c - 为什么内存中的下一个短路具有另一个变量的值？

Question

在主方法中我有以下代码:

int main (void)
{
    unsigned short array[3];
    *array = 65535;
    *(array + 1) = 12453;
    *(array + 2) = 45055;

    int bitposition = 12;

    int bitsperSample = 8;

    unsigned short * pointer_array = & array[0];
    unsigned short result = readSample(pointer_array, bitposition, bitsperSample);


}

我的目的是创建一个 3 元素短数组，其中可以存储 3 个短数组，以便在函数 readSample 中读取它们。该函数如下，内部代码旨在从先前声明的数组中读取两个short:

unsigned short readSample( unsigned short * track, int bitpos, int bitsPerSample )
    {
    int res = 1;
    int short_to_read = bitpos / 16; 

    int pos_bit_in_the_short = bitpos % 16; // local position of the bit inside the short_to_read  

    unsigned short found = *(track + short_to_read);
    unsigned short * pointer_to_found  = & found;
    unsigned short copy_found = * pointer_to_found;
    printf("first short to read is %d \n", copy_found);

    unsigned short second_short_to_read = * (pointer_to_found + 1);
    printf("second short to read is %d \n", second_short_to_read);


    return res;
} 

我希望程序打印:

first short to read is 65535
second short to read is 12453

但是程序输出:

first short to read is 65535
second short to read is 12

这是错误的，因为要读取的第二个短路的值是 pos_bit_in_the_short = bitpos % 16，我不知道为什么。

我还尝试打印要读取的short的十六进制地址，看看它是否访问内存的其他部分，现在，第二个short在第一个short之后2个字节，正如我在中声明数组的方式所预期的那样主要方法。地址是:

0xbfc120f2 for the first short and 
0xbfc120f4 for the second short.

任何人都可能知道为什么会发生这种情况？

Answer 1

问题出在这里:

unsigned short found = *(track + short_to_read);
unsigned short * pointer_to_found  = & found;

found 是一个局部变量(在 readSample 中)，以下逻辑将它用作数组，但事实并非如此。