Why does constructor choose type INT instead of SHORT when invoked with a parameter of type CHAR?

The result of integral promotion is int (not short) for char; and promotions (e.g. char -> int) have higher ranking than other conversions (e.g. char -> short) in overload resolution.

prvalues of small integral types (such as char) may be converted to prvalues of larger integral types (such as int).

• signed char or signed short can be converted to int;
• unsigned char, char8_t (since C++20) or unsigned short can be converted to int if it can hold its entire value range, and unsigned int otherwise;
• char can be converted to int or unsigned int depending on the underlying type: signed char or unsigned char (see above);

and (emphasis mine)

Note that all other conversions are not promotions; for example, overload resolution chooses char -> int (promotion) over char -> short (conversion).

The compiler always chooses the best matching overloading resolution.

in your case:

Type promotion is:

1. A char, unsigned char or short can be promoted to an int. For example void f(int); can be a match for f('a');
2. A float can be promoted to a double.
3. A bool can be promoted to an int (FALSE counts as 0, TRUE as 1).

When casting implicitly, the compiler follows this ranking:

1. Exact match
2. Promotion
3. Conversion

Since, char to int is integral promotion, it takes precedence over char to short which is conversion.

From here (emphasis mine):

char can be converted to int or unsigned int depending on the underlying type: signed char or unsigned char