Question

目前正在加载我的声音

NSString * strName = [NSString stringWithFormat:@"soundpack%ld", (long) (value+1)];

获取每个声音的值。

现在我想将它从数组中拉出来。看起来像这样：

NSMutableArray * sounds = [NSMutableArray arrayWithObjects: @"soundpack1", @"soundpack2", @"soundpack3", @"soundpack4", @"soundpack5", @"soundpack6", @"soundpack7", @"soundpack8", @"soundpack9", nil];
    NSUInteger count = [sounds count];

    for (NSUInteger i = 0; i < count; ++i) {
        NSInteger nElements = count - i;
        NSInteger n = arc4random_uniform((u_int32_t )nElements) + i;
        [sounds exchangeObjectAtIndex:i withObjectAtIndex:n];
    }
    NSString * strName0 = [sounds objectAtIndex:1];

好吧，现在我已经改组了声音。现在我想像以前一样用特定的值初始化。

类似的东西：

NSString * strName = [sounds objectAtIndex:%ld, (long) (value+1)];

或

NSString * strName = [sounds objectAtIndex:@"%ld", (long) (value+1)];

没用。但我不知道，现在该怎么做......

Answer 1

尝试更改此行

NSString * strName = [sounds objectAtIndex:@"%ld", (long) (value+1)];

到

NSString * strName = [sounds objectAtIndex:(long) (value+1)];

Answer 2

为什么不要在洗牌后使用相同的片段？

NSString * strName = [NSString stringWithFormat:@"soundpack%ld", (long) (value+1)];

如何用值初始化nsstring？

2 个答案: