CODEWITHSUNDEEP
cbinarybit-manipulationbit-shift
xdevel2000
xdevel2000

Reputation: 21444

Left shift operator

If I have the following:

char v = 32; // 0010 0000

then I do: 

v << 2

the number becames negative. // 1000 0000 -128

I read the standard but it is only written:

If E1 has a signed type and nonnegative value, and E1 × 2 E2 is representable in the result type, then that is the resulting value; otherwise, the behavior is undefined.

so I don't understand if is a rule that if a bit goes on most left bit the number must begin negative.

I'm using GCC.

Upvotes: 1

Views: 692

Answers (3)

dede9714
dede9714

Reputation: 73

Try using unsigned char instead char uses less bit for representing your character, by using unsigned char you avail more bits for representing your character

unsigned char var=32;
v=var<<2;

Upvotes: 0

ballard26
ballard26

Reputation: 37

Signed data primitives like char use two's complement(http://en.wikipedia.org/wiki/Twos_complement) to encode value. You probably are looking for is unsigned char which won't encode the value using two's complement(no negatives).

Upvotes: 1

legends2k
legends2k

Reputation: 33004

Left shifting it twice would give 1000 0000)2 = 128)10.

If 128 is representable in char i.e. you're in some machine (with a supporting compiler) that provides a char of size > 8 bits then 128 would be the value you get (since it's representable in such a type).

Otherwise, if the size of a char is just 8 bits like most common machines, for a signed character type that uses two's complement for negative values, [-128, 127] is the representable range. You're in undefined behaviour land since it's not representable as-is in that type.

Upvotes: 1

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