I came across this C code, compiled and executed it online on an online GCC compiler and received -5 as output. I changed the values of the constants and received different results but couldn't figure out the logic behind the outputs. Please help...
#include<stdio.h>
int main()
{
int a = -10;
a = a>>1;
printf("%d", a);
return 0;
}
In binary notation with two's complement, a looks like this:
111..1110110
Now, a right shift is done. For signed integers it is implementation specific what value the filled bits have; GCC promises "sane behavior", that is, to do an arithmetic right shift - where the sign bit (here 1) is extended.
An arithmetic right shift on an integer is dividing that integer by 2^n (n is the shift size) - no matter the sign.
Therefore the shift produces the new value:
111..1111011
Which is -5. Flipping all bits and adding one yields 000...00101, which is 5.
A logical shift would have produced
011..1111011
which has the value 2147483643 for a 32-bit integer. Note how the value even depends on the size of the integer you performed the operation on.
>> is right shift operator.
>>1 is shifting right by 1 bit. You can find a detailed discussion here. For the results, the right-shift is equivalent to a division by 2.
The >> Is the right shift operator. It shifts the bits to the right .
And in this occasion ,it shifts by 1 bit.
Check here and here to study more.
Also , you can find here a calculator.
Finally , as "Alter Mann" mentioned , a bitwise right-shift will be the equivalent of integer division by 2.
According to the C Standard (6.5.7 Bitwise shift operators)
5 The result of E1 >> E2 is E1 right-shifted E2 bit positions. If E1 has an unsigned type or if E1 has a signed type and a nonnegative value, the value of the result is the integral part of the quotient of E1 / 2E2. If E1 has a signed type and a negative value, the resulting value is implementation-defined.
it seems that your implementation promotes the sign bit..
So -10 can be represented as (for simplicity I assume that sizeof( int ) is equal to 4)
-10
11111111 11111111 11111111 11110110
10 >> 1
11111111 11111111 11111111 11111011
The last value is equal to -5. To be sure that it is equal to 5 you can get its two-complement value that is calculated as
~a + 1
a
11111111 11111111 11111111 11111011
~a
00000000 00000000 00000000 00000100
~a + 1
00000000 00000000 00000000 00000101
0101 is a binary representation of 5.
The '>>' operator is a bitwise right shift operator (http://en.wikipedia.org/wiki/Bitwise_operation#Bit_shifts), which effectively does a 'divide-by-2'.
The
a = a >> 1;
statement shifts the number one bit to the right.
this is a bit operation a = 01010 when you make a>>1 you say to your program shift one bit to right and yoour result is 00101.
>> means "right shift" of the bits of an integer value. It is equivalent to "/ 2" (divide by 2).
The binary representation of 10 is 1010. The binary representation of -10 is it's 2-complement: 11110110 (assuming 8 bits). Right shifting 1 bit becomes 11111011, which is the 2-complement of 101, or 5 in decimal. So, (-10 >> 1) == -5.
-10 = 11110110
If you do a>>1 then a = -5
-5 = 11111011
If you do a>>2 then a = -3
-3 = 11111101
Hence, >> is shifting the binary number to one bit right with adding 1 at the starting
