It storesPOC_P_Status intodata.

i = a, b;   // stores a into i.

This is equivalent to

(i = a), b;

because the comma operator has lower precedence than assignment.

It's equivalent to the following code:

data = POC_P_Status;TE_OK;

In other words, it assignsPOC_P_Status todata and evaluates toTE_OK.In your first case, the expression stands alone, soTE_OK is meaningful only if it's a macro with side effects. In the second case, the expression is actually part of anif statement, so it always evaluates to the value ofTE_OK. The statement could be rewritten as:

data = POC_P_Status;if (TE_OK) { ... }

From the C11 draft (http://www.open-std.org/jtc1/sc22/wg14/www/docs/n1124.pdf) :

The left operand of a comma operator is evaluated as a void expression; there is a sequence point after its evaluation. Then the right operand is evaluated; the result has its type and value. If an attempt is made to modify the result of a comma operator or to access it after the next sequence point, the behavior is undefined.

That means that in the expression:

a, b

Thea is evaluated and thrown away, and thenb is evaluated. The value of the whole expression is equal tob:

(a, b) == b

Comma operator is often used in places where multiple assignments are necessary but only one expression is allowed, such asfor loops:

for (int i=0, z=length; i < z; i++, z--) {    // do things}

Comma in other contexts, such as function calls and declarations, isnot a comma operator:

int func(int a, int b) {...}              ^              |              Not a comma operatorint a, b;     ^     |     Not a comma operator

• c
• comma-operator