What will be the output-
for(i=1;i<=3;i++)
{
printf("%d",i);
continue;
i++;
}
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Answer / gourab
the o/p will be 123
bcz after printing 1 for d first time then continue will
take to the for statement.thus i will take value 3.and i++
after d continue statement will never get executed,but it
will not give an error.
| Is This Answer Correct ? | 32 Yes | 2 No |
Answer / chetan
This will not result in a Compiler-error, because "Unreachable-code" is a warning in C, not a compilation error.
The output will be:
123
| Is This Answer Correct ? | 6 Yes | 2 No |
Answer / harshit
This will result in compiler error as the code i++; inside
the for loop is not reachable.
If continue statement is enclosed within an if construct
then only the program will compile.
| Is This Answer Correct ? | 14 Yes | 13 No |
It will results in a compile time error, stating that
unreachable code for the second increment statement(i++) after
continue statement.
| Is This Answer Correct ? | 11 Yes | 10 No |
Answer / sakshi arora
The output will be : 123
bcozZ unreachable code is a warning in c , not a compilation error..
| Is This Answer Correct ? | 1 Yes | 0 No |
What is the time complexity T(n) of the following program? a) int n, d, i, j; cin >> n; for (d=1; d<=n; d++) for (i=1; i<=d; i++) for (j=1; j<=n; j += n/10) cout << d << " " << i << " " << j << endl; b) void main() { int n, s, t; cin >> n; for (s = 1; s <= n/4; s++) {t = s; while (t >= 1) { cout << s << " " << t << endl; t--; } } } c) void main() { int n, r, s, t; cin >> n; for (r = 2; r <= n; r = r * 2) for (s = 1; s <= n/4; s++) { t = s; while (t >= 1) { cout << s << " " << t << endl; t--; } } }
3 Answers CTS, IBM, Infosys, Qatar University,
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