Const arrays in C | 易学教程

问题

Original question: If I define:

const int z[5] = {10, 11, 12, 13, 14};

does it mean:

it's a constant array of integers i.e. the address which z points to is always constant and can never change, but the elements of z can change.

Each element of z is a constant i.e. their value can never change.

Edit:

More info:

There is another variable:

const int *y = z;
func((int *) y);

where func is defined as:

void func(int y[]) {
    int i;
    for(i = 0; i < 5; i++) {
        y[i] = i; //y[i] can be set to any integer; used i as example
    }
}

where in func, using y, the array is traversed and each element is changed. Is this is valid even though all elements of z are const?

回答1:

It means that each element of z is read-only.

The object z is an array object, not a pointer object; it doesn't point to anything. Like any object, the address of z does not change during its lifetime.

Since the object z is an array, the expression z, in most but not all contexts, is implicitly converted to a pointer expression, pointing to z[0]. That address, like the address of the entire array object z, doesn't change during the object's lifetime. This "conversion" is a compile-time adjustment to the meaning of the expression, not a run-time type conversion.

To understand the (often confusing) relationship between arrays and pointers, read section 6 of the comp.lang.c FAQ.

It's important to understand that "constant" and const are two different things. If something is constant, it's evaluated at compile time; for example, 42 and (2+2) are constant expressions.

If an object is defined with the const keyword, that means that it's read-only, not (necessarily) that it's constant. It means that you can't attempt to modify the object via its name, and attempting to modify it by other means (say, by taking its address and casting to a non-const pointer) has undefined behavior. Note, for example, that this:

const int r = rand();

is valid. r is read-only, but its value cannot be determined until run time.

回答2:

In your case the answer is:

Each element of z is a constant i.e. their value can never change.

You can't create a const array because arrays are objects and can only be created at runtime and const entities are resolved at compile time.

So, the const is interpreted as in the first example below, i.e. applied for the elements of the array. Which means that the following are equivalent: ^{The array in your example needs to be initialized.}

 int const z[5] = { /*initial (and only) values*/};
 const int z[5] = { /*-//-*/ };

It is some type commutative property of the const specifier and the type-specifier, in your example int.

Here are few examples to clarify the usage of constant:

1.Constant integers definition: (can not be reassigned). In the below two expression the use of const is equivalent:

int const a = 3;  // after type identifier
const int b = 4;  // equivalent to before type qualifier

2.Constant pointer definition (no pointer arithmetics or reassignment allowed):

int * const p = &anInteger;  // non-constant data, constant pointer

and pointer definition to a constant int (the value of the pointed integer cannot be changed, but the pointer can):

const int *p = &anInteger;  // constant data, non-constant pointer

来源：https://stackoverflow.com/questions/32705419/const-arrays-in-c

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