A question About the difference of %hd and %d in scanf() in C

[yyda]

	int y;
	short s;

	scanf("%d %d", &y, &s);
	printf("%d,%d\n", y, s);

the output is gonna be"0,short(s) ".
Obviously the printf() doesn't care %hd or %d,but why y presents "0" instead of s in the case of s received the wrong type and s presents rightly

[dev7060]

        int y;
        short s;
     
        scanf("%d %d", &y, &s);
        printf("%d,%d\n", y, s);

the output is gonna be"0,short(s) ".
Obviously the printf() doesn't care %hd or %d,but why y presents "0" instead of s in the case of s received the wrong type and s presents rightly

short s may or may not alter the bytes occupied by int y (which could be present in adjacent memory locations) and may cause potential corruption. Kind of undefined behavior; the exact outcome depends on the system and compiler. Compilers may pad memory differently or handle misaligned memory accesses in various ways.

Try inputting a value that doesn't fall within the range of short; this might affect both variables. Additionally, consider storing one on the heap and the other on the stack. Do they both appear correct?

[yyda]

Thank you！
So this is an undefined,r?
I tried to store y in heap,it works as I expected.Looks like it's about stack.

[dev7060]

On my machine

Within short range
Input: 5 55
Output: 0 55

Outside short range
Input: 5 5555555
Output: 84 -15005

y on stack, s on heap
Input: 5 5555555
Output: 5 5555555

[yyda]

On my machine

Within short range
Input: 5 55
Output: 0 55

Outside short range
Input: 5 5555555
Output: 84 -15005

y on stack, s on heap
Input: 5 5555555
Output: 5 5555555

Yes！It's the same here.I mean the value is random when out short range.Whether s or y is stored in heap,it works.They may be constant in memory so there is a unexpected interference?