String s = "Hello"; long l = 99; double d = 1.11; int i = 1; int j = 0; 1) j= i <<s;> 2) j= i<<j;> 3) j=i<<d;> 4)j=i<<l;> Answer to Question 46) and it has given the answer is 2,3 compile correctly. how come shifting a number(i=1) 'o' times will give '0' value.
Question doesn't make sense, or maybe you wrote the answers down wrong. I don't see anything with shifting involved. There is not ">>", ">>>", or "<<" anywhere. Bill
what warbled stuff do you write, rajani? pls reframe your question properly. what am i to make of: i<how come it is true?> sorry for being rude but it's the second time you're doing it!
it's impossible to get a 0 value if you shift bits by a 0 00000001 event if you shift by zero you are not moving bits to the left they stay the same as they were. so the answer is incorrect the answer stays 1 as it was.
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