Question

我最近将c数组更改为NSMutable数组，因为我遇到了很多内存问题。但它起作用，但从

开始

/* If two starttimes are within interval millisec, make them the same */
for (int i = 0; i < [starttimes count] - 1; i++) {
   if ([[starttimes objectAtIndex:i+1] integerValue] - [[starttimes objectAtIndex:i] integerValue] <= interval) {
        [starttimes insertObject: [starttimes objectAtIndex:i] atIndex:(i+1)];
    }
}

CPU使用率接近100％。我不知道发生了什么事？逻辑很简单。它只是从这个NSMutable数组中获取两个元素，然后进行比较和插入。顺便说一句，这段代码永远不会停止。它持续运行3秒，然后我的应用程序崩溃。

(TimeSignature*)time {
/* Get all the starttimes in all tracks, in sorted order */
NSInteger initsize = 1;
if ([tracks count] > 0) {
    MidiTrack *track = [tracks objectAtIndex:0];
    initsize = [track.notes count];
    initsize = initsize * [tracks count]/2;
}
NSMutableArray* starttimes = [[NSMutableArray alloc]initWithCapacity:initsize];

for (int tracknum = 0; tracknum < [tracks count]; tracknum++) {

    NSLog(@"tracknum is %d",tracknum);
    MidiTrack *track = [tracks objectAtIndex:tracknum];
     NSLog(@"WE ARE HERE %ld",(long)initsize);
    for (int j = 0; j < [track.notes count]; j++) {
        MidiNote *note = [track.notes objectAtIndex:j];

        NSLog(@"%@",note);
        [starttimes addObject:[NSNumber numberWithInteger:note.startTime]];
    }
}


/* Notes within "millisec" milliseconds apart should be combined */
int interval = time.quarter * millisec * 1000 / time.tempo;

/* If two starttimes are within interval millisec, make them the same */
for (int i = 0; i < [starttimes count] - 1; i++) {
    if ([[starttimes objectAtIndex:i+1] integerValue] - [[starttimes objectAtIndex:i] integerValue] <= interval) {
        [starttimes insertObject: [starttimes objectAtIndex:i] atIndex:(i+1)];
    }

}

非常感谢

Answer 1

一般Objective-C建议：阅读方法名称。他们告诉你想要这些方法。

您误解了-insertObject:。 insertObject做的是... 插入对象。它没有做什么：替换一个对象。每次拨打insertObject时，您都会增加数组的大小。实际上，如果if条件通过一次，你就会创建一个无限循环。

你可能想要-replaceObjectAtIndex:withObject:。与insertObject:一样，其名称将与名称相同。

NSMutableArray高CPU使用率

1 个答案: