Q:

Arithmetic conversion VS integral promotion

char cval;
short sval;
long lval;
sval + cval; // sval and cval promoted to int
cval + lval; // cval converted to long

This is a piece of code on C++ Primer.
I know sval+cval generates an int type according to 

convert the small integral types to a larger integral type. The types
  bool, char, signed char, unsigned char, short, and unsigned short are
  promoted to int if all possible values of that type fit in an int.

But for the last one I couldn't understand why it uses "converted". Why is cval not promoted to int first and then the int converted (or maybe promoted I'm not sure whether promoted can be used from int to long because I only see definition of promotion on smaller type to int) to long. I didn't see any explanation or examples on char straightly to long in that part of the book.   Is there any thing wrong with my understanding?  
I'm quite new at C++, someone please enlighten me! Many thanks in advance!

A:

The additive operators perform what is called the usual arithmetic conversion on their operands which can include integral promotions and then after that we can have further conversions. The purpose is to yield a common type and if the promotions do not accomplish that then a further conversion is required.
This is covered in section 5 [expr] of the draft C++ standard which says (emphasis mine):

Many binary operators that expect operands of arithmetic or enumeration type cause conversions and yield
  result types in a similar way. The purpose is to yield a common type, which is also the type of the result.
  This pattern is called the usual arithmetic conversions, which are defined as follow

and includes the following bullet:

Otherwise, the integral promotions (4.5) shall be performed on both operands.61 Then the following
  rules shall be applied to the promoted operands:

which has the following bullets:

If both operands have the same type, no further conversion is needed
Otherwise, if both operands have signed integer types or both have unsigned integer types, the
  operand with the type of lesser integer conversion rank shall be converted to the type of the
  operand with greater rank.
Otherwise, if the operand that has unsigned integer type has rank greater than or equal to the
  rank of the type of the other operand, the operand with signed integer type shall be converted to
  the type of the operand with unsigned integer type.
Otherwise, if the type of the operand with signed integer type can represent all of the values of
  the type of the operand with unsigned integer type, the operand with unsigned integer type shall
  be converted to the type of the operand with signed integer type.
Otherwise, both operands shall be converted to the unsigned integer type corresponding to the
  type of the operand with signed integer type.

So in the first case after promotions they both have the same type(int) so no further conversion is needed.
In the second case after promotions they do not(int and long) so a further conversion is required.

