1.NSMutableArray和NSArray實例執(zhí)行mutableCopy和copy方法的區(qū)別毁兆？

    NSMutableArray *arrM = [NSMutableArray arrayWithArray:@[@1,@2]];

    id cp1 = [arrM mutableCopy];
    id cp2 = [arrM copy];

    NSLog(@"arrM %p, class %@", arrM, NSStringFromClass([arrM class]));
    NSLog(@"cp1  %p, class %@", cp1, NSStringFromClass([cp1 class]));
    NSLog(@"cp2  %p, class %@", cp2, NSStringFromClass([cp2 class]));
    
    NSArray *arrI = @[@1,@2];

    id cp3 = [arrI mutableCopy];
    id cp4 = [arrI copy];

    NSLog(@"arrI %p, class %@", arrI, NSStringFromClass([arrI class]));
    NSLog(@"cp3  %p, class %@", cp3, NSStringFromClass([cp3 class]));
    NSLog(@"cp4  %p, class %@", cp4, NSStringFromClass([cp4 class]));

執(zhí)行結(jié)果：

arrM 0x600002e562e0, class __NSArrayM
cp1  0x600002e56610, class __NSArrayM
cp2  0x60000206af40, class __NSArrayI

arrI 0x60000206c380, class __NSArrayI
cp3  0x600002e6d920, class __NSArrayM
cp4  0x60000206c380, class __NSArrayI

上面的問題分兩種情況：

1. 當可變對象mutablecopy和copy之后，都會生成新的對象垃僚，只是新的對象的類型一個是可變的一個是不可變的勋又。
2. 當不可變對象mutablecopy和copy之后田轧，只有mutablecopy會生成新的對象，copy則會只復制指針。

2.下面函數(shù)的執(zhí)行結(jié)果及發(fā)生錯誤的原因

    dispatch_semaphore_t signal;
    signal = dispatch_semaphore_create(1);
    __block long x = 0;
    NSLog(@"0_x:%ld",x);
    dispatch_async(dispatch_get_global_queue(DISPATCH_QUEUE_PRIORITY_DEFAULT, 0), ^{
        sleep(1);
        NSLog(@"waiting");
        x = dispatch_semaphore_signal(signal);
        NSLog(@"1_x:%ld",x);
 
        sleep(2);
        NSLog(@"waking");
        x = dispatch_semaphore_signal(signal);
        NSLog(@"2_x:%ld",x);
    });
    //    dispatch_time_t duration = dispatch_time(DISPATCH_TIME_NOW, 1*1000*1000*1000); //超時1秒
    //    dispatch_semaphore_wait(signal, duration);
     
    x = dispatch_semaphore_wait(signal, DISPATCH_TIME_FOREVER);
    NSLog(@"wait 1");
    NSLog(@"3_x:%ld",x);
 
    x = dispatch_semaphore_wait(signal, DISPATCH_TIME_FOREVER);
    NSLog(@"wait 2");
    NSLog(@"4_x:%ld",x);
 
    x = dispatch_semaphore_wait(signal, DISPATCH_TIME_FOREVER);
    NSLog(@"wait 3");
    NSLog(@"5_x:%ld",x);

結(jié)果：

0_x:0
wait 1
3_x:0
waiting
1_x:1
wait 2
4_x:0
waking
2_x:1
wait 3
5_x:0

最后會出現(xiàn)一個Thread 1: EXC_BAD_INSTRUCTION (code=EXC_I386_INVOP, subcode=0x0)的錯誤

首先看一下信號量的幾個函數(shù)：

dispatch_semaphore_create(M)
創(chuàng)建一個值為M的信號量

dispatch_semaphore_wait(信號量饭玲，等待時間)
如果該信號量的值大于0羊壹，則使其信號量的值-1蓖宦，否則，阻塞線程直到該信號量的值大于0或者達到等待時間油猫。

dispatch_semaphore_signal(信號量)
釋放信號量稠茂，使得該信號量的值加1

所以上面的結(jié)果不難推斷出，但是為什么會有最后的錯誤呢情妖？

在蘋果的開發(fā)文檔上有這么一段話

Calls to dispatch_semaphore_signal must be balanced with calls to wait(). Attempting to dispose of a semaphore with a count lower than value causes an EXC_BAD_INSTRUCTION exception.

對dispatch_semaphore_signal的調(diào)用必須與wait()的調(diào)用相匹配睬关。試圖銷毀一個value低于初始化value值的信號將會導致一個EXC_BAD_INSTRUCTION錯誤。