具有枚举类型的多维数组，并将其转换为无符号char数组

Question

here is the code.

#include <stdio.h>
#define N 3
#define COPY(a, i) (a[(i)]) = (a[((i)+1)])

enum course {BTP300 = 1, OOP244, OOP344, OOP444, BTP400 = 8, BTP500};
typedef enum course Course;

void display(void* a, int n) {
    int i;
    unsigned char* c = (unsigned char*)a;

    for (i = 0; i < n; i++)
        printf("%d ", c[i]);
    printf("\n");
}

void process(void *c, int n, int s) {
    int i, j;
    unsigned char* a = (unsigned char*)c;

    for (i = 0; i < s * n; i++) {
        unsigned char x = a[i];
        for (j = 1; j < s - 1; j++, i++)
            COPY(a, i);
        a[++i] = x;
    }
}

int main() {
    Course array[2][N] = {BTP300, BTP400, BTP500, OOP244, OOP344, OOP444};

    display(array[1], sizeof(Course)*N);
    display(array[0], sizeof(Course)*N);
    process(array[0], N, sizeof(Course));
    process(array[1], N, sizeof(Course));
    display(array[1], sizeof(Course)*N);
    display(array[0], sizeof(Course)*N);
    return 0;
}

The out put is:

2 0 0 0 3 0 0 0 4 0 0 0
1 0 0 0 8 0 0 0 9 0 0 0
0 0 0 2 0 0 0 3 0 0 0 4
0 0 0 1 0 0 0 8 0 0 0 9

now what it seems like when casting the pointer the size comes into play. I initially thought that though memory is created, in arrays you just jump over. so i would still get 234. but no. I get 1byte char.

0 2
1 0
2 0
3 0

And this gets printed too.

whats going on?

Answer 1

Enumeration values are of type int in C, and int is commonly four bytes (32 bits) on most platforms. So trying to access the values as char will not give you the expected result.

For the display function you do not need to multiply with sizeof(Course) , the number of entries is N so that is the size you should provide to the function:

display(array[1], N);

And of course you should be using either int for the separate values, or Course .

And you need to rethink your process function as well.