char* to time_t

[crh0872]

Hello,

I'm writing a program in C++ and I have a char string, "1240717949684", which I need to convert into the time_t type. It is apparently far too big for functions like atoi(), unless I'm doing something wrong. I wrote my own atoi() using time_t instead of int but it seems that the problem is that the processor can't multiply with a number that is bigger than 32 bits? It always gives me junk values.

Thanks,
crh

[helios]

Yes, the value is too large for 32-bit integers. Since your version of the library uses 32-bit integers, you won't be able to convert it to a time_t, but you can probably still convert it to a long long.

[PanGalactic]

Are you compiling the application 64-bit or 32-bit?

Check out the Boost Conversion Library, specifically lexical_cast.

If your time_t can hold that value, you can do:

char* bar = "1230717949684";
try
{
    time_t foo = boost::lexical_cast<time_t>(bar);
}
catch (boost::bad_lexical_cast& ex)
{
    // conversion failed
}

If you want to do what lexical_cast is doing behind the scenes, you can do:

std::string bar = "1230717949684";
std::istringstream ss(bar);
time_t foo;
ss >> foo;

[crh0872]

OK, I tried helios's solution first and it worked until I tried to convert the long long into a time_t. As soon as I did that I got the junk values. Any ideas?

I've heard of Boost, PanGalactic, but haven't looked into it yet. That looks refreshingly easy, so I may try that if I can't get this method going. I'm compiling for 64-bit but it would be nice if it could work on 32 bit as well.

[helios]

Since your version of the library uses 32-bit integers, you won't be able to convert it to a time_t

You cannot store that value in a time_t. There's no solution other than to use a 64-bit version of the C library or use a different time library that can take bigger UNIX time integers.

I'm compiling for 64-bit but it would be nice if it could work on 32 bit as well.

That's impossible, as well.

[PanGalactic]

atoi() does integer conversion, atol() does long conversion, atoll() does long long conversion

Doing this portably using the ato*() conversion routines requires some work. time_t bar = atol(foo) will work in most cases, but there are certainly instances where time_t is not 32-bit on 32-bit platforms.

time_t bar;
if (sizeof(time_t) == sizeof(int))
{
    bar = atoi(foo);
}
else if (sizeof(time_t) == sizeof(long))
{
    bar = atol(foo);
}
else if (sizeof(time_t) == sizeof(long long))
{
    bar = atoll(foo);
}

One thing to note is that long long is not yet part of the C++ standard is therefore is not yet fully portable. I prefer lexical_cast.

And, as helios says, you cannot convert the value you have to a time_t on a platform where time_t is 32-bits.

[crh0872]

Ok, thanks for the help guys, I'm going to download the boost libraries and do it that way.