Pass an array to a function by value

问题

Below is a snippet from the book C Programming Just the FAQs. Isn\'t this wrong as Arrays can never be passed by reference?

VIII.6: How can you pass an array to a function by value?

Answer: An array can be passed to a function by value by declaring in the called function the array name with square brackets ([ and ]) attached to the end. When calling the function, simply pass the address of the array (that is, the array’s name) to the called function. For instance, the following program passes the array x[] to the function named byval_func() by value:

The int[] parameter tells the compiler that the byval_func() function will take one argument—an array of integers. When the byval_func() function is called, you pass the address of the array to byval_func():
byval_func(x);
Because the array is being passed by value, an exact copy of the array is made and placed on the stack. The called function then receives this copy of the array and can print it. Because the array passed to byval_func() is a copy of the original array, modifying the array within the byval_func() function has no effect on the original array.

回答1:

Because the array is being passed by value, an exact copy of the array is made and placed on the stack.

This is incorrect: the array itself is not being copied, only a copy of the pointer to its address is passed to the callee (placed on the stack). (Regardless of whether you declare the parameter as int[] or int*, it decays into a pointer.) This allows you to modify the contents of the array from within the called function. Thus, this

Because the array passed to byval_func() is a copy of the original array, modifying the array within the byval_func() function has no effect on the original array.

is plain wrong (kudos to @Jonathan Leffler for his comment below). However, reassigning the pointer inside the function will not change the pointer to the original array outside the function.

回答2:

Burn that book. If you want a real C FAQ that wasn't written by a beginner programmer, use this one: http://c-faq.com/aryptr/index.html.

Syntax-wise, strictly speaking you cannot pass an array by value in C.

void func (int* x); /* this is a pointer */

void func (int x[]); /* this is a pointer */

void func (int x[10]); /* this is a pointer */

However, for the record there is a dirty trick in C that does allow you to pass an array by value in C. Don't try this at home! Because despite this trick, there is still never a reason to pass an array by value.

typedef struct
{
  int my_array[10];
} Array_by_val;

void func (Array_by_val x);

回答3:

In C and C++ it is NOT possible to pass a complete block of memory by value as a parameter to a function, but we are allowed to pass its address. In practice this has almost the same effect and it is a much faster and more efficient operation.

To be safe, you can pass the array size or put const qualifier before the pointer to make sure the callee won't change it.

回答4:

#include<stdio.h>
void  fun(int a[],int n);
int main()
{
    int a[5]={1,2,3,4,5};
    fun(a,5);
}
void  fun(int a[],int n)
{
    int i;
    for(i=0;i<=n-1;i++)
        printf("value=%d\n",a[i]);
}

By this method we can pass the array by value, but actually the array is accessing through the its base address which actually copying in the stack.

来源：https://stackoverflow.com/questions/4774456/pass-an-array-to-a-function-by-value

标签

arrays

pass-by-reference

pass-by-value