Data type promotion or demotion in C

Question

Which of these are data type promotion and demotion :

char ch = 'C';
int i = 65;
float fl = 2.2;

ch = ch + 1;
i = fl + 2 * ch;
fl = 2.0 * ch + i;
ch = 5212205.17;

Am I correct if i say :

Promotion :

• i = fl + 2 * ch;

• fl = 2 * ch + i;

demotion :

  ch = 5211205.17;

not sure if

  ch = ch + 1;

Is also demotion.. please help :)

Answer 1

char ch = 'C'; 

The literal 'C' is of type int and demoted to type char (conversion to the left operator during assignment).

float fl = 2.2;

The literal 2.2 is of type double and demoted to type float (conversion to the left operator during assignment).

ch = ch + 1;

The variable ch is of type char and promoted to type int (integer promotions).

The result of the addition ch + 1 is of type int and demoted to type char (conversion to the left operator during assignment).

i = fl + 2 * ch;

The variable ch is of type char and promoted to type int (integer promotions).

The result of 2 * ch is of type int and promoted to type float (balancing).

The result of fl + 2 * ch is of type float. The float is demoted to an int (conversion to the left operator during assignment). This is a dangerous conversion because of loss of precision and a good compiler should give a warning for attempting to store a float inside an int without an explicit cast.

fl = 2.0 * ch + i;

The variable ch is of type char and first promoted to type int (integer promotions) and then promoted to type double (balancing).

The result of 2.0 * ch is of type double.

The result of 2.0 * ch + i is of type double and demoted to type float (conversion to the left operator during assignment).

ch = 5212205.17;

The literal 5212205.17 is of type double and demoted to type char (conversion to the left operator during assignment). This is a dangerous conversion and possibly also undefined behavior, since the signedness of char is implementation-defined, and also the number cannot fit inside a char.

Attempting to store a signed floating point number inside a type that cannot represent it (such as an unsigned int) is undefined behavior, i.e. a severe bug.

Answer 2

In ch = ch+1; Here 1 is an integer so ch is promoted to integer and add to 1 then result is demoted to character and stored in ch.