Multiple assignment in one line

Remember that assignment is done right to left, and that they are normal expressions. So from the compilers perspective the line

sample1 = sample2 = 0;

is the same as

sample1 = (sample2 = 0);

which is the same as

sample2 = 0;
sample1 = sample2;

That is, sample2 is assigned zero, then sample1 is assigned the value of sample2. In practice the same as assigning both to zero as you guessed.

Formally, for two variables t and u of type T and U respectively

T t;
U u;

the assignment

t = u = X;

(where X is some value) is interpreted as

t = (u = X);

and is equivalent to a pair of independent assignments

u = X;
t = (U) X;

Note that the value of X is supposed to reach variable t "as if" it has passed through variable u first, but there's no requirement for it to literally happen that way. X simply has to get converted to type of u before being assigned to t. The value does not have to be assigned to u first and then copied from u to t. The above two assignments are actually not sequenced and can happen in any order, meaning that

t = (U) X;
u = X;

is also a valid execution schedule for this expression. (Note that this sequencing freedom is specific to C language, in which the result of an assignment in an rvalue. In C++ assignment evaluates to an lvalue, which requires "chained" assignments to be sequenced.)

There's no way to say whether it is a good or bad programming practice without seeing more context. In cases when the two variables are tightly related (like x and y coordinate of a point), setting them to some common value using "chained" assignment is actually perfectly good practice (I'd even say "recommended practice"). But when the variables are completely unrelated, then mixing them in a single "chained" assignment is definitely not a good idea. Especially if these variables have different types, which can lead to unintended consequences.