C：二維數組的大小

Question

我需要一些幫助來計算二維數組的行和列。看來我不能列數列？

#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); 

} 

輸出：
行數：10
列數：0

Answer 1

這是整數除法的問題！

int column = sizeof(result[0])/row; 

應該是

int column = 7/10; 

和整數除法，7/10==0。

你想要做的是劃分一行的長度，例如。例如，該行的一個元素的大小。 sizeof(result[0][0])：

int column = sizeof(result[0])/sizeof(result[0][0]); 

Answer 2

這對我的作品（評論解釋爲什麼）：

#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); 

} 

和這個輸出是：

Total of fields: 70 
Number of rows: 10 
Number of columns: 7 

編輯：

如指出的通過@AnorZaken，將數組作爲參數傳遞給一個函數並在其上打印sizeof的結果，將會輸出另一個total。這是因爲當你傳遞一個數組作爲參數時（不是指向它的指針），C會將它作爲副本傳遞，並且會在它們之間應用一些C魔法，所以你沒有像你認爲的那樣傳遞完全一樣的東西。爲了確定你在做什麼以及爲了避免一些額外的CPU工作和內存消耗，最好通過引用傳遞數組和對象（使用指針）。所以，你可以使用這樣的事情，有相同的結果。原：

#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; 
} 

Answer 3

它更方便（且不易出錯）使用數組長度的宏：

#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; 
}