Compare with size_t, return int?
I'm writing some code examples from "How to Think Like a Computer Scientist in C++", and this one is about handling playing-card type objects and decks. I'm facing this situation:
int Card::find(const std::vector<Card>& deck) const {
size_t deckSize = deck.size();
for (size_t i=0; i<deckSize; i++)
if (equals(*this, deck[i])) return i;
return -1;
}
I couldn't use ".length()" on a vector in C++ in Visual Studio 2010 as in the text, and instead had to use .size() which returns (I believe) std::size_type. I figured I could use size_t and get away with it in order to avoid problems on different architectures, as I've been reading, but I'm wondering if I return i
, but it's bigger than an integer, will I crash the program?
[Edited to be more specific in my question:] Once I start using vectors for larger things than cards, I considered using unsigned int because of a compiler mismatch warning, but I feel returning an unsigned int or int has a few issues: 1) int will not take a sufficiently large vector index. 2) returning unsigned int will not let me return -1. 3) unsigned int isn't equal to size_t on all architectures (I'm also doing microcontroller programming on an ARM Cortex-M3).
What should I d开发者_运维技巧o if I ever have a large enough vector?
Casting from size_t
to int
will not "crash" your program, but it's a bad bad practice. On the other hand, STL includes nice find
algorithm for what you are doing.
int is 32 bit on 32 / 64 bit Windows and Linux. i will get truncated if greater than two at the 31st. you could use unsigned int and your program will be fine unless storing more than 4 G elements in the vector :)
size_t
is typically an unsigned int
but you can't rely on that. If it's bigger than an int
you won't crash, you'll just overflow into a (probably negative) number.
Assuming you're not going to have several tens of thousands of cards in one vector, I'd be happy returning the int
.
You can also return std::pair<size_t, bool>
, similar to std::map insert()
. Second template argument means success or fail.
If you ok with this, you could also use boost::optional
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