How you can store 10 digits to 8 chars? Probably you mean "char TIMESTAMP[10]" or "char TIMESTAMP[11]" (because you need a null-terminated symbol for c-strings). So if you have not a c-string, but just array with digit's symbols - I suggest you to write simple convert function like this int number = 0; for (int i = 0; i < num_of_digits; ++i) { number = number*10 + TIMESTAMP[i]; } In case with c-strings use Igor R solution. On Fri, Nov 8, 2013 at 7:54 PM, Rahul Mathur <srivmuk@gmail.com> wrote:
YES, do have the TIMESTAMP from string viz 1383889129. Since, the declaration of TIMESTAMP is "char TIMESTAMP[8]", so looking how to convert this?
On Fri, Nov 8, 2013 at 11:49 AM, Igor R <boost.lists@gmail.com> wrote:
I have a string Timestamp of 1383889129 which on ONLINE conversion from string to hex gives a 10 bytes as "31 33 38 33 38 38 39 31 32 39".
The above is not "conversion from string to hex", but merely a hex representation of every ascii character in "1383889129" string.
But, the need is to have the TIMEStamp in 8 bytes only.
It's unclear what you want exactly. If you need a time-stamp as a number, use boost::lexical_cast<int>("1383889129") or any other ascii-to-int conversion. _______________________________________________ Boost-users mailing list Boost-users@lists.boost.org http://lists.boost.org/mailman/listinfo.cgi/boost-users
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