将字符串转换为int（C ++）

Question

I looked everywhere and can't find an answer to this specific question :(

I have a string date, which contains the date with all the special characters stripped away. (ie : yyyymmddhhmm or 201212031204).

I'm trying to convert this string into an int to be able to sort them later. I tried atoi, did not work because the value is too high for the function. I tried streams, but it always returns -858993460 and I suspect this is because the string is too large too. I tried atol and atoll and they still dont give the right answer.

I'd rather not use boost since this is for a homework, I dont think i'd be allowed.

Am I out of options to convert a large string to an int ? Thank you!

What i'd like to be able to do :

int dateToInt(string date)
{
date = date.substr(6,4) + date.substr(3,2) + date.substr(0,2) + date.substr(11,2) + date.substr(14,2);
int d;
d = atoi(date.c_str());
return d;

}

Answer 1

You get negative numbers because 201212031204 is too large to fit int . Consider using long long s

BTW, You may sort strings as well.

Answer 2

You're on the right track that the value is too large, but it's not just for those functions. It's too large for an int in general. int s only hold up to 32 bits, or a maximum value of 2147483647 (4294967295 if unsigned). A long long is guaranteed to be large enough for the numbers you're using. If you happen to be on a 64-bit system, a long will be too.

Now, if you use one of these larger integers, a stream should convert properly. Or, if you want to use a function to do it, have a look at atoll for a long long or atol for a long . (Although for better error checking, you should really consider strtoll or strtol .)

Completely alternatively, you could also use a time_t . They're integer types under the hood, so you can compare and sort them. And there's some nice functions for them in <ctime> (have a look at http://www.cplusplus.com/reference/ctime/ ).

Answer 3

typedef long long S64;

S64 dateToInt(char * s) {
    S64 retval = 0;
    while (*s) {
         retval = retval * 10 + (*s - '0');
         ++s;
    }
    return retval;
}

Note that as has been stated, the numbers you're working with will not fit into 32 bits.