How do you initialise a dynamic array in C++?

Question

How do I achieve the dynamic equivalent of this static array initialisation:

char c[2] = {};  // Sets all members to \'\\0\';

In other word

Answer 1

C++ has no specific feature to do that. However, if you use a std::vector instead of an array (as you probably should do) then you can specify a value to initialise the vector with.

std::vector <char> v( 100, 42 );

creates a vector of size 100 with all values initialised to 42.

Answer 2

You can't do it in one line easily. You can do:

char* c = new char[length];
memset(c, 0, length);

Or, you can overload the new operator:

void *operator new(size_t size, bool nullify)
{
    void *buf = malloc(size);

    if (!buf) {
             // Handle this
    }

    memset(buf, '\0', size);

    return buf;
}

Then you will be able to do:

char* c = new(true) char[length];

while

char* c = new char[length];

will maintain the old behavior. (Note, if you want all news to zero out what they create, you can do it by using the same above but taking out the bool nullify part).

Do note that if you choose the second path you should overload the standard new operator (the one without the bool) and the delete operator too. This is because here you're using malloc(), and the standard says that malloc() + delete operations are undefined. So you have to overload delete to use free(), and the normal new to use malloc().

In practice though all implementations use malloc()/free() themselves internally, so even if you don't do it most likely you won't run into any problems (except language lawyers yelling at you)

Answer 3

Since c++11 we could use list initialization:

char* c = new char[length]{};

For an aggregate type, then aggregate initialization will be performed, which has the same effect like char c[2] = {};.

Answer 4

The array form of new-expression accepts only one form of initializer: an empty (). This, BTW, has the same effect as the empty {} in your non-dynamic initialization.

---

The above applies to pre-C++11 language. Starting from C++11 one can use uniform initialization syntax with array new-expressions

char* c = new char[length]{};
char* d = new char[length]{ 'a', 'b', 'c' };

Answer 5

Maybe use std::fill_n()?

char* c = new char[length];
std::fill_n(c,length,0);

Answer 6

Two ways:

char *c = new char[length];
std::fill(c, c + length, INITIAL_VALUE);
// just this once, since it's char, you could use memset

Or:

std::vector<char> c(length, INITIAL_VALUE);

In my second way, the default second parameter is 0 already, so in your case it's unnecessary:

std::vector<char> c(length);

[Edit: go vote for Fred's answer, char* c = new char[length]();]