Conversion of C Style Casting to C++ Style Casting

Question

I'm not familiar in C++ casting and I want to convert my C style casting to a C++ casting. Here is my code,

typedef unsigned char u8;
u8 sTmp[20] = {0};

//.. code to put string data in sTmp

char* sData;
sData = (char*)&(sTmp[0]);

Here, I want to convert (char*)&(sTmp[0]) to a C++ casting.

Many thanks.

http://stackoverflow.com/questions/28002/regular-cast-vs-static-cast-vs-dynamic-cast — chris, Nov 05 '12 at 23:44
Why is the cast needed? (genuine question, I'm not saying it's wrong) — effeffe, Nov 05 '12 at 23:48
@effeffe: It's needed because sTmp is an unsigned char but sData is a signed char. — Carey Gregory, Nov 06 '12 at 00:06

score 3 · Accepted Answer · answered Nov 05 '12 at 23:58

Your cast is unnecessarily complicated. You get the first element of the array and then the address of that element. On expressions, arrays decay into pointers, so you can get the address of the array by its name alone:

sData = (char*)sTmp;

Like @Richard said above, the best way to do the cast on C++ is using reinterpret_cast, like this:

sData = reinterpret_cast<char*>(sTmp);

Finally, sTemp (like I already mentioned) will decay to a pointer on expressions, specifically an unsigned char* (which is the usual way of addressing raw memory), so it is very likely that you don't actually need to cast it to char* at all. (unless you have to print it, which doesn´t seem right anyway)

I like taking the address of the first element. As it makes it easy to convert the code to using vectors without any further changes. — Martin York, Nov 06 '12 at 01:03

Conversion of C Style Casting to C++ Style Casting

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