How Can I Pass an Array by Reference in C to Avoid Copying?

Passing an Array by Reference in C

Passing an array by reference allows you to access the original array within a function without having to copy it. In C , you can use the & operator to pass an array by reference.

Consider the following code snippet:

void foo(int (&myArray)[100])
{
    // Access and manipulate the elements of myArray here
}

int main()
{
    int a[100];
    foo(a);
}

In this example, the foo function takes an array of 100 integers as a reference. The syntax (&myArray)[100] specifies that myArray is being passed by reference and that it is an array of 100 integers.

The (&) operator is used to obtain the address of the array. Parentheses are necessary to disambiguate the expression: the & operator has higher precedence than the [] operator, so &myArray[100] would refer to the address of the 100th element of the array, not to the address of the array itself.

Passing an array by reference instead of by value is beneficial because it avoids the overhead of copying the entire array into the function. This is especially important for large arrays, as it can significantly improve performance.

Note that the syntax (&myArray)[100] only accepts arrays of 100 integers. If you want to pass an array of a different size, you can use the syntax (&myArray), which accepts arrays of any size. However, using this syntax means that you cannot use sizeof on myArray within the function.

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