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void main()
{
int i=5;
printf("%d",i+++++i);
}
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Answer / vinay,
:) its just how the Compiler parses things..
the..maximum matching (of a token) principle... from the
left..
1. i++ is a valid maximum match. Good, next
2. + match, next (expects a + or a identifier,for furthur
match)
3. + (this is not a identifier, but a + will do so:
match=++). Next the parser wants an indentifier.. else
compiler flags an error..
4. + (not an identifier.. so.. fails)
| Is This Answer Correct ? | 0 Yes | 0 No |
Answer / ashwini
This gives an error. The error is as below:
error C2105: '++' needs l-value
if we correct the code to printf("%d", i++ + ++i);
We get the answer as 12...
The above printf becomes a statement when it sees the
semicolon. Unary operators have right to left associativity.
So, ++i is evaluated first to 6. Then, i++ is evaluated. i++
contributes 6 to addition and then increments i. So, we get
the answer as 12.
| Is This Answer Correct ? | 1 Yes | 1 No |
Answer / shivam shukla
all of you are breaking the statement i+++++i correctly as
i++ + ++i and your answer 12 is also write but you are
explaining it the wrong way......first i++ will make the
value 6 and again a ++i will make it 7 but inn actual
calculation of i++ + ++i....5+7.....will be calculated
hence we get the value 12
| Is This Answer Correct ? | 0 Yes | 0 No |
Answer / ashvin solanki (srimca college
There will be a compilation error .....
| Is This Answer Correct ? | 0 Yes | 0 No |
Answer / sulochana
This statement is just i=i++ + ++i;
Initially i=5
i++ increments after the statement completed For now its
value is 5.
++i increments before its execution.so it is 6
It executes like
i=5+6; i.e. i=11
| Is This Answer Correct ? | 2 Yes | 8 No |
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