-2
[a characterAtIndex:0] = [b characterAtIndex:0];

我想将 charAt a 分配给 charAt b。

但是失败了,怎么办?

4

2 回答 2

0
  NSMutableString *a = [NSMutableString stringWithString:@"HOUSE" ];

  NSString *b= @"MUSIC";

 [a replaceCharactersInRange:NSMakeRange(0, 1) withString:[b substringWithRange:NSMakeRange(0, 1)]];

结果你得到了鼠标;-)

于 2013-02-28T08:18:05.190 回答
0

如果您正在寻找这样的方法:setCharacter:atIndex:

然后看下面的代码:

NSMutableString *string=[NSMutableString stringWithString:@"abcdefgh"];

[string setCharacter:'X' atIndex:0]; //Note X is character not string so "X" and its is primitive therefore even @ is not prefixed.

NSLog(@"%@",string);

这是通过NSMutableString其方法实现的类别来实现的

@implementation NSMutableString (SetCharacterAtIndex)

- (void)setCharacter:(char)character atIndex:(NSUInteger)index;{

    [self replaceCharactersInRange:NSMakeRange(index, 1) withString:[NSString stringWithFormat:@"%c",character]];

}

编辑:

有了上述方法后,您可以使用:

[a characterAtIndex:0] = [b characterAtIndex:0];

[a setCharacter:[b characterAtIndex:0] atIndex:0];
于 2013-02-28T08:57:57.233 回答