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使用函数在C中创建和释放2D数组

[英]Use function to create and free 2D array in C

I practice the 2D dynamic array with reference to the URL below: https://thispointer.com/allocating-and-deallocating-2d-arrays-dynamically-in-c-and-c/ 我参考以下URL来练习2D动态数组: https : //thispointer.com/allocating-and-deallocating-2d-arrays-dynamically-in-c-and-c/

My code: 我的代码:

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

int** create_2d_arr(int row_size,int colum_size)
{
    int** array = (int**)malloc(sizeof(int*)*row_size);
    for (int i = 0; i < row_size; i++)
        array[i] = (int*)malloc(sizeof(int)*colum_size);  
    return array;
}

void free_2d_arr(int** matrix,int row_size, int colum_size) {
    for (int i = 0; i < row_size; i++) {
        free(matrix[i]);
    }
    free(matrix);
}

int main(int argc, char const *argv[])
{
    int row=3,cloum=2;
    int** arr_2d = create_2d_arr(row,cloum);

    arr_2d[0,0]=4;
    arr_2d[0,1]=5;

    arr_2d[1,0]=6;
    arr_2d[1,1]=7;

    arr_2d[2,0]=8;
    arr_2d[2,1]=9;

    for(int i=0;i<row;i++)
    for(int j=0;j<cloum;j++)
    printf("arr_2d[%d,%d] = %d \n",i,j,arr_2d[i,j]);

    free_2d_arr(arr_2d,row,cloum);

    return 0;
}

However, there are errors when executing after compilation: 但是,编译后执行时会出现错误:

arr_2d[0,0] = 8 
arr_2d[0,1] = 9 
arr_2d[1,0] = 8 
arr_2d[1,1] = 9 
arr_2d[2,0] = 8 
arr_2d[2,1] = 9 
[1]    9300 segmentation fault (core dumped)  ./t

Only arr_2d[2,0]=8 arr_2d[2,1]=9 are correct. 仅arr_2d [2,0] = 8 arr_2d [2,1] = 9是正确的。 I don't understand where my code is wrong. 我不明白我的代码在哪里错误。 Does anyone help me? 有人帮我吗?


renew 更新

thanks for your replies. 感谢您的回复。 but after I modify arr_2d[2,0]=8 to a rr_2d[2][0]=8 ... 但在我将arr_2d[2,0]=8修改为rr_2d[2][0]=8 ...

result of printf is
arr_2d[0][0] = -267545984
arr_2d[0][1] = -267545952
arr_2d[1][0] = -267545984

... ...

warning of compiler 编译器警告

t.c:38:47: warning: expression result unused [-Wunused-value]
    printf("arr_2d[%d,%d] = %d \n",i,j,arr_2d[i,j]);
                                              ^
t.c:38:40: warning: format specifies type 'int' but the argument has type
      'int *' [-Wformat]
    printf("arr_2d[%d,%d] = %d \n",i,j,arr_2d[i,j]);
                            ~~         ^~~~~~~~~~~
2 warnings generated.

my compiler is clang,even if I use gcc 我的编译器是clang,即使我使用gcc

========= =========

Solved 解决了

After modify: 修改后:

printf("arr_2d[%d,%d] = %d \n",i,j,arr_2d[i,j]);

=> =>

printf("arr_2d[%d,%d] = %d \n",i,j,arr_2d[i],[j]);

It work normally. 它正常工作。 Thank everyone very much. 非常感谢大家。

you can't access array's value like this 你不能像这样访问数组的值

arr_2d[x,y];

you have to use this type of syntax 您必须使用这种类型的语法

arr_2d[x][y];

arr_2d[x,y] is equivalent to arr_2d[y] arr_2d[x,y]等同于arr_2d[y]

use arr_2d[x][y] in assignment and access 在分配和访问中使用arr_2d[x][y]

In C (or C++) a,b,c .., z compute a then b .. then z and returns the value of z 在C(或C ++) a,b,c .., z计算A然后B ..然后z和返回z的值


So : 因此:

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

int** create_2d_arr(int row_size,int colum_size)
{
    int** array = (int**)malloc(sizeof(int*)*row_size);
    int i;
    for ( i = 0; i < row_size; i++)
        array[i] = (int*)malloc(sizeof(int)*colum_size);  
    return array;
}

void free_2d_arr(int** matrix,int row_size, int colum_size) {
  int i;
    for ( i = 0; i < row_size; i++) {
        free(matrix[i]);
    }
    free(matrix);
}

int main(int argc, char const *argv[])
{
    int row=3,cloum=2;
    int** arr_2d = create_2d_arr(row,cloum);

    arr_2d[0][0]=4;
    arr_2d[0][1]=5;

    arr_2d[1][0]=6;
    arr_2d[1][1]=7;

    arr_2d[2][0]=8;
    arr_2d[2][1]=9;

    int i,j;
    for( i=0;i<row;i++)
    for( j=0;j<cloum;j++)
    printf("arr_2d[%d][%d] = %d \n",i,j,arr_2d[i][j]);

    free_2d_arr(arr_2d,row,cloum);

    return 0;
}

The result is : 结果是:

arr_2d[0][0] = 4 
arr_2d[0][1] = 5 
arr_2d[1][0] = 6 
arr_2d[1][1] = 7 
arr_2d[2][0] = 8 
arr_2d[2][1] = 9 

The correct syntax is 正确的语法是

arr_2d[i][j] = n;

I have never seen something like arr_2d[i,j] and would be great-full if someone explained it to us. 我从未见过像arr_2d[i,j]这样的东西arr_2d[i,j]如果有人向我们解释它,我会感到非常arr_2d[i,j]

My guess is that it's equivalent to arr_2d[j] but I'm not sure. 我的猜测是它等同于arr_2d[j]但我不确定。

EDIT: precision: you should change ALL your arr_2d[i,j] (like arr_2d[0,0] ) to the new syntax 编辑:精度:您应将所有arr_2d[i,j] (如arr_2d[0,0] )更改为新语法

the following proposed code: 以下建议的代码:

  1. cleanly compiles 干净地编译
  2. performs the desired functionality 执行所需的功能
  3. incorporates my comments to the OPs question 将我的评论纳入运营问题
  4. for readability, inserts an appropriate space: inside parens, inside braces, after semicolons, after commas, around C operators 为了提高可读性,在C运算符周围插入一个适当的空格:括号内,括号内,分号后,逗号后
  5. used size_t rather than int to avoid compiler warnings about conversions between int and size_t , especially in the calls to calloc() and malloc() 使用size_t而不是int以避免编译器警告intsize_t之间的转换,尤其是在对calloc()malloc()的调用中

And now, the proposed code: 现在,建议的代码:

#include <stdio.h>
#include <stdlib.h>   // exit(), EXIT_FAILURE


// prototypes
void free_2d_arr( int** matrix, size_t row_size, size_t colum_size );


int** create_2d_arr( size_t row_size, size_t colum_size )
{
    int** array = calloc( row_size, sizeof( int* ) );
    if( !array )
    {
        perror( "malloc failed" );
        exit( EXIT_FAILURE );
    }

    // implied else, malloc successful

    for ( size_t i = 0; i < row_size; i++ )
    {
        array[ i ] = malloc( sizeof(int) * colum_size );  
        if( !array[ i ] )
        {
            perror( "malloc failed" );
            free_2d_arr( array, row_size, colum_size ); 
            exit( EXIT_FAILURE );
        }
    }
    return array;
}


void free_2d_arr( int** matrix, size_t row_size, size_t colum_size ) 
{
    (void)colum_size;
    for ( size_t i = 0; i < row_size; i++ ) 
    {
        free( matrix[ i ] );
    }
    free( matrix );
}


int main( void )
{
    size_t row = 3;
    size_t cloum = 2;
    int** arr_2d = create_2d_arr( row, cloum );

    arr_2d[ 0 ][ 0 ] = 4;
    arr_2d[ 0 ][ 1 ] = 5;

    arr_2d[ 1 ][ 0 ] = 6;
    arr_2d[ 1 ][ 1 ] = 7;

    arr_2d[ 2 ][ 0 ] = 8;
    arr_2d[ 2 ][ 1 ] = 9;

    for( size_t i=0; i < row; i++ )
    {
        for( size_t j=0; j < cloum; j++ )
        {
            printf( "arr_2d[%lu,%lu] = %d \n", i, j, arr_2d[ i ][ j ] );
        }
    }

    free_2d_arr( arr_2d, row, cloum );

    return 0;
}

a run of the proposed code results in: 建议代码的运行导致:

arr_2d[0,0] = 4 
arr_2d[0,1] = 5 
arr_2d[1,0] = 6 
arr_2d[1,1] = 7 
arr_2d[2,0] = 8 
arr_2d[2,1] = 9 

Note, however, that the statements: 但是请注意,以下语句:

size_t row = 3;
size_t cloum = 2;

are really parameters to the program and the 2 and 3 are constants and 'magic' numbers. 是程序的真正参数,而23是常数和“魔术”数。 'magic' numbers are numbers with no basis. “魔术”数字是没有基础的数字。 'magic' numbers make the code more difficult to understand, debug, etc. Suggest using #define statements or a enum statement to give those 'magic' numbers meaningful names, then use those meaningful names throughout the code “魔术”数字使代码更难以理解,调试等。建议使用#define语句或enum语句为这些“魔术”数字赋予有意义的名称,然后在整个代码中使用这些有意义的名称

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