CODEWITHSUNDEEP
ccastinginteger-promotion
golark
golark

Reputation: 27

integer promotion and unsigned interpretation

int_8 int8     = ~0;
uint_16 uInt16 = (uint_16) int8;

Regarding the typecast above; where in C standard can I find reference to an indication for the following behaviour? - sign extension to the larger type before the unsigned interpretation (uInt16=0xFFFF) rather than unsigned interpretation followed by 0 extension to the larger type (uInt16=0xFF).

From C99 6.3.1.8

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.

Above statement is clear about which variable needs to be converted however it is not very clear about how the conversation should actually be performed hence my question asking for a reference from the standard. Thanks

Upvotes: 1

Views: 300

Answers (2)

Eugene Sh.
Eugene Sh.

Reputation: 18299

As per the standard:

6.3.1.3 Signed and unsigned integers

......
2. Otherwise, if the new type is unsigned, the value is converted by repeatedly adding or subtracting one more than the maximum value that can be represented in the new type until the value is in the range of the new type.

And the footnote to avoid the confusion when interpreting the above:

The rules describe arithmetic on the mathematical value, not the value of a given type of expression.

I.e. if your int8 has a value of -1 (assuming the negatives representations is 2's complement, it does in your example), when converted into uint16_t, the value (0xFFFF + 1) will be added to it (which one more than the max value that can be represented by uint16_t), which yields the result of 0xFFFF + 1 - 1 = 0xFFFF.

Upvotes: 3

golark
golark

Reputation: 27

Answer I believe is actually part of 6.3.1.8 as well:

Otherwise, the integer promotions are performed on both operands. Then the following rules are applied to the promoted operands: .... Otherwise, if the operand that has unsigned integer type has rank greater or equal to the rank of the type of the other operand.....

meaning that integer promotions are performed first before the conversion to unsigned using the rule 6.3.1.3.

Upvotes: 0

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