Why is the syntax "int (*)[*]" necessary in C?

Question

Just was looking something up in the ISO/IEC9899 When I stumbled on this:

6.7.6 Type names

[...]

Semantics

2 In several contexts, it is necessary to specify a type. This is accomplished using a type name, which is syntactically a declaration for a function or an object of that type that omits the identifier.128) 3 EXAMPLE The constructions

(a) int
(b) int *
(c) int *[3]
(d) int (*)[3]
(e) int (*)[*]
(f) int *()
(g) int (*)(void)
(h) int (*const [])(unsigned int, ...)

name respectively the types (a) int, (b) pointer to int, (c) array of three pointers to int, (d) pointer to an array of three ints, (e) pointer to a variable length array of an unspecified number of ints, (f) function with no parameter specification returning a pointer to int, (g) pointer to function with no parameters returning an int, and (h) array of an unspecified number of constant pointers to functions, each with one parameter that has type unsigned int and an unspecified number of other parameters, returning an int.

What most confused me was:

(e) pointer to a variable length array of an unspecified number of ints

The others I can understand more or less. But what is the use of a pointer to a VLA of unspecified number of 'ints'?

And is there even a need for compiler's to support the syntax of

int foo[*];

?

EDIT for clarification

This Question primaly aims on "Is it even neccessary to support this for a compiler?". Whilest this post ANSI-C grammar - array declarations like [*] et alii clearly improved my knowledge. There is still no answer for: Why does the compiler need to know if the parameter of the prototype just is a address containing unknown size. as with simply doing int foo[] or it will be unspecified size?

So is this realy neccessary to be supported? And if not so, why the standard even is implementing this semantic?

Answer 1

Why does the compiler need to know if the parameter of the prototype just is a address containing unknown size. as with simply doing int foo[] or it will be unspecified size?

The compiler doesn't need to "know" anything, it's a tool.

The difference between int (*)[*] and int[] is about the same as between int (*)[5] and int[]. If you agree that the latter pair is not interchangeable, then the former isn't either.

In pre-C99, the way to specify an array of unknown number of T elements is T[]. This is an incomplete type, which means you cannot have an array of T[]. There is no T[][]. Inside a function declarator, T[] means the same as T*. OTOH T[*] is a variable-length array, which is different from an array of unknown number of elements. You can have an array of variable-size arrays, i.e. there is T[*][*]. The syntax you are asking about is necessary to support this variable-size-array type. Luckily you are not asking why we need different types, because the answer would be really long-winded, but here's my stab at it.

The purpose of types is two-fold. First, types are needed for object code generation (things like a++ typically generate different object code, depending on the type of a). Second, types are needed for type-checking (things like a++ may be allowed or not depending on the type of a).

The [*] types are only allowed in function declarators that are not parts of function definitions. So code generation and is not relevant here. This leaves us with type checking. Indeed,

int foo(int, int (*)[*]);
int bar(int, int (*)[5]);

int main ()
{
    int a;
    int aa[5];
    int aaa[5][5];

    foo(1, &a);    // incorrect, `&a` is `int*`, `int*` and `int (*)[*]` are different
    bar(1, &a);    // incorrect, `&a` is `int*`, `int*` and `int (*)[5]` are different
    foo(5, aa);    // incorrect, `aa` is `int*` (!), `int*` and `int (*)[*]` are different
    bar(5, aa);    // incorrect, `aa` is `int*` (!), `int*` and `int (*)[5]` are different
    foo(5, &aa);   // correct
    bar(5, &aa);   // correct
    foo(5, aaa);   // correct
    bar(5, aaa);   // correct
}

If we are agree on which calls to bar are correct and which are not, we must agree also on calls to foo.

The only remaining question is, why int foo(int m, int (*)[m]); is not enough for this purpose? It probably is, but the C language does not force the programmer to name formal parameters in function declarators where parameter names are not needed. [*] allows this small freedom in case of VLAs.

Answer 2

I am going to answer your question strictly as asked:

This Question primaly aims on "Is it even neccessary to support this for a compiler?"

For a C99 compiler, yes: it is part of the standard so a C99-conforming compiler must support it. The question of what int foo[*]; is useful for is quite orthogonal to the question of whether it must be supported. All compilers claiming to conform to C99 that I tested supported it (but I am not sure what it is useful for, either).

---

For a C11 compiler, good news! Variable-Length Arrays have been made a “conditional feature”. You can implement a C11-compliant without Variable-Length Arrays as long as it defines __STDC_NO_VLA__:

6.10.8.3 Conditional feature macros

…

__STDC_NO_VLA__ The integer constant 1, intended to indicate that the implementation does not support variable length arrays or variably modified types.

Answer 3

If I pass an array with more than one dimension to a function, and if the function parameters used to express the number of elements in a given dimension of the array come after the array parameter itself, the [*] syntax may be used. In the case of an array with more than two dimensions, and if the array parameter, again, precedes the element count parameters, this syntax must be used, as array decay only ever occurs once. After all, you can't very well use int (*)[][] or int [][][] because the standard requires that in int [A][B] and int [A][B][C][D], only A may be omitted due to the array decaying to a pointer. If you use pointer notation in the function parameter, you're allowed to use int (*)[], but this makes very little sense to me, especially since:

• sizeof ptr[0] and sizeof *ptr are both illegal -- how should the compiler determine the size of an array that has an indeterminate element count? Instead you must find it at runtime using N * sizeof **ptr or sizeof(int (*)[N]). This also means that any arithmetic operations on ptr, such as the usage of ++ptr, are illegal since they rely upon the size information, which cannot be calculated. Type casts may be used to get around this, but it is easier just to use a local variable with the proper type information. Then again, why not just use the [*] syntax and include the proper type information from the start?
• sizeof ptr[0][0] is illegal, but sizeof (*ptr)[0] is not -- array indexing is still performed even when simply getting info like size, so it is like writing sizeof (*(ptr + 0))[0], which is illegal because you cannot apply arithmetic operations to an incomplete type as previously mentioned.
• Someone who has never encountered this issue before might think [] can be replaced by *, yielding int ** instead of int (*)[], which is incorrect because that sub-array has not decayed. Array decay only occurs once.

I noted that the [*] syntax is unnecessary if the parameters used as element counts came first, which is true, but when is the last time anybody saw any of the following?

void foo (int a, int b, int c, int arr[a][b][c]);

void bar (int a, int b, int c, int arr[][b][c]);

void baz (int a, int b, int c, int (*arr)[b][c]);

So to answer your question:

• if a function is able to operate upon multidimensional arrays of various lengths (or a pointer to a 1-D array),

and

• the parameters denoting element count are listed after the array parameter itself,

the [*] syntax may be required. I actually encourage usage of [*] since [] comes with problems when size information is required.

Answer 4

In C99 it is possible to declare arrays using variable dimensions, providing the variable has a ( positive integer ) value at the time the declaration is made. It turns out that this carries over to the declaration of arrays in function parameters as well, which can be particularly useful for multi-dimensional arrays.
For example, in the prototype:

int arrayFunction( int nRows, int nCols, double x[ nRows ], 
                   double y[ nRows ][ nCols ] );

the variable dimension on x is informative to the human but not necessary for the computer, since we could have declared it as x[ ]. However the nCols dimension on y is very useful, because otherwise the function would have to be written for arrays with pre-determined row sizes, and now we can write a function that will work for arrays with any row length.

For two array types to be compatible, both must have compatible element types, and if both size specifiers are present and are integer constant expressions, then both sizes must have the same value. A VLA is always compatible with another array type if they both have the same element type. If the two array types are used in a context that requires them to be compatible, it is undefined behavior if the dimension sizes are unequal at run time

Answer 5

It might be useful if you want to work with "jagged" arrays, when the size of the rows of that matrix, while unknown at compile-time, will be initialized in run-time and the remain the same during the whole execution time. But to make sure you will stay in bounds for each row, you will have to store actual sizes of that array somehow, separately for each row if want it "jagged", because sizeof operator will not work properly for run-time initialized arrays (it will return the size of the pointer at best, since it's a compile-time operator).