循环执行无限次
你能告诉我错误可能是什么吗？
for (int r=[list count]-1; r>=0;r--) {
NSMutableArray *temp;
temp=[list objectAtIndex:r];
[list insertObject:temp atIndex:r++];
}
第一次调用时，[列表计数] 为 2。
Could you please tell me what the error could be?
for (int r=[list count]-1; r>=0;r--) {
NSMutableArray *temp;
temp=[list objectAtIndex:r];
[list insertObject:temp atIndex:r++];
}
At the first call, [list count] is 2.
如果你对这篇内容有疑问，欢迎到本站社区发帖提问 参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。
绑定邮箱获取回复消息
由于您还没有绑定你的真实邮箱，如果其他用户或者作者回复了您的评论，将不能在第一时间通知您！
发布评论
评论（3）
您在循环内递增循环变量
r，因此它永远不会递减到零。您需要将此行：更改
为：
You're incrementing your loop variable
rinside the loop, so it can never decrement to zero.You need to change this line:
to perhaps:
您有一个无限循环，因为循环开始通过测试 (r == 1 && 1 >= 0) 并且从那时起， r 永远不会改变。您只需在 r (1) 处获取对象并将其插入到 r (1) 处，然后递增 r (r == 2)。最后，循环结束， r 递减 (r == 1) 并且您再次运行测试 (1 >= 0 )，因此它运行循环并发生完全相同的事情。
您可能想在下一个索引 (r + 1) 处插入 temp，但这会导致崩溃，因为您的数组只有 2 个位置。您需要执行 addObject: 来增加数组的大小（并在本例中将项目插入到 r + 1 处）。
You have an infinite loop because the loop starts passing the test (r == 1 && 1 >= 0) and from that point, r never changes. You simply grab the object at r (1) and insert it at r (1) and then increment r (r == 2). Finally, the loop ends, r gets decremented (r == 1) and you run your test again (1 >= 0 ) so it runs the loop and the exact same thing happens.
You probably want to insert temp at the next index (r + 1) but that will cause a crash since your array has only 2 places. You would need to do an addObject: in order to increase the size of the array (and to insert the item at r + 1 in this case).
保罗是对的。也许您认为
r++是r+1的简写？不是，它的意思是r=r+1。Paul’s right. Maybe you think that
r++is a shorthand forr+1? It’s not, it meansr=r+1.