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Interpretation of int (*a)[3]

Asked 2010-02-12T08:12:28.647
26

When working with arrays and pointers in C, one quickly discovers that they are by no means equivalent although it might seem so at a first glance. I know about the differences in L-values and R-values. Still, recently I tried to find out the type of a pointer that I could use in conjunction with a two-dimensional array, i.e.

int foo[2][3];
int (*a)[3] = foo;

However, I just can't find out how the compiler "understands" the type definition of a in spite of the regular operator precedence rules for * and []. If instead I were to use a typedef, the problem becomes significantly simpler:

int foo[2][3];
typedef int my_t[3];
my_t *a = foo;

At the bottom line, can someone answer me the questions as to how the term int (*a)[3] is read by the compiler? int a[3] is simple, int *a[3] is simple as well. But then, why is it not int *(a[3])?

EDIT: Of course, instead of "typecast" I meant "typedef" (it was just a typo).

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2 Answers

23

First, you mean "typedef" not "typecast" in your question.

In C, a pointer to type T can point to an object of type T:

int *pi;
int i;
pi = &i;

The above is simple to understand. Now, let's make it a bit more complex. You seem to know the difference between arrays and pointers (i.e., you know that arrays are not pointers, they behave like them sometimes though). So, you should be able to understand:

int a[3];
int *pa = a;

But for completeness' sake: in the assignment, the name a is equivalent to &a[0], i.e., a pointer to the first element of the array a. If you are not sure about how and why this works, there are many answers explaining exactly when the name of an array "decays" to a pointer and when it does not:

I am sure there are many more such questions and answers on SO, I just mentioned some that I found from a search.

Back to the topic: when we have:

int foo[2][3];

foo is of type "array [2] of array [3] of int". This means that foo[0] is an array of 3 ints, and foo[1] is an array of 3 ints.

Now let's say we want to declare a pointer, and we want to assign that to foo[0]. Th

answered 2010-02-12T08:27:37.897
2

In situations when you don't know what is declaration for, I found very helpful so called Right-left rule.

BTW. Once you've learned it, such declarations are peace of cake:)

answered 2010-04-15T13:21:40.500

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