C: 警告 returning void * from a function with return type in makes integer from pointer without cast

Question

我正在编写一个程序来使用回溯算法解决数独谜题，当我编译我的代码时，编译器 returns 一条警告指出以下内容在函数 find_empty_box()

warning: returning ‘void *’ from a function with return type ‘int’ makes integer from pointer without a cast [-Wint-conversion]

我已经尝试将函数初始化为 void *（我也尝试将它初始化为 int 但这会导致关于返回 NULL 的警告，这表明我应该将函数声明为 void * 它确实解决了之前的警告但是创建了两个其他警告第一个警告是在函数 find_empty_box() 中引起的第二个警告是在 Solver()

中引起的

warning: returning ‘int’ from a function with return type ‘void *’ makes pointer from integer without a cast [-Wint-conversion]
   13 |                                 return x, y;

warning: assignment to ‘int’ from ‘void *’ makes integer from pointer without a cast [-Wint-conversion]
   78 |         row, col = find_empty_box(sudoku);

程序的代码是

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>

int find_empty_box(int sudoku)
{
    for (int x = 0; x <= 9; ++x)
    {
        for (int y = 0; y < 9; ++y)
        {
            if (sudoku[&x][&y] == 0)
            {
                return x, y;
            }
        }
        return NULL, NULL;
    }
}

int Answer_Valid(int sudoku, int guess, int row, int col)
{
    int row_values = sudoku[&row];

    for (int i = 0; i < sudoku[&row]; ++i)
    {
        if (guess == sudoku[&row[&i]])
        {
            return false;
        }
    }

    int column_values = sudoku[&col];

    for (int t = 0; t <= 9; ++t)
    {
        for (int n = 0; n < sudoku[&col]; ++n)
        {
            if (&guess == &sudoku[&col[&n]])
            {
                return false;
            }
        }
    }

    int row_start = (row / 3) * 3;
    int col_start = (col / 3) * 3;

    for (int x = 0; x <= row_start && row_start + 3; ++x)
    {
        for (int y = 0; y < col_start && col_start + 3; ++y)
        {
            if (sudoku[&x][&y] == guess)
            {
                return false;
            }
            return true;
        }
    }

}


int Solver(int sudoku, int guess, int row, int col)
{
    int grid[9][9] = 
    {
        {8,0,0,0,0,0,0,0,0},
        {0,0,3,6,0,0,0,0,0},
        {0,7,0,0,9,0,2,0,0},
        {0,5,0,0,0,7,0,0,0},
        {0,0,0,0,4,5,7,0,0},
        {0,0,0,1,0,0,0,3,0},
        {0,0,1,0,0,0,0,6,8},  
        {0,0,8,5,0,0,0,1,0},
        {0,9,0,0,0,0,4,0,0}
    };

    row, col = find_empty_box(sudoku);

    if (&row == NULL)
    {
        return true;
    }

    for (int i = 0; (i < 1)||(i < 2)||(i < 3)||(i < 4)||(i < 5)||(i < 6)||(i < 7)||(i < 8)||(i < 9); ++i)
    {
        if(Answer_Valid(sudoku, guess, row, col))
        {
            sudoku[&row][&col] = guess;

            if(Solver(sudoku, guess, row, col))
            {
                return true;
            }
            return false;
        }
    }

    int LineNum = 9;
    int RowAmount = 9;

    for (int i = 0; i < LineNum; ++i)
    {
        for (int t = 0; t < RowAmount; ++t)
        {
            printf("%d ", grid[i][i]);
        }

        printf("\n");   
    }
}

int main()
{
    int grid[9][9] = 
    {
        {8,0,0,0,0,0,0,0,0},
        {0,0,3,6,0,0,0,0,0},
        {0,7,0,0,9,0,2,0,0},
        {0,5,0,0,0,7,0,0,0},
        {0,0,0,0,4,5,7,0,0},
        {0,0,0,1,0,0,0,3,0},
        {0,0,1,0,0,0,0,6,8},  
        {0,0,8,5,0,0,0,1,0},
        {0,9,0,0,0,0,4,0,0}
    };

    int LineNum = 9;
    int RowAmount = 9;

    for (int i = 0; i < LineNum; ++i)
    {
        for (int x = 0; x < RowAmount; ++x)
        {
            printf("%d ", &grid[i][x]);
        }

        printf("\n");   
    }
}

Answer 1

虽然错误是不言自明的（你在 void* 中将函数声明为 return 但你试图 return 一对整数），底层问题是 C 不支持像这样的 'pairs' 或 'tuples'，所以你最好使用额外的指针参数来 return 这些值，并且只使用 return succeed/fail 布尔标志的值。所以你的 find_empty_box 函数变成：

bool find_empty_box(int sudoku[9][9], int *row, int *col)
{
    for (int x = 0; x <= 9; ++x)
    {
        for (int y = 0; y < 9; ++y)
        {
            if (sudoku[x][y] == 0)
            {
                *row = x;
                *col = y;
                return true;
            }
        }
    }
    return false;
}

然后您可以称其为

if (!find_empty_box(grid, &row, &col)) {
    ... could not find an empty box, done?

这带来了另一个问题——您想传递 grid（不清楚其他地方的 sudoku 是什么），但它是一个数组，所以它实际上是通过引用传递的（隐式指针），因此您在函数中所做的任何更改都会影响调用者的 grid。这可能是也可能不是你想要的。您对 sudoku[&x][&y] 的引用完全是胡说八道——它们很可能会崩溃并且不会做任何有用的事情。

C: 警告 returning void * from a function with return type in makes integer from pointer without cast

C: warning returning void * from a function with return type in makes integer from pointer without cast

c

integer

void