Comma as separator in variable initialization (not as operator)

user_906839 0
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This seems such a simple question, but something I've not examined for ages in my own style... When initializing variables separated by a comma, I've assumed the following to be an unsafe practice:

unsigned int n = foo.size, nxn = n * n;

Since I don't really ever use the comma operator as such for syntactic sugar, etc; but rather to indicate that two expressions are independent - as a kind of implicit commentary on 'fine-grained parallelism' (or expression independence), that often makes code a bit more terse, e.g.,

if (<some condition>)
    a = true, b = value;

rather than requiring {} scope for semi-colon separated expressions.

But my question is really in re-examining the variable initialization case. Have I been incorrect in my assumption that nxn can't be relied on to be initialized as expected? Or have I been laboring under a misinterpretation all this time?

comma-operator initialization c variables c++
2 Answers

Per [dcl.decl]/3

Each init-declarator in a declaration is analyzed separately as if it was in a declaration by itself. [...]

we get that

unsigned int n = foo.size, nxn = n * n;

is the same as

unsigned int n = foo.size;
unsigned int nxn = n * n;

There is a note with exceptions for other rules like auto or if a name shadows the type but those don't apply in this case.

Be very wary with pointers if you put multiple variables on a single line

int * foo, bar;

does not give you two pointers. Instead, foo is a pointer and bar is an int. You would need

int * foo, * bar;

to get two pointers. For this reason I would prefer to use

int * foo;
int * bar;

and pay the extra keystorkes for safeties sake.

user_43424980

nxn will be initialized properly, since n has been defined and initialized at the point where nxn is defined.

For clarity however, it would be better to put the variables on separate lines. Doing so avoids ambiguity, making your intent more clear to anyone who reads your code.

user_16871190
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