Initializing structure object using std::fill

Question

I have below a structure and need to initialize the object of it using std::fill.

typedef struct _test
{
    char name[32];
    char key[4];
    int count;

}test;

As of now, I am using memset. I need to change it to std::fill. I have tried below but std::fill throws compiler error for the structure object.

test t;
char a[5];
std::fill(a, a + 5, 0);
std::fill(t, sizeof(t), 0);

Note: I don't want to initialize using this way. char a[5] = {0};

Answer 1

You don't need std::fill (or std::memset, on which you should read more here). Just value initialize the structure:

test t{};

This will in turn zero initialize all the fields. Beyond that, std::fill accepts a range, and t, sizeof(t) is not a range.

And as a final note, typedef struct _test is a needless C-ism. The structure tag in C++ is also a new type name. So what this does is pollute the enclosing namespace with a _test identifier. If you need C compatibility on the off-chance, the way to go is this

typedef struct test
{
    char name[32];
    char key[4];
    int count;

} test;

This sort of typedef is explicitly valid in C++, despite both the struct declaration and the typedef declaring the same type name. It also allows both C and C++ code to refer to the type as either test or struct test.