Re: [RFA/m32r] Fix breakpoint bug

["Kei Sakamoto" <sakamoto.kei@renesas.com>]

> I'm curious about the little endian case:
> 
> +  else /* little-endian */
> +    {
> +      if ((addr & 3) == 0)
> + {
> +   buf[0] = contents_cache[0];
> +   buf[1] = contents_cache[1] & 0x7f;
> +   buf[2] = bp_entry[1];
> +   buf[3] = bp_entry[0];
>   }
> 
> Shouldn't the breakpoint be placed at buf[0] here rather than buf[2]?

For most of architectures - yes, it shold be at buf[0]. But the little endian
mode of M32R is a kind of unique.

In other architectures, two 16-bit instructions, A and B, are placed as
the following:

Big endian:
A0 A1 B0 B1

Little endian:
A1 A0  B1 B0

In M32R, they are placed like this:

Big endian:
A0 A1 B0 B1

Little endian:
B1 B0 A1 A0

This is because M32R always fetches instructions in 32-bit.
So the breakpoint should be placed at buf[2].

Kei Sakamoto