Data type promotions during arithmetic operations: -1 < (unsinged int) 1 == false

问题

main()  {   
  if ( -1 < (unsigned char) 1 )
    printf("less than");
  else        
    printf("NOT less than");
} 

Prints less than. Because, (unsigned char) 1 is converted to (signed char) 1 and then: (signed) -1 < (signed) 1, thus output is less than.

But if I change the above code to if ( (-1 < (unsigned int) 1 )

then the output is NOT less than.

So it's obvious that when I change unsigned char to unsigned int:

• (signed) -1 is converted to unsigned int [exactly opposite is happening]
• since -1 is stored as 2's compliment of 1; the bit-pattern is evaluated as 255 (probably)
• thus 255 < 1 will evaluate to false and else will execute.
• even if you substitute int a = -1; in place of '-1' same result

Questions:

1. during signed and unsigned arithmetic...how to be sure if signed will be converted to unsigned or vice versa.

2. why is conversion different for arithmetic between unsigned char and char : apparently unsigned is converted to signed and unsigned int and int : apparently signed is converter to unsigned

PS: I know this is not compiler dependent..so don't say it is.

回答1:

The rules are as follows:

6.3.1.8 Usual arithmetic conversions

...

Otherwise, the integer promotions are performed on both operands. Then the following rules are applied to the promoted operands:

1. If both operands have the same type, then no further conversion is needed.
2. Otherwise, if both operands have signed integer types or both have unsigned integer types, the operand with the type of lesser integer conversion rank is converted to the type of the operand with greater rank.
3. Otherwise, if the operand that has unsigned integer type has rank greater or equal to the rank of the type of the other operand, then the operand with signed integer type is converted to the type of the operand with unsigned integer type.
4. 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, then the operand with unsigned integer type is converted to the type of the operand with signed integer type.
5. Otherwise, both operands are converted to the unsigned integer type corresponding to the type of the operand with signed integer type.

The rules then work as follows:

• -1 < (unsigned char) 1

First both operands are converted to ints (because an int can represent all values of unsigned char). Then the comparison is made on these signed types. Rule 1 is then used. The comparison succeeds.

• -1 < (unsigned int) 1

An int cannot represent all the values of an unsigned int so rule 3 is used and the signed integer is converted to an unsigned integer (UINT_MAX - 1). The comparison now fails.

回答2:

This is due to integer promotions. Both arguments can be represented as an int, so they are converted to an int.

ISO C 6.3.1.1, paragraph 2:

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.

来源：https://stackoverflow.com/questions/3900385/data-type-promotions-during-arithmetic-operations-1-unsinged-int-1-fals