来自不兼容指针类型的分配

Question

I'm not sure why this is happening, I think I'm doing everything correctly.. Maybe someone can help point me in the right direction.

    unsigned short* x;

    int textLeft[16];

    x = shm->textLeft;

These are spaced out in the program so I didn't want to copy a bunch of code but if more is needed please let me know.

Shouldn't this work correctly without giving me the incompatible pointer type?

Answer 1

No, this should not work, because you're assigning an int* value to an unsigned short* variable, which causes undefined behavior per the C strict aliasing rule .

The way to make this work without changing the types is to

1. cast the pointer, x = (unsigned short *)(shm->textLeft); , and
2. compile with GCC's -fno-strict-aliasing to turn the aliasing rule off.

But really, I strongly recommend you change the types to be compatible, since otherwise you're tying yourself to a single compiler's extensions to the C standard.

Answer 2

`unsigned short`

is not

`int`

So define

• x as int * or
• textLeft[16] as unsigned short

and things are ok.

Answer 3

In your case x is a unsigned short pointer and textLeft is a signed integer. You are trying to assign signed integer address to unsigned short pointer.