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I expected that below code will output 10 because (~port) equal to 10100101 So, when we right shift it by 4 we get 00001010 which is 10. But the output is 250! Why?
int main() { uint8_t port = 0x5a; uint8_t result_8 = (~port) >> 4; //result_8 = result_8 >> 4; printf("%i", result_8); return 0; } 
806
The right shift operator ( >> ) returns the signed number represented by the result of performing a sign-extending shift of the binary representation of the first operand (evaluated as a two's complement bit string) to the right by the number of bits, modulo 32, specified in the second operand.
The bitwise shift operators move the bit values of a binary object. The left operand specifies the value to be shifted. The right operand specifies the number of positions that the bits in the value are to be shifted.
The bitwise shift operators are the right-shift operator ( >> ), which moves the bits of an integer or enumeration type expression to the right, and the left-shift operator ( << ), which moves the bits to the left.
A bit-shift moves each digit in a number's binary representation left or right. Within right-shifts, there are two further divisions: logical right-shift and arithmetic right-shift. A left-shift is represented by the << operator, while a right-shift is represented by the >> operator.
In this article we will see the use of Bitwise Unsigned Right Shift operator in Java programming language. Bitwise Unsigned Right Shift which is represented by >>> symbol. It shifts the bits of a number towards right with specified position.
The bitwise shift operators move the bit values of a binary object. The left operand specifies the value to be shifted. The right operand specifies the number of positions that the bits in the value are to be shifted. The result is not an lvalue.
Table of contents Table of contents Syntax Description Examples Specifications Browser compatibility See also Unsigned right shift (>>>) The unsigned right shift operator (>>>)(zero-fill right shift) shifts the first operand the specified number of bits to the right. Excess bits shifted off to the right are discarded.
Excess bits shifted off to the right are discarded. Zero bits are shifted in from the left. The sign bit becomes 0, so the result is always non-negative. Unlike the other bitwise operators, zero-fill right shift returns an unsigned 32-bit integer. Syntax a >>>b Description
C promotes uint8_t to int before doing operations on it. So:
port is promoted to signed integer 0x0000005a.~ inverts it giving 0xffffffa5.0xfffffffa.uint8_t giving 0xfa == 250.To fix that, either truncate the temporary result:
uint8_t result_8 = (uint8_t)(~port) >> 4; mask it:
uint8_t result_8 = (~port & 0xff) >> 4; or xor it (thanks @Nayuki!):
uint8_t result_8 = (port ^ 0xff) >> 4; If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
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