c++ const_cast<char*> same binary different results on 2 64bit ms os's

Question

I have the following function (shortened for brevity) : (PortNo = 12345)

void startparser(){
    std::ostringstream convert;
    convert.str("");
    convert << '"' << "c:\\some\\file path\\a_program.exe" << '"' << " " << PortNo;
    std::cout << "DEBUG2 " << convert.str() <<std::endl;
    char *cmd = const_cast<char*> ( convert.str().c_str() );
    std::cout << "DEBUG3 " << cmd <<std::endl;
}

I compile on w10 64bit with eclipse MinGW64 and the code outputs

DEBUG2 "c:\\some\\file path\\a_program.exe" 12345

DEBUG3 "c:\\some\\file path\\a_program.exe" 12345

Now if I copy the binary and the 3 dll's it uses to a win7 64 machine and run the same code I get

DEBUG2 "c:\\some\\file path\\a_program.exe" 12345

DEBUG3 .

I don't get it? I need cmd to be an LPTSTR . Can someone please clarify? Another method to get my LPTSTR is fine too.

Answer 1

You're using a temp object: ostringstream::str() returns an object which is destroyed immediately. Therefore after this line

char *cmd = const_cast<char*> ( convert.str().c_str() );

the cmd points to memory which was used for a temp object and is therefore no longer valid.

If you really need to operate with const char* then you need to create an intermediate string tmp = convert.str() . Then the pointer to tmp.c_str() will be valid within the whole scope of tmp .