c - 如何将数组传递给函数？指针？

Question

试图自学如何将数组从 int main() 传递给函数，但我就是不明白。

这是我的代码。

#include <stdio.h>

void printboard( char *B ) {
 /*Purpose: to print out the tic tac toe board B
  */
    int i,j;

    printf("\n");
    for (i=0;i<3;i++) {
        for (j=0;j<3;j++) {
            printf( "      %c ",B[i][j]);
        }
        printf("\n\n");
    }
}

int main( void ) {
    char B[3][3] = { '-','-','-',
                     '-','-','-',
                     '-','-','-' };

    printboard(B);
}

我收到这个错误:

test.c: In function 'printboard':
test.c:12: error: subscripted value is neither array nor pointer
test.c: In function 'main':
test.c:25: warning: passing argument 1 of 'printboard' from incompatible pointer type

只需要了解指针的工作方式和传递方式，这样我就可以继续我的工作。

Answer 1

二维数组衰减为指向一维数组的指针(即 (*)[] )

所以，从 -

void printboard( char *B ) 
{
   // ....
}

到

void printboard( char B[][3] )
                 // Compiler is not bothered about first index because it 
                 // converts to (*)[3]. So, is the reason it left empty.
{
    // ...
}

---

还有，二维数组的初始化应该是这样的——

char B[3][3] = { {'-','-','-'},
                 {'-','-','-'},
                 {'-','-','-'}
               };

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