Question

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Converting a four character string to a long

I want to convert a char array to a long and back again but I’m a bit stuck.

This is a fragment of the code I’ve got so far:

char mychararray[4] = {'a', 'b', 'c', 'd'};
unsigned long * mylong = (unsigned long *)&mychararray;
cout << *mylong << endl;

Which should take the char array, and represent the first 4 bytes (the length of a long) as a long (I think).

Is this correct? And how would I undo it to get the char array back?

Thanks for your help.

EDIT: The third line was a typo – *mychararray should have been *mylong

Answer 1

1. You’re assuming sizeof(long) == 4, which can be wrong. (Especially on 64 bit platforms). If that assumption is broken, you’re in undefined behavior territory: manipulating mylong will read/write beyond mychararray‘s allocated memory.
2. Your second line has no effect if you don’t use *mylong on the third line

You can get back a char* with something like:

char *thing = reinterpret_cast<char*>(mylong);

(You should be using that type of cast in the first case also, it’s more explicit than the C-type cast.)