C puzzle; multiplying Char's

Question

Recently my C professor gave us the following puzzle:

char c1, c2, c3;

c1 = 'a';
c2 = 'e';
c3 = c1 * c2;
printf("%c",c3);

Answer: E

However, I'm a bit confused as to how this would be solved intuitively other than having the product already memorized. From what I've researched the int values of the characters 'a' and 'b' are 97 and 101 respectively and 'E' being 69 which is what I'm having a rough time understanding how that result was achieved.

Answer 1

first, 97 * 101 is 9797. (In hex, 0x2645).

and char is 1byte. so 0x2645 become 0x45.

0x45 == 69 == 'E'. That's it.

Answer 2

char c1, c2, c3;

c1 = 'a';
c2 = 'e';
c3 = c1 * c2;

The result is implementation-defined, but likely to be 'E'.

The C language does not specify the numeric values of the characters 'a', 'e', and 'E' (or of any character other than the null character '\0'). For a system that uses an ASCII-based character set, the values are 97, 101, and 69, respectively, but on a system that uses EBCDIC the values will be quite different.

Assuming an ASCII-compatible character set, the values of c1 and c2 will be promoted to int before the multiplication. The result of the multiplication is 9797 decimal, or 0x2645 in hex.

The assignment converts that result from int to char. If char is a signed type (as it commonly is), the result of the conversion is implementation-defined, but it's typically done by discarding all but the low-order bits. If char is an unsigned type, the conversion is well defined, and is reduced modulo CHAR_BIT + 1 (very likely to be module 256).

The value assigned to c3 is probably going to be 0x45, or 69 in decimal, or 'E' as a character (again assuming an ASCII-compatible character set).

So the output is E if a number of assumptions are satisfied:

• The implementation uses an ASCII-based character set (could be ASCII, or Latin-1, or Windows-1252, or Unicode, or ...);
• Either plain char is unsigned, or it's signed and the conversion from int to char discards the high-order bits; and
• CHAR_BIT == 8

And finally:

• Somebody was willing to write silly code that multiplies char values, stores the product back in a char object, and prints the result. Such code may be useful for testing your understanding of C, but it has no practical use (something that's true of a lot of class exercises).

Answer 3

$ bc
bc 1.06.95
Copyright 1991-1994, 1997, 1998, 2000, 2004, 2006 Free Software Foundation, Inc.
This is free software with ABSOLUTELY NO WARRANTY.
For details type `warranty'. 
97 * 101
9797
9797 % 256
69
quit

char is 8 bits, allowing for a maximum number 256. When you multiply 97 and 101, you get 9797. The significant bits in the binary representation of 9797 are lost due to 8-bit arithmetic giving you 69 which is E.