Why the static_cast on time(0) on this code

Question

I'm currently reading a beginner book on C++ and it has this example:

int main()
{
    srand(static_cast<unsigned int>(time(0))); //seed random number generator
    int randomNumber = rand(); //generate random number
    int die = (randomNumber % 6) + 1; // get a number between 1 and 6
    cout << "You rolled a " << die << endl;
    return 0;
}

I just want to know the purpose of the cast. I tried

cout << time(0);

and

cout << static_cast<unsigned int>(time(0));

it produces the same result so I'm not sure why the cast in the code.

score 5 · Accepted Answer · answered Feb 09 '17 at 13:56

The type returned by std::time which std::time_t is an implementation defined numeric type.

Numeric types are implicitly convertible to each other, so srand(time(0)); is guaranteed to be well defined, but on some implementation, std::time_t could be a type whose all values are not representable by unsigned int, so a helpful compiler might produce a warning about a narrowing conversion that might be have been accidental.

The explicit static cast tells the compiler that a conversion is intentional, so no warning would be produced regardless of how std::time_t is defind.

score 4 · Answer 2 · edited May 23 '17 at 12:24

4

srand() expects an unsigned int and time() returns std::time_t which is an implementation defined type. It might not be unsigned but signed. The static_cast is there to suppress the possibility of an error/warning if std::time_t is defined as something else as unsigned int. Note that this isn't actually good practice.

P.S If you have a compiler that supports C++11 then stop using rand()/srand() all together and start using the new header <random>.

edited May 23 '17 at 12:24

Community

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answered Feb 09 '17 at 13:52

Hatted Rooster

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Would you like to explain why suppressing the warning(in this case) isn't good practise? – Martin Bonner supports Monica Feb 09 '17 at 13:56
1

@MartinBonner Whoops, I was referring to the way of generating those numbers. Not the cast itself. Now having re-read my answer it doesn't sound like that. Thanks. – Hatted Rooster Feb 09 '17 at 13:59

Why the static_cast on time(0) on this code

2 Answers2