Does Unary + operator do type conversions?

Till now I was believing that there is no use of unary + operator.

But then I came across with following example:

char ch;
short sh;
int i;

printf("%d %d %d",sizeof(ch),sizeof(sh),sizeof(i)); // output: 1 2 4

printf("%d %d %d",sizeof(+ch),sizeof(+sh),sizeof(i)); // output: 4 4 4

Does it mean + is doing type conversion here?

Because it is behaving same as following

printf("%d %d %d",sizeof((int)ch),sizeof((int)sh),sizeof(i)); // output: 4 4 4

This forces me to think + is doing type conversion.

But then I try it on double

double f;
printf("%d %d %d", sizeof(+f), sizeof((int)f), sizeof(f)); // output: 8 4 8

This forces me to rethink about unary + operator.

So my second question is: does unary + operator has special effect in sizeof operator?

Unary + performs integer promotions on its operand, we can see this by going to the draft C99 standard section 6.5.3.3 Unary arithmetic operators which says (emphasis mine going forward):

The result of the unary + operator is the value of its (promoted) operand. The integer promotions are performed on the operand, and the result has the promoted type.

and section 6.3.1 Arithmetic operands says:

If an int can represent all values of the original type, the value is converted to an int; otherwise, it is converted to an unsigned int. These are called the integer promotions.⁴⁸⁾ All other types are unchanged by the integer promotions.

Note that all other types are unchanged by the integer promotions and thereofore double stays a double. This would also hold for float as well, which would not be promoted to double.

Also note that using %d for the result of sizeof is undefined behavior since the result is size_t. the proper format specifier would be %zu.