Question

我 st着这种法典:

static NSMutableSet* test_set;

-(void)foo1
{
  test_set=[NSMutableSet setWithObject:[NSNumber numberWithInt:1]];
  NSLog(@"count:%d",[test_set count]);
}


-(void)foo2
{
  NSLog(@"pointer:%p",test_set);
  NSLog(@"count:%d",[test_set count]); // here I get EXC_BAD_ACCESS
}

I calling foo2 only after foo1. My debug out is like:

count:1
pointer:0x262790
Program received signal:  “EXC_BAD_ACCESS”.

What s wrong? __ Intresting note: it fails only when foo2 is calling in schedule.__ Sorry, I missed details. Both works perfect. Thank you all

Answer 1

您没有掌握被分配到<条码>测试_set的物体,这意味着该物体可在<条码>-foo2前处理。一般来说,如果在采用某种方法后需要一个物体才能生存,那么你就应当拥有该物体,例如通过<代码>+alloc或-retain<>/code>:



-(void)foo1
{
  test_set=[[NSMutableSet alloc] initWithObjects:[NSNumber numberWithInt:1], nil];
  NSLog(@"count:%d",[test_set count]);
}


或

-(void)foo1
{
  test_set=[[NSMutableSet setWithObject:[NSNumber numberWithInt:1]] retain];
  NSLog(@"count:%d",[test_set count]);
}


The rules of taking and relinquishing ownership of objects are discussed in the Mem或y Management Programming Guide.

Answer 2


You haven t retained test_set.  The set returned by setWithObject: will be autoreleased. If you add

[test_set retain];


在从<条码>查询后: 载于<条码>foo1(),并添加

[test_set release];

至<代码>foo2(>>末,它应当工作。

或许应该读到Cocoa 。《记忆管理方案规划指南》()。

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