我需要一些帮助来计算二维数组的行和列。好像我无法计算列数？
#include <stdio.h>
int main() {
char result[10][7] = {
{'1','X','2','X','2','1','1'},
{'X','1','1','2','2','1','1'},
{'X','1','1','2','2','1','1'},
{'1','X','2','X','2','2','2'},
{'1','X','1','X','1','X','2'},
{'1','X','2','X','2','1','1'},
{'1','X','2','2','1','X','1'},
{'1','X','2','X','2','1','X'},
{'1','1','1','X','2','2','1'},
{'1','X','2','X','2','1','1'}
};
int row = sizeof(result) / sizeof(result[0]);
int column = sizeof(result[0])/row;
printf("Number of rows: %d\n", row);
printf("Number of columns: %d\n", column);
}
输出：
行数：10
列数：0
答案 0 :(得分：9)
这是整数除法的问题！
int column = sizeof(result[0])/row;
应该是
int column = 7 / 10;
并以整数除法7/10==0。
你想要做的是划分一行的长度，例如。 sizeof(result[0])按该行的一个元素的大小，例如。 sizeof(result[0][0])：
int column = sizeof(result[0])/sizeof(result[0][0]);
答案 1 :(得分：9)
使用数组长度宏更方便（并且更不容易出错）：
#include <stdio.h>
#define LEN(arr) ((int) (sizeof (arr) / sizeof (arr)[0]))
int main(void)
{
char result[10][7];
printf("Number of rows: %d\n", LEN(result));
printf("Number of columns: %d\n", LEN(result[0]));
return 0;
}
答案 2 :(得分：8)
这对我有用（评论解释为什么）：
#include <stdio.h>
int main() {
char result[10][7] = {
{'1','X','2','X','2','1','1'},
{'X','1','1','2','2','1','1'},
{'X','1','1','2','2','1','1'},
{'1','X','2','X','2','2','2'},
{'1','X','1','X','1','X','2'},
{'1','X','2','X','2','1','1'},
{'1','X','2','2','1','X','1'},
{'1','X','2','X','2','1','X'},
{'1','1','1','X','2','2','1'},
{'1','X','2','X','2','1','1'}
};
// 'total' will be 70 = 10 * 7
int total = sizeof(result);
// 'column' will be 7 = size of first row
int column = sizeof(result[0]);
// 'row' will be 10 = 70 / 7
int row = total / column;
printf("Total fields: %d\n", total);
printf("Number of rows: %d\n", row);
printf("Number of columns: %d\n", column);
}
这个的输出是：
Total of fields: 70
Number of rows: 10
Number of columns: 7
修改
正如@AnorZaken指出的那样，将数组作为参数传递给函数并在其上打印sizeof的结果，将输出另一个total。这是因为当你传递一个数组作为参数（而不是指向它的指针）时，C会将它作为副本传递，并在它们之间应用一些C魔法，所以你没有像你想象的那样传递完全相同的东西。为了确保您正在做什么并避免一些额外的CPU工作和内存消耗，最好通过引用（使用指针）传递数组和对象。所以你可以使用这样的东西，结果和原作相同：
#include <stdio.h>
void foo(char (*result)[10][7])
{
// 'total' will be 70 = 10 * 7
int total = sizeof(*result);
// 'column' will be 7 = size of first row
int column = sizeof((*result)[0]);
// 'row' will be 10 = 70 / 7
int row = total / column;
printf("Total fields: %d\n", total);
printf("Number of rows: %d\n", row);
printf("Number of columns: %d\n", column);
}
int main(void) {
char result[10][7] = {
{'1','X','2','X','2','1','1'},
{'X','1','1','2','2','1','1'},
{'X','1','1','2','2','1','1'},
{'1','X','2','X','2','2','2'},
{'1','X','1','X','1','X','2'},
{'1','X','2','X','2','1','1'},
{'1','X','2','2','1','X','1'},
{'1','X','2','X','2','1','X'},
{'1','1','1','X','2','2','1'},
{'1','X','2','X','2','1','1'}
};
foo(&result);
return 0;
}
答案 3 :(得分：0)
// gets you the total size of the 2d array
printf("Arrays Total size: %ld\n",sizeof(result));
// gets you the cumulative size of row which is 5 columns * sizeof(int)
printf("1 row cumulative size: %ld\n",sizeof(result[0]));
// division of total array size with cumulative size of row gets you total number of rows
printf("total number of rows: %ld\n",sizeof(result)/sizeof(result[0]));
// and total number of columns you get by dividing cumulative row size with sizeof(char)
printf("total number of columns: %ld\n",sizeof(result[0])/sizeof(char));
答案 4 :(得分：0)
使用以下代码中显示的宏来获取1D，2D或3D数组的任何尺寸大小。可以类似地编写更多宏，以获取4D数组及其以后的尺寸。 （我知道对于Wickerman来说来不及看，但是对于其他访问此页面的人来说都是这样）
// Output of the following program
// [
/*
Demo of the advertised macros :
----------------------------------------------
sizeof(int) = 4
sizeof(Array_1D) = 12
ELEMENTS_IN_1D_ARRAY(Array_1D) = 3
sizeof(Array_2D) = 24
ELEMENTS_IN_2D_ARRAY(Array_2D) = 6
ROWS_IN_2D_ARRAY(Array_2D) = 2
COLUMNS_IN_2D_ARRAY(Array_2D) = 3
sizeof(Array_3D) = 96
ELEMENTS_IN_3D_ARRAY(Array_3D) = 24
MATRICES_IN_3D_ARRAY(Array_3D) = 4
ROWS_IN_3D_ARRAY(Array_3D) = 2
COLUMNS_IN_3D_ARRAY(Array_3D) = 3
Array_3D[][][] Printed :
----------------------------------------------
001 002 003
011 012 013
---------------
101 102 103
111 112 113
---------------
201 202 203
211 212 213
---------------
301 302 303
311 312 313
---------------
Wickerman's problem solved :
----------------------------------------------
sizeof(result) = 70
ELEMENTS_IN_2D_ARRAY(result) = 70
ROWS_IN_2D_ARRAY(result) = 10
COLUMNS_IN_2D_ARRAY(result) = 7
*/
// ]
// ====================================================================================================
// Program follows
// ====================================================================================================
// Array Size Macros
// [
#define ELEMENTS_IN_1D_ARRAY(a1D) ( sizeof( a1D ) / sizeof( a1D[0] )) // Total no. of elements in 1D array
#define ELEMENTS_IN_2D_ARRAY(a2D) ( sizeof( a2D ) / sizeof( a2D[0][0] )) // Total no. of elements in 2D array
#define ROWS_IN_2D_ARRAY(a2D) ( sizeof( a2D ) / sizeof( a2D[0] )) // No. of Rows in a 2D array
#define COLUMNS_IN_2D_ARRAY(a2D) ( sizeof( a2D[0] ) / sizeof( a2D[0][0] )) // No. of Columns in a 2D array
#define ELEMENTS_IN_3D_ARRAY(a3D) ( sizeof( a3D ) / sizeof( a3D[0][0][0] )) // Total no. of elements in 3D array
#define MATRICES_IN_3D_ARRAY(a3D) ( sizeof( a3D ) / sizeof( a3D[0] )) // No. of "Matrices" (aka "Slices"/"Pages") in a 3D array
#define ROWS_IN_3D_ARRAY(a3D) ( sizeof( a3D[0] ) / sizeof( a3D[0][0] )) // No. of Rows in each "Matrix" of a 3D array
#define COLUMNS_IN_3D_ARRAY(a3D) ( sizeof( a3D[0][0] ) / sizeof( a3D[0][0][0] )) // No. of Columns in each "Matrix" of a 3D array
// ]
#define PRINTF_d(s) (printf(#s " = %d\n", (int)(s))) // Macro to print a decimal no. along with its corresponding decimal expression string,
// while avoiding to write the decimal expression twice.
// Demo of the Array Size Macros defined above
// [
main()
{
// Sample array definitions
// [
int Array_1D[3] = {1, 2, 3}; // 1D array
int Array_2D[2][3] = // 2D array
{
{1, 2, 3},
{11, 12, 13}
};
int Array_3D[4][2][3] = // 3D Array
{
{
{1, 2, 3},
{11, 12, 13}
},
{
{101, 102, 103},
{111, 112, 113}
},
{
{201, 202, 203},
{211, 212, 213}
},
{
{301, 302, 303},
{311, 312, 313}
}
};
// ]
// Printing sizes and dimensions of arrays with the advertised Array Size Macros
printf(
"Demo of the advertised macros :\n"
"----------------------------------------------\n");
PRINTF_d(sizeof(int));
PRINTF_d(sizeof(Array_1D));
PRINTF_d(ELEMENTS_IN_1D_ARRAY(Array_1D));
PRINTF_d(sizeof(Array_2D));
PRINTF_d(ELEMENTS_IN_2D_ARRAY(Array_2D));
PRINTF_d(ROWS_IN_2D_ARRAY(Array_2D));
PRINTF_d(COLUMNS_IN_2D_ARRAY(Array_2D));
PRINTF_d(sizeof(Array_3D));
PRINTF_d(ELEMENTS_IN_3D_ARRAY(Array_3D));
PRINTF_d(MATRICES_IN_3D_ARRAY(Array_3D));
PRINTF_d(ROWS_IN_3D_ARRAY(Array_3D));
PRINTF_d(COLUMNS_IN_3D_ARRAY(Array_3D));
// Printing all elements in Array_3D using advertised macros
// [
int x, y, z;
printf(
"\nArray_3D[][][] Printed :\n"
"----------------------------------------------\n");
for(x = 0; x < MATRICES_IN_3D_ARRAY(Array_3D); x++)
{
for(y = 0; y < ROWS_IN_3D_ARRAY(Array_3D); y++)
{
for(z = 0; z < COLUMNS_IN_3D_ARRAY(Array_3D); z++)
printf("%4.3i", Array_3D[x][y][z]);
putchar('\n');
}
printf("---------------\n");
}
// ]
// Applying those macros to solve the originally stated problem by Wickerman
// [
char result[10][7] = {
{'1','X','2','X','2','1','1'},
{'X','1','1','2','2','1','1'},
{'X','1','1','2','2','1','1'},
{'1','X','2','X','2','2','2'},
{'1','X','1','X','1','X','2'},
{'1','X','2','X','2','1','1'},
{'1','X','2','2','1','X','1'},
{'1','X','2','X','2','1','X'},
{'1','1','1','X','2','2','1'},
{'1','X','2','X','2','1','1'}
};
printf(
"\nWickerman's problem solved :\n"
"----------------------------------------------\n");
PRINTF_d(sizeof(result)); // radha_SIZEOF_2D_ARRAY
PRINTF_d(ELEMENTS_IN_2D_ARRAY(result)); // radha_SIZEOF_2D_ARRAY
PRINTF_d(ROWS_IN_2D_ARRAY(result));
PRINTF_d(COLUMNS_IN_2D_ARRAY(result));
// ]
}
// ]