在 C 中声明一个带有指针的数组

Question

Why is this syntax illegal in C?

int *a = {1,2,3};

Aren't arrays and pointers the same thing? Why would this work and not the above?

int a[] = {1,2,3};
int *b = a;

Answer 1

Why is this syntax illegal in C?

In int a[] = {1,2,3}; , the {1,2,3} is a syntax for providing initial values of the things being initialized. A list of values in braces is a general notation for listing initial values. It is not a notation for representing an array.

The declaration says a is an array, but that is just in its int a[] part. The {1,2,3} does not indicate an array—it just provides a list of values, and it could be used for non-array objects, such as providing values for the members of a structure.

In int *a = {1,2,3}; , the int *a part defines a to be a pointer. A pointer just has one value in it—the address of something (or null). So it does not make sense to initialize it with multiple values. To initialize a pointer, you need to provide an address for it to point to.

There is a notation in C for representing an array of given values. It is called a compound literal, and can be used for types in general, not just arrays. Its syntax is “( type ) { initial values }”. So we can create an array with (int []) {1, 2, 3} . A pointer could be defined and initialized to point to this array with int *a = (int []) {1, 2, 3}; . However, this should be done with care. Such an array is temporary (has automatic storage duration that ends when execution of its associated block ends) when defined inside a function, so the pointer will become invalid when the function returns (or before, if it is in an inner block).