指向int数组，从另一个方法传递和使用它 [英] Pointer to int array, passing and using it from another method

问题描述

我没有巩固我对C ++数组的学习，并忘记了如何正确地这样做。我已经使用char数组前面，但它不能正常工作为int数组。

我声明一个新的空白数组：

  int myIntArray [ 10]; 
  

那么这应该是一个空数组，暂时是正确的？

然后我为这个数组指定一个指针：

  int * pMyArray = myIntArray 
  
 
 
 
希望正确的。
 
 
 
另一种方法在其他地方：
  anotherMethod（pMyArray）
   
 
其中我想将此指针分配给一个局部变量（这是我真的不知道的）：
  anotherMethod（int * pMyArray）{
 int myLocalArray []; 
 myLocalArray [0] = * pMyArray; 
} 
  
 
 
我没有收到任何编译错误，但我不确定这是就在几条战线上。 
 
 
 
编辑：
 
 
 
我应该说我在做什么
 
 
 
真的很简单，我只想从另一个方法修改一个局部数组。 
 
 
 
所以我有：
 
 
 
方法1会包含：
 
 $ b b 
 int myArray1 [10] = {0}; 
 
 
 
方法2将传递指向myArray的指针：
 
 $ b b 
然后一些代码修改数组中的变量myArray。 
解决方案
 
  int myIntArray [10]; 
  
这是一个未初始化的数组。它不一定包含 0 。
  int * pMyArray = myIntArray 
  
好的， pMyArray   myIntArray 中的第一个元素。
  anotherMethod（int * pMyArray）{
 int myLocalArray [10]; 
 myLocalArray [0] = * pMyArray; 
} 
  
这不会将指针指定为任何东西，由 pMyArray 指向的 int 的本地数组，其中，记住，未初始化。我添加了 10 ，因为在C ++中不能有一个未知大小的数组。
 
 
 
  pMyArray 指向，您需要通过引用传递它：
  anotherMethod （int *& pMyArray）
  
此外，如果将其分配给自动存储中的某些值将导致未定义的行为，因为该函数退出时该内存不再有效。
 
I haven't cemented my learning of C++ arrays and have forgotten how to do this properly. I've done it with char array before but its not working as well for int array. 

I declare a new blank int array:
int myIntArray[10];
So this should be an array of nulls for the moment correct?

Then I assign a pointer to this array:
int *pMyArray = myIntArray
Hopefully thats correct to there.

Then I pass this to another method elsewhere:
anotherMethod(pMyArray)
where I want to assign this pointer to a local variable (this is where I'm really not sure):
anotherMethod(int *pMyArray){    
    int myLocalArray[];    
    myLocalArray[0] = *pMyArray;    
}
I'm not getting any compilation errors but I'm not sure this is right on a few fronts. Any and all help and advice appreciated with this.

Edit:

I should have said what I am trying to do.

Very simple really, I'd just like to modify a local array from another method. 

So I have:

Method 1 would contain: 

int myArray1[10] = {0};

Method 2 would be passed the pointer to myArray:

Then some code to modify the variables in the array myArray. 
解决方案
int myIntArray[10];
This is an uninitialized array. It doesn't necessarily contain 0's.
int *pMyArray = myIntArray
Okay, pMyArray points to the first element in myIntArray.
anotherMethod(int *pMyArray){    
    int myLocalArray[10];    
    myLocalArray[0] = *pMyArray;    
}
This doesn't assign your pointer to anything, it assigns the first value of the local array to the int pointed to by pMyArray, which, remember, was uninitialized. I added the 10 there because you can't have an array of unknown size in C++.

To modify what pMyArray points to, you need to pass it by reference:
anotherMethod(int *& pMyArray)
Also, if you assign it to some values in automatic storage, it will result in undefined behavior, as that memory is no longer valid when the function exits.

                        
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