's the same as if you created a string and autoreleased it. So you create an object by using @"a String"
.
如果你写@“一个字符串”,它就像你创建一个字符串并自动释放它一样。因此,您使用@“a String”创建对象。
But an array can only store objects (more precise: pointers to object). So you have to create objects which store your integer.
但是数组只能存储对象(更精确:指向对象的指针)。所以你必须创建存储整数的对象。
NSNumber *anumber = [NSNumber numberWithInteger:4];
[yourArray addObject:anumber];
To retrive the integer again, do it like this
要再次检索整数,请执行此操作
NSNumber anumber = [yourArray objectAtIndex:6];
int yourInteger = [anumber intValue];
I hope my answer helps you to understand why it doesn't work. You can't cast an integer to a pointer. And that is the warning you get from Xcode.
我希望我的回答可以帮助你理解为什么它不起作用。您不能将整数强制转换为指针。这是你从Xcode得到的警告。
EDIT:
编辑:
It is now also possible to write the following
现在也可以写下面的内容
[yourArray addObject:@3];
which is a shortcut to create a NSNumber. The same syntax is available for arrays
这是创建NSNumber的快捷方式。数组的语法相同
@[@1, @2];
will give you an NSArray containing 2 NSNumber objects with the values 1 and 2.
将为您提供一个NSArray,其中包含2个值为1和2的NSNumber对象。
#2
7
You have to use NSNumbers I think, try adding these objects to your array: [NSNumber numberWithInteger:myInt];
您必须使用NSNumbers,尝试将这些对象添加到您的数组中:[NSNumber numberWithInteger:myInt];
NSMutableArray *trArray = [[NSMutableArray alloc] init];
NSNumber *yourNumber = [[NSNumber alloc] numberWithInt:5];
[trArray addObject: yourNumber];
#3
4
You can also use this if you want to use strings:
如果要使用字符串,也可以使用它:
NSMutableArray *array = [[NSMutableArray alloc] init];
[array addObject:[NSString stringWithFormat:@"%d",1]];
[[array objectAtIndex:0] intValue];