#include #include int main() {
int number = 1234567;
int digits = log10(number)+1;
printf("%d has %d digits\n",number,digits);
}

Try it here[^]

I recently used ConvertXtoDVD[^] and I thought it worked excellently. I haven't used the other ones that everyone else is suggesting so I can't compare...

or sscanf() ... He did say the string always had the exact same format ...

Not only that. Think of the SQL injection you could do...

Ok, Maybe you can tell me what is wrong with this. When I run this, it dies with memory corruption in the find statement in the Execute function. I pass the pointer to keep the map from being copied (since it might get huge in a real project) To me there is no obvious reason why. // The DLL #include #include using namespace std; typedef map<string,string> param_t; __declspec(dllexport) bool Execute(param_t *params){ params->find("xyz"); return true; } // The EXE #include #include using namespace std; typedef map<string,string> param_t; __declspec(dllimport) bool Execute(param_t *params); int main() { param_t params; params["xyz"]="abc"; params.find("xyz"); // test it here before we call the other function return Execute(¶ms); } I was using Visual C++ 6.0. Compiling it with any other compiler seems to work fine (including newer versions of Visual C++)

You can tell when a company goes to a lot of effort to make their compiler easy to use. The most expensive part of a project is the amount of debugging and fixing bugs. A compiler that can save this much time is a Good Thing.

Take a look at this article from the FAQ of Dr. Math: http://www.mathforum.org/dr.math/faq/faq.interest.html[^] In an installment loan, a lender loans a borrower a principal amount P, on which the borrower will pay a yearly interest rate of i (as a fraction, e.g. a rate of 6% would correspond to i=0.06) for n years. The borrower pays a fixed amount M to the lender q times per year. At the end of the n years, the last payment by the borrower pays off the loan. The amount of the fixed payment is determined by: M = P * i / [ q (1 - [1+( i / q)]^(-n* q) ) ] M = 2500 * 2.399961 / [26 (1 - [1+(2.399961/26)]^(-1*26) ) ] 674653127756771870877187207 (which, when rounded to 2 dp is $256.61) The total amount paid by the borrower is Mnq, and the total amount of interest paid is I = Mnq - P: I = M * n * q - P I = 256.60 * 1 * 26 - 2500 I = 4171.7754098132167606864280686738 (which, when rounded is $4171.78) That is a lot of interest! Jerry http://www.mathforum.org/dr.math[^]