ios 不兼容的整数到指针转换将“NSInteger”（又名“int”）发送到“NSInteger *”（又名“int *”）类型的参数
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Incompatible Integer to pointer conversion sending 'NSInteger' (aka 'int') to parameter of type 'NSInteger *' (aka 'int *')
提问by JamEngulfer
I'm trying to parse an integer from a NSDictionary using the code
我正在尝试使用代码从 NSDictionary 解析一个整数
[activeItem setData_id:[[NSString stringWithFormat:@"%@", [dict valueForKeyPath:@"data_id"]] integerValue]];
[activeItem setData_id:[[NSString stringWithFormat:@"%@", [dict valueForKeyPath:@"data_id"]] integerValue]];
However, this is giving me this error: Incompatible integer to pointer conversion sending 'NSInteger' (aka 'int') to parameter of type 'NSInteger *' (aka 'int *')
但是，这给了我这个错误： Incompatible integer to pointer conversion sending 'NSInteger' (aka 'int') to parameter of type 'NSInteger *' (aka 'int *')
setData_id takes an integer as a parameter. If I want to parse to a string, [NSString stringWithFormat:@"%@", [dict valueForKeyPath:@"data_id"]]works perfectly.
setData_id 接受一个整数作为参数。如果我想解析为字符串，[NSString stringWithFormat:@"%@", [dict valueForKeyPath:@"data_id"]]效果很好。
What I'm doing here is parsing the result of valueForKeyPath to a String, then parsing an integer from that.
我在这里做的是将 valueForKeyPath 的结果解析为一个字符串，然后从中解析一个整数。
回答by Macmade
How is your setData_id:method declared?
你的setData_id:方法是如何声明的？
Looks like it expect an NSInteger *rather than a NSInteger...
看起来它期望一个NSInteger *而不是一个NSInteger......
Either it is declared as:
要么声明为：
- ( void )setData_id: ( NSInteger )value;
And you can use your code.
你可以使用你的代码。
Otherwise, it means it is declared as:
否则，这意味着它被声明为：
- ( void )setData_id: ( NSInteger * )value;
It might be a typo... If you really need an integer pointer, then you may use (assuming you know what you are doing in terms of scope):
这可能是一个错字......如果你真的需要一个整数指针，那么你可以使用（假设你知道你在范围方面做什么）：
NSInteger i = [ [ NSString stringWithFormat: @"%@", [ dict valueForKeyPath: @"data_id" ] ] integerValue ];
[ activeItem setData_id: &i ];
But I think you just made a typo, adding a pointer (NSInteger *), while you meant NSInteger.
但我认为您只是打了一个错字，添加了一个指针 ( NSInteger *)，而您的意思是NSInteger.
Note: If setData_idis a property, the same applies:
注意：如果setData_id是属性，同样适用：
@property( readwrite, assign ) NSInteger data_id;
versus:
相对：
@property( readwrite, assign ) NSInteger * data_id;
I guess you wrote the second example, while meaning the first one...
我猜你写了第二个例子，而意思是第一个......
回答by Leo Natan
The property is defined incorrectly.
属性定义不正确。
It should be:
它应该是：
@property (readwrite) NSInteger data_id;
instead of
代替
@property (readwrite) NSInteger *data_id;
You are attempting to pass an integer value to a format that expects to have a pointer type.
您正在尝试将整数值传递给期望具有指针类型的格式。
Either use
要么使用
[activeItem setData_id:[NSString stringWithFormat:@"%@", [dict valueForKeyPath:@"data_id"]]];
or
或者
[activeItem setData_id:[NSString stringWithFormat:@"%d", [[dict valueForKeyPath:@"data_id"] integerValue]]];
If you need to set an integer, drop the [NSString stringWithFormat:@"%@"]- this creates a string.
如果您需要设置一个整数，请删除[NSString stringWithFormat:@"%@"]- 这将创建一个字符串。
[activeItem setData_id:[[dict valueForKeyPath:@"data_id"] integerValue]];
回答by zeeawan
Use integerValueand intValuewhere appropriate.
在适当的地方使用integerValue和intValue。
