CODEWITHSUNDEEP
c++bit-manipulation
Deohboeh
Deohboeh

Reputation: 37

C++ NOT bitwise operator binary char conversion

I am studying C++ from Primer 5th edition. One of the questions is:

What is the value of ~'q' << 6 on a machine with 32-bit ints and 8 bit chars, that uses Latin-1 character set in which 'q' has the bit pattern 01110001?

What i am confused with is will the compiler convert the 'q' to 32-bit int before ~ or before <

As far as I understand it should work like so:

'q' = 01110001

~'q' = 10001110

~'q' << 6 = 00000000000000000010001110000000

Am I right?

Upvotes: 2

Views: 708

Answers (2)

Angew is no longer proud of SO
Angew is no longer proud of SO

Reputation: 171117

The operator ~ performs integral promotions on its operand (C++11, [expr.unary.op]§10). Which means it will convert the char to int before doing the complement. So it will go like this:

'q' == 01110001

~ 'q' == ~ 00000000000000000000000001110001

~ 'q' == 11111111111111111111111110001110

Then comes a problem. We see the value of ~ 'q' is negative. Left-shifting a negative value has undefined behaviour ([expr.shift]§2). (Thanks to @colombien's answer for pointing out this last part)

Upvotes: 8

colombien
colombien

Reputation: 149

I guess the answer is here: Bitwise operators and signed types

The expression ~'q' << 6 results in undefined behavior.

Upvotes: 2

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