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General Tech Learning Aids/Tools . 2 years ago
Posted on 16 Aug 2022, this text provides information on Learning Aids/Tools related to General Tech. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.
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Answers (2)
Actually i am a kickbhut in C. I just started to learn Java. And directly preparing for OCJP6 Certification. In Kathy-Sierra book, and as well as in exam syllabus also, there's no Operator's precedence matter. But i was in fully confused when i saw the Java Operators precedence table from orablce-sun Documentation.
From my C-Language: Operators(44) precedence table 1) () [] .(dot) 2)Unary: ++pre/--pre, -, +, (cast) 3)Arithmetic: 4)bitshift 5)relational: 6)bitwise: 7)logical: 8)ternary Operator ?: 9)Assignment operators = += -== etc 10) unary post++/post-- //In C if i say.. int main() { int i = 1, j; j = i++; /* The above expression is solved based on the operators precedence! 2 operators i have used! one is unary post, another is assignment. here, assignment is higher precedence then unary post. So, first i value is assigned to j. then, i value is incremented because of post increment operator! */ printf("i = %d, j = %d", i, j);// i = 2, j = 1 return 0; } Java Operators precedence table 1)postfix expr++ expr-- 2)unary ++expr --expr +expr -expr ~ ! 3) arithmetic 4)shift << >> >>> 5)relational < > <= >= instanceof equality == != 6)bitwise AND & bitwise exclusive OR ^ bitwise inclusive OR | 7)logical AND && logical OR || 8)ternary ? : 9)assignment = += -= *= /= %= &= ^= |= <<= >>= >>>= //What my biggest doubt is...! class Test { public static void main(String[] args) { int i = 1, j; j = i++; /* according to the operators precedence table! unary post operator is 1st precedence than assignment operator! This way first i value should be increment! then after assignment should happen! How come here also i am getting the same values as in C language? */ //System.out.printf("i = %d, j= %d", i, j); //i = 2, j = 1 System.out.println("i = "+i+", j = "+j); // i = 2, j = 1 } }
Please some one clarify me!
I think you are confused as to what precedence means. It doesn't mean order of execution in all cases. It means order of nested or association e.g. it means that
j = i++
is the same as
j = (i++)
not
(j = i)++
like
a = b + c * d
is
(a = (b + (c * d)))
As Jacob notes: the ++ means increment this value after using or saving the original value. In Java this is always done at the end, whereas in C and C++ it is not defined as to when this will happen.
++
EDIT: A more complex example is as follows
int i = 3; int j = 4; int k = i-- * j++; // same as int k = i * j; i--; j++; System.out.println(k);
prints
12
and it is not the same as
int i = 3; int j = 4; int k = (i = i - 1) * (j = j + 1); System.out.println(k);
which prints
10
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