obj-c 中的参数通过引用传递

发布于 2024-11-04 08:34:33 字数 402 浏览 1 评论 0原文

我刚刚做了一项研究，但我遇到了一个问题

- (BOOL)application:(UIApplication *)application didFinishLaunchingWithOptions:(NSDictionary *)launchOptions
{
  int r1=40,r2=5;
  swap(r1,r2);

  NSLog(@" temp is %d",r1);
}

void swap(int r1, int r2)
{
  NSLog(@" m i 1st");
  int temp;
  temp=r1;
  r1=r2;
  r2=temp;

  NSLog(@" temp is %d",r1);
}

，我遇到了 swap 类型冲突；这是正确的做法吗？谢谢！

原文

I just did one research, but I am having a problem

- (BOOL)application:(UIApplication *)application didFinishLaunchingWithOptions:(NSDictionary *)launchOptions
{
  int r1=40,r2=5;
  swap(r1,r2);

  NSLog(@" temp is %d",r1);
}

void swap(int r1, int r2)
{
  NSLog(@" m i 1st");
  int temp;
  temp=r1;
  r1=r2;
  r2=temp;

  NSLog(@" temp is %d",r1);
}

I am getting conflicting type of swap; is this correct way of doing this? Thanks!

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半衾梦 2024-11-11 08:34:33

如果您希望交换 r1 和 r2，则必须传递指针或使用 C++ 引用。请注意，使用 C++ 引用将要求您深入研究 Objective-C++，在本例中这意味着将文件命名为 .mm 而不是 .m。

void swap_with_pointers(int *r1, int *r2)
{
  int temp;
  temp = *r1;
  *r1 = *r2;
  *r2 = temp;
}

void swap_with_references(int &r1, int &r2)
{
  int temp;
  temp = r1;
  r1 = r2
  r2 = temp;
}

然后，使用您的实现之一，如下所示：

- (BOOL)application:(UIApplication *)application didFinishLaunchingWithOptions:(NSDictionary *)launchOptions
{
  int x = 3;
  int y = 4;
  swap_with_pointer(&x, &y); // swap_with_references(x,y);
  printf("x = %d, y = %d", x, y);
  return 0;
}

输出，无论哪种方式：

x = 4，y = 3

If you want r1 and r2 to be swapped, you either have to pass pointers, or use C++ references. Note that using C++ references will require you to dive into Objective-C++, which in this case means naming your file .mm instead of .m.

void swap_with_pointers(int *r1, int *r2)
{
  int temp;
  temp = *r1;
  *r1 = *r2;
  *r2 = temp;
}

void swap_with_references(int &r1, int &r2)
{
  int temp;
  temp = r1;
  r1 = r2
  r2 = temp;
}

Then, use one of your implementations like so:

- (BOOL)application:(UIApplication *)application didFinishLaunchingWithOptions:(NSDictionary *)launchOptions
{
  int x = 3;
  int y = 4;
  swap_with_pointer(&x, &y); // swap_with_references(x,y);
  printf("x = %d, y = %d", x, y);
  return 0;
}

The output, either way: