Sequence Point - Xor Swap on Array get wrong result

I learned using Xor operator to swap two integers,like:
int a = 21;
int b = 7;
a^=b^=a^=b;
I would finally get a=7 and b=21.

I try to use xor operator on array like this way:

int main() { int a[] = {7,21}; a[0]^=a[1]^=a[0]^=a[1]; cout << a[0] <<',' <<a[1]; return 0; } 

The output is 0,7
I compile the code on Xcode and g++, they have the same issue.

Xor swap on array works fine with multiple lines:

int main() { int a[] = {7,21}; a[0]^=a[1]; a[1]^=a[0]; a[0]^=a[1]; cout << a[0] <<',' <<a[1]; return 0; } 

I would get output as 21,7

Here is the information what I already find:
- the issue is about sequence point: Array + XOR swap fails
- even for simple integers, they may have side affect to this undefined behavior: Why is this statement not working in java x ^= y ^= x ^= y;
- some other issue on xor swap: Weird XOR swap behavior while zeroing out data

So I should avoid using xor swap, instead, swap with temp would guarantee correct result.

But I still not very clear about what happen on a[0]^=a[1]^=a[0]^=a[1]; what is the sequence point issue with it?

I could not figure out what's the different on compiler between a[0]^=a[1]^=a[0]^=a[1]; and a^=b^=a^=b; ?

My doubt is:
" How does compiler output 0,7 for a[0]^=a[1]^=a[0]^=a[1];. "

I know this is sequence pointer issue, I could understand why printf("%d,%d",i++, i++); is undefined as some compiler parse parameter of function from left to right, and some do it from right to left.

But I do not know what is the problem on a[0]^=a[1]^=a[0]^=a[1];, it looks just the same as a^=b^=a^=b;. So I'd like to know how it works with array. So that I would know more about kind like "sequence pointer on index of array"