How to zero array members when my compiler isn't standard conform

Question

My compiler (C++Builder6) syntactically allows array member initialization (at least with zero), but actually it doesn't really do it. So the assert in the example given below fails depending from the context.

#include <assert.h>

struct TT {
    char b[8];
    TT(): b() {}
};

void testIt() {
    TT t;
    assert(t.b[7] == 0);
}

Changing the compiler isn't an option at the moment. My question is: what will be the best way to "repair" this flaw with respect to future portability and standard conformance?

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Edit: As it turns out, my first example was too short . It missed the point, that the fill level of the array is so essential, that it has to be stored very close to the array, which is: in the same class.

Even if the original problem remains , my actual problem pattern is usually this:

struct TT2 {
    int size;
    char data[8];
    // ... some more elements
    TT2(): size(0), data() {}
    // ... some more methods
};

Answer 1

I think you may use this:

TT() { std::fill(b, b + 8, char()); }

This way you will solve your problem while nothing is wrong with portability and standard conformance!

Answer 2

You may use fill_n like suggested in: C/C++ initialization of a normal array with one default value

If no fill_n is available, you can always use memset like:

TT() {memset(b, 0, sizeof b);}

Answer 3

I would like to append previous posts that if you are using a character array as a string then it is enough to write in the constructor

TT() { b[0] = '\\0'; }