不。
图像如果答案是肯定的,在哪里NSInvocation
聪明到可以复制块,它应该做这样的事情:
for (/*every arguments*/) {
if (/*arg is object. i.e. @encode(arg) is '@'*/) {
if ([arg isKindOfClss:[NSBlock class]]) {
arg = [arg copy]; // copy block
} else {
[arg retain];
}
}
}
问题是arg
在复制块时被修改,这不应该发生,因为这意味着 callretainArguments
可能会更改NSInvocation
.这将打破许多已经做出的假设。(即参数 get fromNSInvocation
应该与用于创建的参数相同NSInvocation
)
更新
刚刚做了测试以确认答案是否定的,但我之前的观点是不正确的......
@interface Test : NSObject
@end
@implementation Test
- (void)testMethodWithBlock:(void (^)(void))block obj:(id)obj cstr:(const char *)cstr {
NSLog(@"%p %p %p %@", block, obj, cstr, [block class]);
}
@end
@implementation testTests
- (void)test1 {
__block int dummy;
Test *t = [[Test alloc] init];
NSMethodSignature *ms = [t methodSignatureForSelector:@selector(testMethodWithBlock:obj:cstr:)];
NSInvocation *invocation = [NSInvocation invocationWithMethodSignature:ms];
void (^block)(void) = ^ {
dummy++; // stop this become global block
};
id obj = @"object";
char *cstr = malloc(5);
strcpy(cstr, "cstr");
NSLog(@"%@", [ms debugDescription]);
NSLog(@"%p %p %p %@", block, obj, cstr, [block class]);
[invocation setSelector:@selector(testMethodWithBlock:obj:cstr:)];
[invocation setArgument:&block atIndex:2];
[invocation setArgument:&obj atIndex:3];
[invocation setArgument:&cstr atIndex:4];
[invocation invokeWithTarget:t];
[invocation retainArguments];
[invocation invokeWithTarget:t];
free(cstr);
}
@end
输出,ARC 禁用(并崩溃):
2013-04-18 19:49:27.616 test[94555:c07] 0xbfffe120 0x70d2254 0x7167980 __NSStackBlock__
2013-04-18 19:49:27.617 test[94555:c07] 0xbfffe120 0x70d2254 0x7167980 __NSStackBlock__
2013-04-18 19:49:27.618 test[94555:c07] 0xbfffe120 0x70d2254 0x736a810 __NSStackBlock__
启用 ARC:
2013-04-18 19:51:03.979 test[95323:c07] 0x7101e10 0x70d2268 0x7101aa0 __NSMallocBlock__
2013-04-18 19:51:03.979 test[95323:c07] 0x7101e10 0x70d2268 0x7101aa0 __NSMallocBlock__
2013-04-18 19:51:03.980 test[95323:c07] 0x7101e10 0x70d2268 0xe0c1310 __NSMallocBlock__
如您所见,c 字符串被复制retainArguments
但不是块。但是启用 ARC 后,问题应该会消失,因为 ARC 在某个时候为您复制了它。