Re: [PATCH] integer overflow checking builtins in constant expressions

[Jason Merrill <jason at redhat dot com>]

On 05/01/2016 12:39 PM, Martin Sebor wrote:
> +  if (TREE_CODE (arg0) == INTEGER_CST && TREE_CODE (arg1) == INTEGER_CST)
> +    {
> +      if (tree result = size_binop_loc (EXPR_LOC_OR_LOC (t, input_location),
> +					opcode, arg0, arg1))
> +	{
> +	  if (TREE_OVERFLOW (result))
> +	    {
> +	      /* Reset TREE_OVERFLOW to avoid warnings for the overflow.  */
> +	      TREE_OVERFLOW (result) = 0;
> +
> +	      return build_complex (TREE_TYPE (t), result, integer_one_node);
> +	    }
> +
> +	  return build_complex (TREE_TYPE (t), result, integer_zero_node);
> +	}
> +    }

Should this be in the middle-end somewhere, perhaps shared with 
fold_builtin_arith_overflow?  I notice that the comment for that 
function says that it folds into normal arithmetic if the operation can 
never overflow, but I don't see any code that would accomplish that.

Jason