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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 |
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What will be the output- for(i=1;i<=3;i++) { printf("%d",i); continue; i++; }
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