c - 如何在函数中传递和返回整数数组

Question

我有一个函数需要一个数组作为参数，并在修改它后返回相同的数组。 编译给我一个错误，我传递参数并调用函数。 为什么会这样？

int solve(int a[9][9])
{
  ...
  return a;

}
int main()
{
  int a[9][9];

  a = solve(a);    <error here>
}

Answer 1

对于初学者来说，数组没有赋值运算符。数组指示符是不可修改的左值。

来自 C 标准(6.3.2.1 左值、数组和函数指示符)

1. ... A modifiable lvalue is an lvalue that does not have array type, does not have an incomplete type, does not have a const qualified type, and if it is a structure or union, does not have any member (including, recursively, any member or element of all contained aggregates or unions) with a const qualified type.

其次，像数组一样声明的参数被调整为指向其元素类型的指针。

摘自 C 标准(6.7.6.3 函数声明符(包括原型(prototype)))

7 A declaration of a parameter as ‘‘array of type’’ shall be adjusted to ‘‘qualified pointer to type’’, where the type qualifiers (if any) are those specified within the [ and ] of the array type derivation...

例如这个函数声明

void f(int a[9][9]);

调整为

void f(int ( *a )[9]);

第三，函数不能以数组作为返回类型。但它们可能会返回指针。

来自 C 标准(6.9.1 函数定义)

3 The return type of a function shall be void or a complete object type other than array type.

例如，函数solve可以声明为

int ( * solve(int a[9][9]) )[9]
{
    // ...
    return a;
}

如果函数在任何情况下都更改了数组的元素，那么编写就没有意义

int a[9][9];

a = solve(a);    

你可以直接写

int a[9][9];

solve(a);

或

int a[9][9];

int ( *p )[9] = solve(a);

这是一个演示程序

#include <stdio.h>

#define M   2
#define N   3

int ( * f( int ( *a )[N], size_t n ) )[N]
{
    for ( size_t i = 0; i < n; i++ )
    {
        for ( size_t j = 0; j < N; j++ ) a[i][j] *= 10;
    }

    return a;
}

int main(void) 
{
    int a[M][N] =
    {
        { 1, 2, 3 },
        { 4, 5, 6 }
    };

    f( a, M );

    for ( size_t i = 0; i < M; i++ )
    {
        for ( size_t j = 0; j < N; j++ ) printf( "%d ", a[i][j] );
        putchar( '\n' );
    }

    putchar( '\n' );

    int ( *p )[N] = f( a, M );

    for ( size_t i = 0; i < M; i++ )
    {
        for ( size_t j = 0; j < N; j++ ) printf( "%d ", p[i][j] );
        putchar( '\n' );
    }

    putchar( '\n' );

    f( p, M );

    for ( size_t i = 0; i < M; i++ )
    {
        for ( size_t j = 0; j < N; j++ ) printf( "%d ", p[i][j] );
        putchar( '\n' );
    }

    putchar( '\n' );

    return 0;
}

它的输出是

10 20 30 
40 50 60 

100 200 300 
400 500 600 

1000 2000 3000 
4000 5000 6000 

为了简化函数声明，您可以引入 typedef 名称。

例如

typedef int( *PArray )[N];

PArray f( int ( *a )[N], size_t n )
{
    for ( size_t i = 0; i < n; i++ )
    {
        for ( size_t j = 0; j < N; j++ ) a[i][j] *= 10;
    }
    return a;
}

或者像这样

typedef int( *PArray )[N];

PArray f( PArray a, size_t n )
{
    for ( size_t i = 0; i < n; i++ )
    {
        for ( size_t j = 0; j < N; j++ ) a[i][j] *= 10;
    }
    return a;
}