Override the odd sum between 1.

This time I made two changes to this code than the previous one. The first change was defined using the macro command # define. Thanks to @ garbageMan for this syntax.

The second change changes the first I + 2 to be determined using the if statement.

The Code is as follows:

         # include <stdio.h> # define BEGIN  # define END   main(            sum =      (i=BEGIN;i<=END;++           (i%==)          sum +=i;       printf(                

Code running step thinking, the sequence is 16 rows of I = BEGIN is 1, I <= END is 2, if step 3, 1% 2 = 1 of the 18 rows are run, sum + = I is run, and then the 16 rows of ++ I are run to end with one loop. (If 1% 2 = 1 is not valid, directly execute ++ I)

For ease of Self-calculation, change I <= END to I <= 10

1 -----> when I is 1, less than or equal to the END of STEP (2), perform step (3) I % 2 = 1, and run sum + = I. Sum is 0 plus I value 1, sum becomes 1, I is 1 + + I, I is 2. The first cycle ends.

2 -----> when I is 2 and is less than or equal to the END of STEP (2), the execution of STEP (3) I % 2 = 1 is not true, and the execution of ++ I, the second cycle ends. I is 3 at this time

3 -----> when I = 3, if I is less than or equal to the END of STEP (2), perform step (3) I % 2 = 1, and run sum + = I. Sum has changed to 1 in the first loop, and the value of I is 3. The value of sum after the third loop is 4, I is 3, and after ++ I, I is 4, the third cycle ends.

4 -----> when I = 4, if I is less than or equal to the END of STEP (2), the execution of STEP (3) I % 2 = 1 is not true, and the execution of ++ I, the fourth cycle ends. I is 5 at this time

5 -----> when I = 5, less than or equal to the END of STEP (2), perform step (3) I % 2 = 1, and run sum + = I. Sum has changed to 4 in the third loop, and the value of I is 5. The value of sum after the fifth loop is 9, I is 5, and after ++ I, I is 6, the fifth cycle ends.

6 -----> when I = 6, if I is less than or equal to the END of STEP (2), the execution of STEP (3) I % 2 = 1 is not true, and the execution of ++ I, the sixth cycle ends. I is 7 at this time

7 -----> when I = 7, if it is less than or equal to the END of STEP (2), perform step (3) I % 2 = 1 and run sum + = I. Sum has changed to 9 in the fifth cycle, and the value of I is 7. The value of sum after the seventh cycle is 16, and I is 7. After ++ I, I is 8, the seventh cycle ends.

8 -----> when I = 8, if I is less than or equal to the END of STEP (2), the execution of STEP (3) I % 2 = 1 is not true, and the execution of ++ I, the eighth cycle ends. I is 9 at this time

9 -----> when I = 9, if it is less than or equal to the END of STEP (2), perform step (3) I % 2 = 1 and run sum + = I. Sum has changed to 16 in the seventh cycle, with the I value 9, the sum value after the ninth cycle is 25, I is 9, after ++ I, I is 10, the ninth cycle ends.

10 ----> when I = 10, if I is less than or equal to the END of STEP (2), the execution of STEP (3) I % 2 = 1 is not true, and the execution of ++ I, the tenth cycle ends. I is 11 at this time

11 ----> when I = 11, the END less than or equal to step (2) is not valid and the for loop is stopped.

Bytes -----------------------------------------------------------------------------------

The odd number between the end result 1-10 is 25.

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If the above Code is incorrect, please see this blog's friends point out that your dial is the greatest help for me to learn.