Question

我有一个包含对象A，B，A，A，B的数组，其中包含键1,2,3,4,5

但是当A重复时，NSDictionary没有采取'A'2次并且提出错误说

-[NSDictionary initWithObjects:forKeys:]: count of objects (1) differs from count of keys (0)'

那么解决方案是什么

我的代码：

Ar=[[NSArray alloc]initWithObjects:A,A,A,A,A,nil];
Ar2=[[NSArray alloc]initWithObjects:1,2,3,4,5,nil];
    dic2=[[NSMutableDictionary alloc]initWithObjects:Ar forKeys:Ar2];

Answer 1

请尝试以下代码片段

NSMutableDictionary *data = [[NSMutableDictionary alloc]init];
 [data setValue:@"A" forKey:@"1"];
 [data setValue:@"B" forKey:@"2"];
 [data setValue:@"A" forKey:@"3"];
 [data setValue:@"A" forKey:@"4"];
 [data setValue:@"B" forKey:@"5"];

Answer 2

使用新语法使代码更具可读性，并且在大多数情况下可以避免此类问题

NSDictionary *data = @{@"1" : @"A", @"2" : @"B", @"3" : @"A", @"4" : @"A", @"5" : @"B"};

Answer 3

应该是这样的。

    NSArray *Ar=[[NSArray alloc]initWithObjects:@"A",@"A",@"A",@"A",@"A",nil];
NSArray *Ar2=[[NSArray alloc]initWithObjects:@1,@2,@3,@4,@5,nil];
NSMutableDictionary *dic2=[[NSMutableDictionary alloc]initWithObjects:Ar forKeys:Ar2];

最诚挚的问候。

NSDictionary没有重复的字符

3 个答案: