6

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;     
}
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1
  • @Everyone, Thank you guys for clearing my doubts.. Commented Jul 25, 2011 at 10:33

5 Answers 5

Reset to default
5

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;     
}
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4 Comments

does prefix or postfix in for loop will make any difference.
No pre-incrementing vs post-incrementing shouldn't make any difference in this case. I'm used to writing the former since the latter involves making a temporary copy of the object being incremented. This can be significant in case of C++ and custom classes that define operator++. But here it should be identical in either case.
Last time I did C was several months ago. If I have a dynamically allocated integer array, I'd have to store the size I passed to malloc in order to make it through the loop without wandering past the edge of that block of memory, right?
@Ungeheuer Yes, there's no (portable) way to retrieve the size of allocated array if you only have the pointer returned by malloc.
3

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.
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7 Comments

something like void(int *ptr, int size){ for int i =0; i<size; ptr++) printf("%d", *(ptr+i));
just for curiousity , is there is any way that pointers can determine size of array in for loop may be using pointers to pointers etc. This is only my wild guess
@samprat: Yes, something like this (but don't forget to also increment i).
@axis, yeah i almost forget to increment i.. Thanks
and *(ptr+i) is equivalent to ptr[i] so it's losing the point of iterating using pointer.
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1

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));
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Comments

0

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;     
}
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1 Comment

Your sample solution not only solved my main problem but also helped me to clear few more doubts. THANKS
0

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;     
}
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