Question

首先，抱歉这个简单的问题。但我需要了解正在发生的事情。

我认为输出应为 upper case string 。但它是 UPPER CASE STRING 。

- (void)test
{
     NSString *stringVar = @"UPPER CASE STRING";
     [self changeString:stringVar];
     NSLog(@"value after changed : %@", stringVar);
}

- (void)changeString:(NSString*)string
{
     string = [string lowercaseString];
}

发生了什么以及如何解决？

Answer 1

[string lowercaseString]调用会创建一个新 NSString对象，并将其分配给局部变量string。这不会更改stringVar函数之外的changeString的值。指针本身按值传递。

执行所需操作的一种方法是将指针传递给指针：

-(void) test
{
     NSString *stringVar = @"UPPER CASE STRING";
     [self changeString:&stringVar];
     NSLog(@"value after changed : %@", stringVar);
}

-(void) changeString:(NSString**)string
{
     *string = [*string lowercaseString];
}

Answer 2

如果您查看对[NSString lowerCaseString] method的引用，您会看到它返回一个新字符串，并带有小写字符：

返回接收器的小写表示。
- (NSString *)lowercaseString

您的代码所做的只是使用lowercaseString调用的输出覆盖对输入字符串的引用，这无效。解决此问题的最佳方法是让您自己返回值，这样可以更容易理解该方法：

-(void) test
{
     NSString *stringVar = @"UPPER CASE STRING";
     stringVar = [self changeString:stringVar];
     NSLog(@"value after changed : %@", stringVar);
}

-(NSString *) changeString:(NSString*)string
{
     string = [string lowercaseString];
     return string;
}

您需要了解NSString是不可变的，因此除了重新分配引用之外，没有办法更改字符串的内容。但是，您可以使用NSMutableString代替，可以对其进行修改。

Answer 3

我指的是上面给出的问题，并帮助你解决错误。找到我的意见

- (void)test
{
     NSString *stringVar = @"UPPER CASE STRING";
     //StringVar is a pointer to integer class.let us assume the address the stringVar be 0x50 and the value it has be 0x100
     //So 0x100 has the string 

     [self changeString:stringVar];
     //above we are calling the method to lowercase and argument is stringVar
     //As the method is called we pass 0x100 in the argument as this is the memory where the value is kept

     NSLog(@"value after changed : %@", stringVar);
    //Our StringVar still points to 0x100 where the value is in upperString
}

- (void)changeString:(NSString*)string
{
     string = [string lowercaseString];
    // Now operation performed on string returns a new value on suppose location 0x200
   //String parameter passed in argument is assigned the new value.But we never pass values as we pass only location under the hood
   //New parameter passed now points to new memory location 0x200

}

---------------------------------------------------------------------------------
With the new solution
-(void) test
{
     NSString *stringVar = @"UPPER CASE STRING";
     //let 0x50 be stringVar memorylocation pointing to 0x100 with above value
     [self changeString:&stringVar];
     //0x50 is passed in argument
     NSLog(@"value after changed : %@", stringVar);
     //On new location stored in address of stringVar it points to new string value
}

-(void) changeString:(NSString**)string
{
     *string = [*string lowercaseString];
     //0x200 is new memory location received.*string gives the 0x100 location and hence the value
   //LHS is assigned to new location .On LHS you find *string which will be assigned 0x200
   //As string with location 0x50 is value 0x200 which will make it to point new      location where lowercase string exist
}

Answer 4

<{1}}中的

字符串是局部变量，修改方法返回值：

-(void) changeString:(NSString*)string

Answer 5

string是一个局部变量（一个指向NSString的指针，它是不可变的），你只是改变了string指向本地函数的内容，但是当你返回它的值时会被抛出程。

您可能想要做的是简单地将字符串作为参数传递，并从函数中返回小写字符串。

Objective C - 按值传递并通过引用传递

5 个答案: