CAKeyframeAnimation *animation = [CAKeyframeAnimation animationWithKeyPath:@"position"];
animation.fillMode = kCAFillModeForwards;
animation.removedOnCompletion = NO;
animation.duration = 1.0;
animation.path = path.CGPath;
[view.layer addAnimation:animation forKey:@"myPathAnimation"];
#include<stdio.h>
int main()
{
int i = 5;
float a = 3.14;
char*ii,*aa;
ii = (char*)&i;
aa = (char*)&a;
printf("address contained in ii=%u\n",ii);
printf("address contained in aa=%u\n",aa);
printf("value at address contained in ii=%d\n",*ii);
printf("value at address contained in aa= %d\n",*aa);
return(0);
}
/*the output for the following program was
address contained in ii=65524
address contained in aa=65520
value at address contained in ii=5
value at address contained in aa=-61*/
和ii = (char*) &i的含义是什么？为什么编译器在aa = (char*) &a的地址打印值是正确的，而ii中的值是错误的？如果您使用aa的任何其他值，例如327，则i中包含的地址值将变为其他值。有人可以向我解释一下吗？我无法从编写代码的书中得到适当的解释。
程序源：了解c中的指针。作者：yashvant kanetkar，第4版。
答案 0 :(得分：1)
The program as given in the book is an example of what not to do:
Don't try to access a value through a pointer when the pointer is not a pointer to the same type as the original value.
In the book, i is an int, but the code deliberately makes a char* point to the integer. So, when ii is dereferenced you just get a value that seems to be a char.
edit: So, the expression (char*)&i; first takes the address of i, which the compiler thinks of as type int* and then converts the type to char* without changing the value of the pointer.
The important thing to note about the different types of pointers is that they often imply a different size of object pointed to.
Also note that the output of all the whole program will depend on the platform you are using.