Code:Start(){ char exit,key; exit =1; while(exit) { key = if ((key & 0x40)==0) exit=0; } }
Code:Start(){ char exit,key; exit =1; while(exit) { key = if ((key & 0x40)==0) exit=0; } }
It has syntax errors.
Look up a C++ Reference and learn How To Ask Questions The Smart WayOriginally Posted by Bjarne Stroustrup (2000-10-14)
exit can't be a variable name since it is similar to return.
It's incomplete.
Code:Start(){ /* what does it return? compiler should warn */ char exit,key; exit =1; while(exit) { key = /* key equals what now? */ if ((key & 0x40)==0) exit=0; /* don't waste time doing a flag */ } } /* horrible indentation all the way through */
It should look something like this:
Code:void Start (char key) { while (!(key & 0x40)) key; /* idk. key++? key--? key*=5? you need to specify */ }
That has to be the ugliest code I've ever seen.