Passing Arrays into an Array of Pointers in C

Question

Please assist with below question on pointers to arrays. I have 20 arrays that are each 350 elements long. I need to pass the address of a select 3 out of the 20 arrays, to an array of pointers. Then later in my code I need to access individual elements within the arrays, within the array of pointers. However I am unsure as to the syntax, please comment as to whether the below is correct.

unsigned short      Graph1[350];
unsigned short      Graph2[350];
unsigned short      Graph3[350];
...       ...          ...
unsigned short      Graph19 [350];
unsigned short      Graph20 [350];
unsigned short      *ptr_Array[3];
...
*ptr_Array[0] = &Graph6;    // Passing the address of array Graph6, into the array of pointers.
*ptr_Array[1] = &Graph11;   // Passing the address of array Graph11, into the array of pointers.
*ptr_Array[2] = &Graph14;   // Passing the address of array Graph14, into the array of pointers.
...
Varriable1 = *ptr_Array[1]+55   // Trying to pass the 55th element of Graph11 into Varriable1.

Not an answer: 20 arrays that are each 350 elements why don't you use 2-d array? — Mohit Jain
– Mohit Jain, Commented Apr 21, 2016 at 4:52
@Rolice No array is not a pointer. It depletes to a pointer if used in expression or passed as argument to function. — Mohit Jain
– Mohit Jain, Commented Apr 21, 2016 at 5:02
Ah, thanks @MohitJain. I saw dereferencing and appeared strage to me. — Rolice
– Rolice, Commented Apr 21, 2016 at 5:03

Some programmer dude · Accepted Answer · 2016-04-21 04:57:02Z

2

The expression *ptr_Array[1]+55 is wrong multiple times, because of operator precedence.

The compiler sees it as (*(ptr_Array[1]))+55, that is it takes the second pointer in ptr_Array and dereferences it to get the first value, and add 55 to that value, which is not what you want. You need to explicitly use parentheses like *(ptr_Array[1]+55). Or simply ptr_Array[1][55].

And you should really consider the comment by Mohit Jain. Instead of having 20 different variables, just use one:

unsigned short Graph[20][350];

answered Apr 21, 2016 at 4:57

Some programmer dude

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Comments

Mohit Jain · Accepted Answer · 2016-04-21 06:30:03Z

2

*ptr_Array[0] = &Graph6; is wrong. It should be:

ptr_Array[0] = Graph6; /* or &Graph6[0] */

Type of ptr_Array is array 3 of pointer to unsigned short. ptr_Array[0] has type of pointer to unsigned short and *ptr_Array has type unsigned short.

Type of Graph6 is array 350 of unsigned short which will deplete to pointer to unsigned short if used in an expression.

Varriable1 = *ptr_Array[1]+55 is also wrong. To pass the 55^th element, use

Varriable1 = ptr_Array[1][55];

edited Apr 21, 2016 at 6:30

answered Apr 21, 2016 at 4:53

Mohit Jain

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2 Comments

Jonathan Leffler Over a year ago

One reason &Graph6 is wrong is that the type of &Graph6 is unsigned short (*)[350] (pointer to an array of 350 unsigned short), whereas each element of ptr_Array is meant to be an unsigned short *, and the two types are visibly quite different — which is why the compiler is complaining about pointer type mismatches.

Rishikesh Raje Over a year ago

Varriable1 = *ptr_Array[1][55]; should be Varriable1 = ptr_Array[1][55];

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