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In C we have two dimensional arrays, i.e. a[m][n].

In one dimensional arrays a is a pointer to the start of the array.

What about two dimensional arrays? Does a[i] hold a pointer to the start of the i row in an array? And thus a[i] is an array of pointers that is passed to a function in the following matter function(int **a, m, n)?

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  • "In one dimensional arrays a is a pointer to the start of the array." No, it isn't. Commented Feb 16, 2015 at 11:27
  • a[m][n] is an array of arrays. Commented Feb 16, 2015 at 11:27
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    @MichałSzydłowski: No, a is an array in all cases the OP mentions. "Arrays are not Pointers!" Commented Feb 16, 2015 at 11:29
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    @Dean: If an array is used without an index it decays to a pointer to its 1st element. Commented Feb 16, 2015 at 11:36
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    Related: stackoverflow.com/q/12674094/694576 Commented Feb 16, 2015 at 11:54

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Does a[i] hold a pointer to the start of the i row in an array?

No. The data of a 2D array in C is a contiguous block of elements plus some clever indexing access. But a 2D array is an array of arrays, not an array of pointers.

Formally, the a[i] holds a 1D array. This may decay to a pointer to the first element of the ith row in certain contexts, but its type is still T[n], for some type T that you have not specified.

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7 Comments

1D array of pointers, each points to the first element of the 'second level' arrays.
@Dean Yes, what about it? What on earth are you trying to say? A black dog on a hill.
I wanted to get is straight what each array contains, it's perfectly clear from the answer now, thanks.
@Dean Each array contains i number of items of the given type, no more, no less. There are no pointers anywhere. (Unless of course you declare an array of "pointers-to-type".)
Unfortunately during a C course they presented a multidimensional array as an array of pointers, which point to the first elements of array, hence the confusion.
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In one dimensional arrays a is a pointer to the start of the array.

Not correct. a is an array. When you use a in an expression, it "decays" into a pointer to the first element. To better understand this, read this chapter of the C FAQ, particularly this one.

What about two dimensional arrays? Does a[i] hold a pointer to the start of the i row in an array?

No. In a 2D array, a[i] is an array, while int a[x][y]; is an array of arrays. There are no pointers anywhere.

You might be confused because C allows this syntax: int a[][N] = ...;, but that syntax merely means that the size of the array of arrays depends on the number of items in the initialization list.

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"When you use a in an expression". Surely it depends on the expression. For example, if the expression is the RHS of an assignment to a pointer to array, then it doesn't decay.
@juanchopanza Well, yes, but it doesn't seem very helpful to drag all the details of that into this answer, given that the OP is a beginner.
Shouldn't a[x][y] be an element, not an array of arrays?
@rcgldr Aah well yeah good point. It depends on the context. int a[x][y]; is an array of arrays, while a[x][y] = 0; indeed refers to one item of the array. Edited my answer for clarification

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