Point out the error line in the following program.1. #include<stdio.h>2. int main()3. {4. int i;5. for(i=5;i<1;i--);6. break;7. for(i=6;;i--)8. {9. if(i=3)10. {11. break;12. }13. }14. for(;;)15. break;16. if(i -= 3)17. printf("%d\n", i);18. return 0;19. }
Question
Point out the error line in the following program.1. #include<stdio.h>2. int main()3. {4. int i;5. for(i=5;i<1;i--);6. break;7. for(i=6;;i--)8. {9. if(i=3)10. {11. break;12. }13. }14. for(;;)15. break;16. if(i -= 3)17. printf("%d\n", i);18. return 0;19. }
Solution
The errors in the program are as follows:
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Line 5: The for loop will not execute because the condition i<1 is false at the beginning. The initial value of i is 5 which is not less than 1.
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Line 6: The break statement is not inside a loop or a switch. It should be inside a loop or a switch to terminate it.
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Line 9: The if statement is using the assignment operator (=) instead of the equality operator (==). It should be if(i==3) to compare i with 3.
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Line 16: The if statement is using the compound assignment operator (-=) which subtracts 3 from i and then checks if i is non-zero. If you want to compare i with 3, it should be if(i == 3).
So, the error lines are 5, 6, 9, and 16.
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