How Does C Pass Arrays by Reference and What are the Implications?

Passing an Array by Reference

In C , arrays are typically passed to functions by reference, allowing the function to modify the elements of the original array. One way to achieve this is by using the syntax:

void foo(int (&myArray)[100]);

Understanding the Syntax

The syntax int(&myArray)[100] specifies that the function foo will receive a reference to an array of 100 integers. The (&myArray) part indicates that the function receives a reference to the array itself, not a copy of it.

Implications of Passing by Reference

By passing the array by reference, the function has direct access to the actual elements of the array. Any changes made to the elements within the function will be reflected in the original array. This eliminates the need for creating a copy of the array, improving both memory usage and performance.

Parsing the Function Parameters

C allows multiple interpretations of function parameter types with arrays, depending on the syntax used. To clarify the intention of passing a reference to an array, the following syntax is employed:

void foo(int *x); // Accepts arrays of any size as int *
void foo(int x[100]); // Accepts arrays of 100 integers as int *
void foo(int[] x); // Accepts arrays of any size as int *

void foo(int (&x)[100]); // Accepts arrays of 100 integers as int (&x)[100]

void foo(int &x[100]); // Invalid syntax, attempting to create an array of references

Therefore, the (&myArray) syntax in void foo(int (&myArray)[100]); explicitly specifies that the function is passing a reference to an array of 100 integers.

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