I have done some searching, but didn't find the answer
The code:
char b = 'b';
char c = 'c';
char a[5] = "";
a[0] = b, c;
What last line means? The b, c part?
Thank you all
Asked
Active
Viewed 85 times
-2
A_S
- 7
- 2
-
`a[0] = b, c;` This will assign `c` to `a[0]`. See also this question: http://stackoverflow.com/questions/52550/what-does-the-comma-operator-do-in-c – πάντα ῥεῖ Jun 11 '14 at 07:12
-
3@πάνταῥεῖ No, it won't. – Jun 11 '14 at 07:12
-
3@πάνταῥεῖ It evaluates `a[0] = b` and discards the result, then evaluates `c` and discards the result. – Jun 11 '14 at 07:14
-
@hvd Ah yes, you're right! Operator precedence ... – πάντα ῥεῖ Jun 11 '14 at 07:15
1 Answers
5
That uses the elusive comma operator to cause confusion.
It evaluates b, and result of that is then asssigned to a[0]. After that, c is evaluated but its value thrown away. At least this is the case in C.
The comma has lower precedence than assignment (see this handy table) which is extra confusing.
unwind
- 391,730
- 64
- 469
- 606
-
-
2
-
I had it wrong first, but I double-checked since I got cautious, corrected and linked in the reference material. – unwind Jun 11 '14 at 07:14
-
Yes b is assigned to a[0] but are you sure c is evaluated last? I think that is undefined if the comma is not operator-overloaded. – Patrick Fromberg Jun 11 '14 at 07:35
-
-
@unwind, I looked in the reference and it is actually left to right in C++ too. I wonder since when or I just dreamed it was the other way round once. Or it was only in parameter lists. – Patrick Fromberg Jun 11 '14 at 07:55