CODEWITHSUNDEEP
cpointers
samprat
samprat

Reputation: 2214

using pointers to display content of array

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.

Below is the sample program

 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;     
}

Upvotes: 6

Views: 58944

Answers (5)

Praetorian
Praetorian

Reputation: 109169

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

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!='\0'; ++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;     
}

Upvotes: 5

Surjit Joshi
Surjit Joshi

Reputation: 3515

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

 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;     
}

Upvotes: 0

unkulunkulu
unkulunkulu

Reputation: 11922

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

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

    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;     
}

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

    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;     
}

Upvotes: 0

NPE
NPE

Reputation: 500495

You have several options:

  • 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.

Upvotes: 3

Anders Abel
Anders Abel

Reputation: 69270

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:

  • 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.

This would give the following for statements:

//Marker value.
for(;*ptr != -1; ++ptr)
    printf("%d", *ptr);  

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

Upvotes: 1

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