我愿用<代码>NSMutableArray,如果我发现一些物体符合某些条件,我就希望从阵列中删除这些物体。
如果没有两个(或更多)阵列,我怎么做?
对于那些想说: 我可以说,如果我把想要从中删除药品的提炼器捆绑起来,我就没有问题去做。 当我发现一个药箱时,我会找下一个药箱检查。 我不需要第二位推手。
我愿用<代码>NSMutableArray,如果我发现一些物体符合某些条件,我就希望从阵列中删除这些物体。
如果没有两个(或更多)阵列,我怎么做?
对于那些想说: 我可以说,如果我把想要从中删除药品的提炼器捆绑起来,我就没有问题去做。 当我发现一个药箱时,我会找下一个药箱检查。 我不需要第二位推手。
我找到了一种超立法的解决办法,它有两阵列,即ok,但在许多局势中却阻止了 par;
NSArray* matchingItems = [mainArray filteredArrayUsingPredicate:[NSPredicate predicateWithFormat:@" attributetockeck MATCHES[cd] %@ ", attributevalue]];
[mainArray removeObjectsInArray:matchingItems];
当我需要把物体从 mu阵列中除去时,我使用后 lo。
for (NSInteger i = arrayCount - 1; i >= 0; i--) {
// remove is OK here
}
如果你有2个阿雷拉。 两种类型的<代码>NSMUtableArray。 你可以做的是:
我将复制你想要保留的新 mu阵,然后将旧阵列分配给新阵列。
for( NSUInter i = 0, j = 0; i < array.count; i++ )
{
if( test )
[array removeObjectAtIndex:j];
else
j++;
}
这里是你可以操作的指挥线路版本中的代码,可以测试它去除每一奇数。
#import <Foundation/Foundation.h>
int main (int argc, const char * argv[])
{
NSAutoreleasePool * pool = [[NSAutoreleasePool alloc] init];
NSMutableArray * array = [NSMutableArray array];
for( NSUInteger i = 0; i < 20; i++ )
[array addObject:[NSNumber numberWithInteger:random()%100]];
NSLog( @"%@", array );
for( NSUInteger i = 0, j = 0; i < array.count; i++ )
{
if( [[array objectAtIndex:j] unsignedIntegerValue] & 1 )
[array removeObjectAtIndex:j];
else
j++;
}
NSLog( @"%@", array );
[pool drain];
return 0;
}
如果你在某一时间重新确认物体,你可以使用:
[mainArray removeObject:objectToRemove];
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