I have this code. I would like to put argv[1] as prefix to fileName. How do I do that?
int _tmain(int argc, char** argv)
{
...
_stprintf(fileName, _T("%04d-%02d-%02d-%02d-%02d-%02d-%03d.jpeg"), lt.wYear, lt.wMonth, lt.wDay, lt.wHour, lt.开发者_Go百科wMinute, lt.wSecond, lt.wMilliseconds);
_stprintf(fileName, _T("%s-%04d-%02d-%02d-%02d-%02d-%02d-%03d.jpeg"), argv[1], lt.wYear, lt.wMonth, lt.wDay, lt.wHour, lt.wMinute, lt.wSecond, lt.wMilliseconds);
That will do what you want, but since you're using C++ you would be best to use a stringstream instead.
#include <stringstream>
int main(int argc, char** argv)
{
std::stringstream ss;
ss << argv[1] << "-" << lt.wYear << "-" << lt.wMonth << "-" << lt.wDay << "-" << lt.wHour << "-" << lt.wMinute << "-" << lt.wSecond << "-" << lt.wMilliseconds << ".jpeg";
}
You can then access the string with ss.str()
. You can also use the same stream format modifiers you would use with any output stream.
Assuming that the wholly non-standard _stprintf()
is closely related to sprintf()
, then:
- Use
%s
at the point in the format string where the name goes. - Pass
argv[1]
at the corresponding point in the argument list.
Note that argv
has the type char **
; therefore, argv[1]
has the type char *
.
精彩评论