C语言 C：二维数组的大小
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原文地址: http://stackoverflow.com/questions/34134074/
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C: Size of two dimensional array
提问by Wickerman
I need some help counting the rows and columns of a two dimensional array. It seems like I can't count columns?
我需要一些帮助来计算二维数组的行和列。好像不能数列？
#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);
}
Output:
Number of rows: 10
Number of columns: 0
输出：
行数：10
列数：0
采纳答案by Marcus Müller
That's a problem of integer division!
那是整数除法的问题！
int column = sizeof(result[0])/row;
should be
应该
int column = 7 / 10;
and in integer division, 7/10==0.
在整数除法中，7/10==0。
What you want to do is divide the length of one row, eg. sizeof(result[0])by the size of one element of that row, eg. sizeof(result[0][0]):
你想要做的是划分一行的长度，例如。sizeof(result[0])通过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]);
回答by August Karlstrom
It's much more convenient (and less error prone) to use an array length macro:
使用数组长度宏更方便（也更不容易出错）：
#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;
}
回答by emi
This works for me (comments explains why):
这对我有用（评论解释了原因）：
#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);
}
And the output of this is:
这个输出是：
Total of fields: 70
Number of rows: 10
Number of columns: 7
EDIT:
编辑：
As pointed by @AnorZaken, passing the array to a function as a parameter and printing the result of sizeofon it, will output another total. This is because when you pass an array as argument (not a pointer to it), C will pass it as copy and will apply some C magic in between, so you are not passing exactly the same as you think you are. To be sure about what you are doing and to avoid some extra CPU work and memory consumption, it's better to pass arrays and objects by reference (using pointers). So you can use something like this, with same results as original:
正如@AnorZaken 所指出的，将数组作为参数传递给函数并sizeof在其上打印结果，将输出另一个total. 这是因为当您将数组作为参数（而不是指向它的指针）传递时，C 会将它作为副本传递并在两者之间应用一些 C 魔法，因此您传递的内容与您认为的完全不同。为了确定您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，与原来的结果相同：
#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;
}
回答by mfaisal
// 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));
回答by ShreeRadheKrishna
Use the macros shown in below code to get any dimension size of 1D, 2D or 3D arrays. More macros can be written similarly to get dimensions for 4D arrays and beyond. (i know its too late for Wickerman to have a look but these're for anyone else visiting this page)
使用下面代码中显示的宏来获取 1D、2D 或 3D 数组的任何维度大小。可以类似地编写更多宏来获取 4D 数组及以后的维度。（我知道 Wickerman 已经太晚了，但这些是给访问此页面的其他人的）
// 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));
// ]
}
// ]
