Assignment operator to assign expressions separated by comma in C++98 to a variable does not assign

Question

Using Android Marshmallow CppDroid app, which uses C++98 but supports some C++11.

I'm following a C++ course, and I'm confused by the output of the following:

int a=0, b=0, c=0;
std::cout << "a:" << a << '\n' << "b:" << b << '\n' << "c:" << c << '\n';
// output is 
// a:0
// b:0
// c:0
a = b++, c++
std::cout << "a:" << a << '\n' << "b:" << b << '\n' << "c:" << c << '\n';
// Output is 
// a:0
// b:1
// c:1

I'm confused as to why the assignment operator did not assign the value of c++ to a.

EDIT

As noted by the comment by @molbdnilo, this is not a matter of precedence, but rather a matter of the values equaling 0 because of the increment operator c++ used instead of ++c.

int a=0, b=0, c=0;
cout<<"a:"<<a<<'\n'<<"b:"<<b<<'\n'<<"c:"<<c<<"\n\n";
// output is 
// a:0
// b:0
// c:0
a = ++b,++c;
cout<<"a:"<<a<<'\n'<<"b:"<<b<<'\n'<<"c:"<<c<<"\n\n";
// output is 
// a:1
// b:1
// c:1

`=` has a higher precedence than `,`. See http://en.cppreference.com/w/cpp/language/operator_precedence — Richard Critten, Apr 08 '17 at 09:32
Thank you very much. Suppose it is best to consider precedence when using operators. — scriptkiddie, Apr 08 '17 at 09:36
@molbdnilo You are correct, because of operator precedence as noted by Richard Critten. The course I was taking sure didn't explain that. So, -1 to them, and +1 to Richard. — scriptkiddie, Apr 10 '17 at 06:16
@envisisoft I think you're missing my point. Precedence has no visible effect on the result in this example – if you write just `a = c++;`, `a` will also be `0`. — molbdnilo, Apr 10 '17 at 07:17
@molbdnilo I see what you mean. If i assign `++c` to `a`, then it increments by 1. — scriptkiddie, Apr 10 '17 at 09:49

Assignment operator to assign expressions separated by comma in C++98 to a variable does not assign

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