为什么在循环中将j初始化为0而不是其递增值？

Question

Consider the following c code:

int main()
{
    int a[5] = {5, 2, 1, 6, 3}, b[5] = {1, 6, 3, 2, 5}, c[10], i = 0, j = 0, k = 0;
    for (i = 0 ; i < 5 ; i++)
    {
        while (a[i] != b[j])
            j++;
        c[k]   = a[i];
        c[k+1] = b[j];
        k      = k + 2;
    }
    for (i = 0 ; i < 10 ; i += 2)
        printf("%d->%d\n", c[i], c[i + 1]);
    getch();
}

The program prints two same numbers each choosen from a[5] and b[5]

Q: j is initialized only once and in the loop the value of j gets incremented, so it may get incremented beyond 5 as no again initialization of j takes place inside the loop, hence the o/p should be some garbage value, but it is not? Why?

Answer 1

j contains 0 when entering the first for loop. It becomes 4 when exiting the while loop as only then will the condition a[i] != b[j] be false.

Then, in the next iteration of the first for loop, j gets incremented and you try to read past the array ( b[5] , b[6] etc) and this invokes Undefined Behavior which means that anything can happen.

The reason that it worked perfectly is by pure luck. But you cannot rely on this.

Answer 2

If you print the address of the array elements whose values match, you can see the truth of @Marian's comment, that j indexes the same array as i does, after the first match.

#include <stdio.h>

int main()
{
    int a[5] = {5, 2, 1, 6, 3}, b[5] = {1, 6, 3, 2, 5}, c[10], i = 0, j = 0, k = 0;
    for (i = 0 ; i < 5 ; i++)
    {
        while (a[i] != b[j])
            j++;
        printf ("%p %p\n", (void*)&a[i], (void*)&b[j]);   // added a cue
        c[k]   = a[i];
        c[k+1] = b[j];
        k      = k + 2;
    }
    for (i = 0 ; i < 10 ; i += 2)
        printf("%d->%d\n", c[i], c[i + 1]);
    getch();
}

Program output

0018FF2C 0018FF24
0018FF30 0018FF30
0018FF34 0018FF34
0018FF38 0018FF38
0018FF3C 0018FF3C
5->5
2->2
1->1
6->6
3->3