Question

如何将一个变量a替换为另一个变量b，以便更改b？

例如：

NSString *a = @"a";
NSString *b = @"b";
NSString *c = @"c";

a = b;
a = c;

在这种情况下，b的值为@"b"，对吗？我想在不使用b的情况下设定@"c" b = c的值。可能我应该尝试理解指针。

请理解我糟糕的解释，并给我任何建议。

Answer 1

你可能会感到困惑，因为坦率地说，指针起初有点令人困惑。它们是保存内存位置的变量。如果有两个指针保持相同的位置，并且使用一个指针更改该位置的内容，则可以通过另一个指针查看这些新内容。但它仍指向同一地点。

int x = 10;

// xp is a pointer that has the address of x, whose contents are 10
int * xp = &x;
// yp is a pointer which holds the same address as xp
int * yp = xp;

// *yp, or "contents of the memory address yp holds", is 10
NSLog(@"%i", *yp);

// contents of the memory at x are now 62
x = 62;

// *yp, or "contents of the memory address yp holds", is now 62
NSLog(@"%i", *yp);
// But the address that yp holds has _not_ changed.

根据您的评论，是的，您可以这样做：

int x = 10;
int y = 62; 

// Put the address of x into a pointer
int * xp = &x;
// Change the value stored at that address
*xp = y;

// Value of x is 62
NSLog(@"%i", x);

你可以用NSString做同样的事情，虽然我想不出这样做的好理由。将示例中的任何int更改为NSString *; int *变为NSString **。根据需要更改分配和NSLog()格式说明符：

NSString * b = @"b";
NSString * c = @"c";

// Put the address of b into a pointer
NSString ** pb = &b;
// Change the value stored at that address
*pb = c;
// N.B. that this creates a memory leak unless the previous
// value at b is properly released.

// Value at b is @"c"
NSLog(@"%@", b);

通过另一个变量更改变量的值

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