Difference between revisions of "c/language/array"

Latest revision as of 16:08, 16 January 2024

Array is a type consisting of a contiguously allocated nonempty sequence of objects with a particular element type. The number of those objects (the array size) never changes during the array lifetime.

[edit] Syntax

In the declaration grammar of an array declaration, the type-specifier sequence designates the element type (which must be a complete object type), and the declarator has the form:

float fa[11], *afp[17]; // fa is an array of 11 floats
                        // afp is an array of 17 pointers to floats

[edit] Explanation

There are several variations of array types: arrays of known constant size, variable-length arrays, and arrays of unknown size.

[edit] Arrays of constant known size

If expression in an array declarator is an integer constant expression with a value greater than zero and the element type is a type with a known constant size (that is, elements are not VLA)(since C99), then the declarator declares an array of constant known size:

int n[10]; // integer constants are constant expressions
char o[sizeof(double)]; // sizeof is a constant expression
enum { MAX_SZ=100 };
int n[MAX_SZ]; // enum constants are constant expressions

Arrays of constant known size can use array initializers to provide their initial values:

int a[5] = {1,2,3}; // declares int[5] initialized to 1,2,3,0,0
char str[] = "abc"; // declares char[4] initialized to 'a','b','c','\0'

void fadd(double a[static 10], const double b[static 10])
{
    for (int i = 0; i < 10; i++)
    {
        if (a[i] < 0.0) return;
        a[i] += b[i];
    }
}
// a call to fadd may perform compile-time bounds checking
// and also permits optimizations such as prefetching 10 doubles
int main(void)
{
    double a[10] = {0}, b[20] = {0};
    fadd(a, b); // OK
    double x[5] = {0};
    fadd(x, b); // undefined behavior: array argument is too small
}

int f(const int a[20])
{
    // in this function, a has type const int* (pointer to const int)
}
int g(const int a[const 20])
{
    // in this function, a has type const int* const (const pointer to const int)
}

void fadd(double a[static restrict 10],
          const double b[static restrict 10])
{
    for (int i = 0; i < 10; i++) // loop can be unrolled and reordered
    {
        if (a[i] < 0.0)
            break;
        a[i] += b[i];
    }
}

#include <stdio.h>
 
int main(void)
{
   int n = 1;
label:;
   int a[n]; // re-allocated 10 times, each with a different size
   printf("The array has %zu elements\n", sizeof a / sizeof *a);
   if (n++ < 10)
       goto label; // leaving the scope of a VLA ends its lifetime
}

void foo(size_t x, int a[*]);
void foo(size_t x, int a[x])
{
    printf("%zu\n", sizeof a); // same as sizeof(int*)
}

extern int n;
int A[n];            // Error: file scope VLA
extern int (*p2)[n]; // Error: file scope VM
int B[100];          // OK: file-scope array of constant known size
void fvla(int m, int C[m][m]); // OK: prototype-scope VLA

void fvla(int m, int C[m][m]) // OK: block scope/auto duration pointer to VLA
{
    typedef int VLA[m][m]; // OK: block scope VLA
    int D[m];              // OK: block scope/auto duration VLA
//  static int E[m]; // Error: static duration VLA
//  extern int F[m]; // Error: VLA with linkage
    int (*s)[m];     // OK: block scope/auto duration VM
    s = malloc(m * sizeof(int)); // OK: s points to VLA in allocated storage
//  extern int (*r)[m]; // Error: VM with linkage
    static int (*q)[m] = &B; // OK: block scope/static duration VM}
}

struct tag
{
    int z[n]; // Error: VLA struct member
    int (*y)[n]; // Error: VM struct member
};

[edit] Arrays of unknown size

If expression in an array declarator is omitted, it declares an array of unknown size. Except in function parameter lists (where such arrays are transformed to pointers) and when an initializer is available, such type is an incomplete type (note that VLA of unspecified size, declared with * as the size, is a complete type)(since C99):

extern int x[]; // the type of x is "array of unknown bound of int"
int a[] = {1,2,3}; // the type of a is "array of 3 int"

struct s { int n; double d[]; }; // s.d is a flexible array member
struct s *s1 = malloc(sizeof (struct s) + (sizeof (double) * 8)); // as if d was double d[8]

[edit] Qualifiers

typedef int A[2][3];
const A a = {{4, 5, 6}, {7, 8, 9}}; // array of array of const int
int* pi = a[0]; // Error: a[0] has type const int*
void* unqual_ptr = a; // OK until C23; error since C23
// Notes: clang applies the rule in C++/C23 even in C89-C17 modes

typedef int A[2];
// _Atomic A a0 = {0};    // Error
// _Atomic(A) a1 = {0};   // Error
_Atomic int a2[2] = {0};  // OK
_Atomic(int) a3[2] = {0}; // OK

[edit] Assignment

Objects of array type are not modifiable lvalues, and although their address may be taken, they cannot appear on the left hand side of an assignment operator. However, structs with array members are modifiable lvalues and can be assigned:

int a[3] = {1,2,3}, b[3] = {4,5,6};
int (*p)[3] = &a; // okay, address of a can be taken
// a = b;            // error, a is an array
struct { int c[3]; } s1, s2 = {3,4,5};
s1 = s2; // okay: can assign structs holding array members

[edit] Array to pointer conversion

Any lvalue expression of array type, when used in any context other than

• as the operand of the address-of operator
• as the operand of sizeof
• as the operand of typeof and typeof_unqual (since C23)
• as the string literal used for array initialization

undergoes an implicit conversion to the pointer to its first element. The result is not an lvalue.

If the array was declared register, the behavior of the program that attempts such conversion is undefined.

int a[3] = {1,2,3};
int* p = a;
printf("%zu\n", sizeof a); // prints size of array
printf("%zu\n", sizeof p); // prints size of a pointer

When an array type is used in a function parameter list, it is transformed to the corresponding pointer type: int f(int a[2]) and int f(int* a) declare the same function. Since the function's actual parameter type is pointer type, a function call with an array argument performs array-to-pointer conversion; the size of the argument array is not available to the called function and must be passed explicitly:

#include <stdio.h>
 
void f(int a[], int sz) // actually declares void f(int* a, int sz)
{
    for (int i = 0; i < sz; ++i)
        printf("%d\n", a[i]);
}
 
void g(int (*a)[10]) // pointer to array parameter is not transformed
{
    for (int i = 0; i < 10; ++i)
        printf("%d\n", (*a)[i]);
}
 
int main(void)
{
    int a[10] = {0};
    f(a, 10); // converts a to int*, passes the pointer
    g(&a);    // passes a pointer to the array (no need to pass the size)
}

[edit] Multidimensional arrays

When the element type of an array is another array, it is said that the array is multidimensional:

// array of 2 arrays of 3 ints each
int a[2][3] = {{1,2,3},  // can be viewed as a 2x3 matrix
               {4,5,6}}; // with row-major layout

Note that when array-to-pointer conversion is applied, a multidimensional array is converted to a pointer to its first element, e.g., pointer to the first row:

int a[2][3]; // 2x3 matrix
int (*p1)[3] = a; // pointer to the first 3-element row
int b[3][3][3]; // 3x3x3 cube
int (*p2)[3][3] = b; // pointer to the first 3x3 plane

int n = 10;
int a[n][2*n];

[edit] Notes

Zero-length array declarations are not allowed, even though some compilers offer them as extensions (typically as a pre-C99 implementation of flexible array members).

If the size expression of a VLA has side effects, they are guaranteed to be produced except when it is a part of a sizeof expression whose result doesn't depend on it:

int n = 5, m = 5;
size_t sz = sizeof(int (*[n++])[m++]); // n is incremented, m may or may not be incremented

[edit] References

[edit] See also