C语言 使用指针显示数组内容

using pointers to display content of array

提问by samprat

I am stuck about how to use pointers to display array. I can easily do this with array using for loop but I am interested in knowing how to use via pointers and I am stuck how to calculate starting and ending point of an array.

我对如何使用指针显示数组感到困惑。我可以使用 for 循环轻松地使用数组执行此操作，但我有兴趣知道如何通过指针使用，并且我被困在如何计算数组的起点和终点。

Below is the sample program

下面是示例程序

 void printArray(int *ptr);            
{         
    //for statement to print values using array             
    for( ptr!=NULL; ptr++) // i know this doesn't work         
    printf("%d", *ptr);        
}         

int main()    
{    
    int array[6] = {2,4,6,8,10};     
    printArray(array);    
    return 0;     
}

采纳答案by Praetorian

The checking for NULL trick only works for NULL terminated strings. For a numeric array you'll have to pass in the size too.

检查 NULL 技巧仅适用于 NULL 终止的字符串。对于数字数组，您也必须传入大小。

void printArray(int *ptr, size_t length);            
{         
    //for statement to print values using array             
    size_t i = 0;
    for( ; i < length; ++i )      
    printf("%d", ptr[i]);        
}   

 void printString(const char *ptr);            
{         
    //for statement to print values using array             
    for( ; *ptr!='
//Marker value.
for(;*ptr != -1; ++ptr)
    printf("%d", *ptr);  

//Length parameter
for(int i = 0; i < length; ++i)
    printf("%d", *(ptr+i));
'; ++ptr)        
    printf("%c", *ptr);        
}         

int main()    
{    
    int array[6] = {2,4,6,8,10};     
    const char* str = "Hello World!";
    printArray(array, 6);    
    printString(str);
    return 0;     
}

回答by NPE

You have several options:

您有多种选择：

• You could pass the size of your array into the function.
• You could have a special "sentinel" value (e.g. -1) as the last element of your array; if you do this, you must ensure that this value cannot appear as part of the array proper.

• 您可以将数组的大小传递给函数。
• 你可以有一个特殊的“哨兵”值（例如-1）作为数组的最后一个元素；如果你这样做，你必须确保这个值不能作为数组的一部分出现。

回答by Anders Abel

When an array is passed as a parameter to a function, it is decayed into a pointer to the first element of the array, loosing the information about the length of the array. To handle the array in the receiving function (printArray) requires a way to know the length of the array. This can be done in two ways:

当数组作为参数传递给函数时，它会衰减为指向数组第一个元素的指针，从而丢失有关数组长度的信息。要在接收函数 ( printArray) 中处理数组，需要知道数组长度的方法。这可以通过两种方式完成：

• A special termination marker used for the last element. For strings this is NULL. In your example it could be -1, if that value will never occur in the real data.
• Passing a length parameter to printArray.

• 用于最后一个元素的特殊终止标记。对于字符串，这是NULL. 在您的示例中-1，如果该值永远不会出现在实际数据中，则可能是。
• 将长度参数传递给printArray.

This would give the following forstatements:

这将给出以下for语句：

void printArray(int *ptr, size_t size)
{
    int *const end = ptr + size;
    while( ptr < end ) {
        printf("%d", *ptr++);        
    }
}

int main()    
{    
    int array[6] = {2,4,6,8,10};     
    printArray(array, sizeof(array) / sizeof(array[0]) );    
    return 0;     
}

回答by unkulunkulu

The function needs to know the size of the array. There are two common approaches:

该函数需要知道数组的大小。有两种常见的方法：

1. Pass the actual size to the function, e.g.

   void printArray(int *ptr);            
   {         
       //for statment to print values using array             
       for( *ptr != 0; ptr++) // i know this doesn't work         
       printf("%d", *ptr);        
   }         
   
   int main()    
   {    
       int array[6] = {2,4,6,8,10, NULL};     
       printArray(array);    
       return 0;     
   }
   

2. Explicitly provide a sentinel 0 (or other appropriate) element for the array:

   void printArray(int *ptr, size_t size)
   {
       int *const end = ptr + size;
       while( ptr < end ) {
           printf("%d", *ptr++);        
       }
   }
   
   int main()    
   {    
       int array[6] = {2,4,6,8,10};     
       printArray(array, sizeof(array) / sizeof(array[0]) );    
       return 0;     
   }
   

1. 将实际大小传递给函数，例如

   void printArray(int *ptr);            
   {         
       //for statment to print values using array             
       for( *ptr != 0; ptr++) // i know this doesn't work         
       printf("%d", *ptr);        
   }         
   
   int main()    
   {    
       int array[6] = {2,4,6,8,10, NULL};     
       printArray(array);    
       return 0;     
   }
   

2. 为数组显式提供哨兵 0（或其他适当的）元素：

    void printArray(int arr[]);            
   {         
      int *ptr;
   
      for(ptr = &arr[0]; ptr <= &arr[5]; ptr++)
      {
          if(ptr != null)       
               printf("%d", *ptr);
   
         ptr++;   // incrementing pointer twice, as there are ‘int' values in array which
                  //are of size 2 bytes, so we need to increment it twice..
      }       
   }         
   
    int main()    
   {    
       int array[6] = {2,4,6,8,10};     
       printArray(array);    
       return 0;     
   }
   

回答by Surjit Joshi

Here is the answer buddy (Non-tested)...

这是朋友的答案（未测试）...

##代码##