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#include<stdio.h>
main()
{
int a=1;
int b=0;
b=++a + ++a;
printf("%d %d",a,b);
}
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Answer / banavathvishnu
let consider the statement
b = ++a + ++a;
++a will be 2
++a again will be 3
now replace its value in the expression
b = a + a = 3+3=6
hence a is 3 and b is 6
| Is This Answer Correct ? | 26 Yes | 12 No |
Answer / aditya gupta
Let me make this more clear... infact if the case is of
pre-increment:
1- find all the variables of pre-increment, and compute them
2- do the assignment.
for example, what I do:
main()
{
int a=1; // initialization
int b=0; // initialization
b=++a + ++a; // find the pre-increment i.e. 2 increments of
'a' so now 'a' in this step will be incremented by 2 so now
'a' will contain 1+2=3. so now a=3. Again before assignment
compute 'a+a' which is '3+3'=6
printf("%d %d",a,b); //3 6
}
Just a trick:- always compute the pre-increments in the same
step...
If I say b= ++a + ++a; answer is 3 and 6
If I say b= ++a + a++; answer is 3 and 4 because in this
line one pre-increment is there. So now '++a + a++'= "2 + 2"
Thanks!!
Aditya Gupta
| Is This Answer Correct ? | 8 Yes | 2 No |
Answer / ashok
initially a=1,b=0
++a=2 //1+1=2
++a=3 //2+1=3
b=2+3=5
answer:a=3 b=5
| Is This Answer Correct ? | 6 Yes | 4 No |
Answer / anand
answer should be 3 5
b = 2 + 3
b = ++a + ++a
here the compiler will work as below
b = ++a + 2
thn
b = 3 + 2
thn
b = 5
| Is This Answer Correct ? | 9 Yes | 10 No |
Answer / vijay r15
ans 3 6
Let me explain
First a=1&b=0
b=++a + ++a;
The operation will be as
b= ++1 + ++a
=2 + ++a
=2 + ++2
=2 + 3=a+a now a=3
Remember here is the trick
Now b= a + a
I.e b=3+3=6
Got it
Vijay r15
For any clarification mail to
raj.vijay55@gmail.com
| Is This Answer Correct ? | 2 Yes | 4 No |
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