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

Answers were Sorted based on User's Feedback



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

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

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

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

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

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

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

Answer / mehtaashish

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

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

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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