以下代码的结果是什么？

Question

Why the result is 7??? I can't find the log if that :(

#include <stdio.h>
main() {
    int i, j, a = 1, b = 1;
    for (i = 1; i < 4; i++)
        for (j = 1; j < 3; j++)
            a = a + b;
    printf("a=%d", a);
}

Answer 1

The structure is like below

- Outer `for` loop
 - inner `for` loop
     - instruction

so the "instruction" (statement/block) will get executed inner for loop count times, for outer for loop count times.

What is basically says is, add the value of b to the latest value of a (in a recurring way) to get the current value of a. Now , do that for "outer" number of time, for which, do the same for "inner" number of times.

Outer for loop ==> 3 times,
inner for loop ==> 2 times

So, final value = 3*2 (increment) + (initial) = (3*2)*1 + 1 = 7 .

Answer 2

The result is 7 because b is initialized as 1 and stays 1 the whole time. The outer loop is run 3 times (1, 2, 3), the inner loop is run 2 times (1 and 2), so there are 6 runs where b is added to a (which is initialized as 1). 1 + 6 = 7.

Answer 3

In such scenarios, you should add a watch and debug your code line by line. I believe the shortcut is F11. Regards to why the output of your code is 7... The inner loop runs six times. J loops twice - 1, 2, (ends when it is 3), and I loops thrice (1, 2, 3, 4 - end)... for a total of 2 X 3 = 6.

Since b is '1', you are basically adding the number 1 to a six times. Since a started with '1', the output is:

a = 1 + 1 + 1 + 1 + 1 + 1 + 1 = 7