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what is the output of this? cout<<(1< <1) & how? please explain....

22nd May 2018, 9:02 AM
Din Mohammad Dohan
Din Mohammad Dohan - avatar
3 Respuestas
+ 2
Mathematically, bitwise left shift ('<<' operator, as previously stated, shift bits of the left hand value by nth position specified at right hand value) is equivalent to multiply the left hand value by 2 raised to the power of the right hand value (because each one digit left shift of number in base 2 multiply it by 2, as one digit left shift of number in base 10 multiply it by 10): a << b ... is equivalent to: a * std::pow(2,b) ... but with more efficiency (quickest). Obviously, you need to include <cmath> header to be able to use (call) the pow() function ^^ So (with 'using namespace std;'): 1 << 1 == 1 * pow(2,1) == 1 * 2 == 2
22nd May 2018, 10:21 AM
visph
visph - avatar
+ 1
It's cout << ( 1 << 1 ) In the machine the 1 is represented as 0000 0001. And the operation << moves all bits on the left by how places you want 1 -> 0000 0001 1 << 1 = [ 1 shift all by one position ] result -> 0000 0010 Which is 2 for the humans
22nd May 2018, 9:13 AM
Andrei Voinea
Andrei Voinea - avatar
0
answer is 2 This is called left shift this is what happened there :- 1 = 0000 0001 (base 2 form of 1) so when you do 1<<1, entire base 2 digits are shifted left once and 0 is added to the end so it becomes, 1 << 1 = 0000 0010 left part is converted into base 2 form, while the right part specifies number of bits to shift and '<<' or '>>' specifies the direction of shift eg: 3 << 2 base 2 form of 3 = 0000 0011 no. of bits to shift = 2 direction of shift = left base 2 form after shift = 0000 1100 numerical form after (base 10) = 12 numerical form of the number after shift was displayed when you ran the program which in this case is 12
22nd May 2018, 9:27 AM
‎ ‏‏‎Anonymous Guy