Invalid char pointer deference behavior

[ChAoSUnItY]

Considering the following code:

#include <stdlib.h>
#include <string.h>

char a[100];

typedef struct {
    char *raw;
} data_t;

void init_data(data_t *data, char *raw) {
    data->raw = raw;
}

int main() {
    strcpy(a, "DATA");
    data_t *data = malloc(sizeof(data_t));
    init_data(data, a);
    
    char *raw = data->raw;
    printf("%c\n", a[0]);
    printf("%c\n", raw[0]);
    printf("%c\n", data->raw[0]);
    
    
    free(data);

    return 0;
}

Using gcc to compile the program and run it, it would output:

D
D
D

But using shecc, it would output:

D
D
�

This happens because of the incorrect deference assembly code.

[DrXiao]

The test code in the issue #181 makes me think of this issue, and I add the following printf() calls to execute the code and observe the potential bug:

int main() {
    strcpy(a, "DATA");
    data_t *data = malloc(sizeof(data_t));
    init_data(data, a);
    
    char *raw = data->raw;
    printf("%c\n", a[0]);
    printf("%c\n", raw[0]);
    printf("%c\n", data->raw[0]);
+   printf("%c\n", *data->raw);

+   printf("raw = %x\n", raw);
+   printf("data->raw = %x\n", data->raw);
+   printf("&raw[0] = %x\n", &raw[0]);
+   printf("&data->raw[0] = %x\n", &data->raw[0]);
+   printf("data = %x\n", data);
+   printf("&data[0] = %x\n", &data[0]);
    
    free(data);

    return 0;
}

• GCC

$ gcc -o test test.c
$ ./test
D
D
D
D
raw = 28863040
data->raw = 28863040
&raw[0] = 28863040
&data->raw[0] = 28863040
data = 5f8d62a0
&data[0] = 5f8d62a0

• shecc

$ qemu-arm out/shecc-stage2.elf -o test test.c
$ qemu-arm test
D
D
�
D
raw = 407ffdcc
data->raw = 407ffdcc
&raw[0] = 407ffdcc
&data->raw[0] = 4080400c
data = 4080400c
&data[0] = 4080400c

It is obvious that &data->raw[0] equals &data[0].

After observing the code snippets in this comment and the comment in the issue #181, I think the potential bug is that using arrow operator and subscripting operator (-> and [ ]) simultaneously may obtain an incorrect object.

In this comment, &data->raw[0] is equivalent to &data[0].

For my comment in the issue (#181) , the statement cur = map->buckets[0] in the for loop equals cur = map->buckets.