Why do we need dynamic initialization of objects in C plus plus?

Answer 1

All objects must be initialised before we can use them. Initialisation can either occur statically (at compile-time) or dynamically (at runtime) however declaring an object static does not imply static initialisation.

Static initialisation relies on constant expressions. A constant expression is any expression that can be computed at compile-time. If a computation involves one or more function calls, those functions must be declared constant expression (constexpr) in order to take advantage of compile-time computation. However, whether or not the computation actually occurs at compile-time or not is implementation-defined. Consider the following code:

constexpr unsigned f (unsigned n) { return n<2?1:n*f(n-1); }

static unsigned num = f (4); // e.g., num = 4 * 3 * 2 * 1 = 24

int main () {

// ...

}

The above example is a potential candidate for compile-time computation. A good compiler will interpret this code as if it had been written:

static unsigned num = 24;

int main () {

// ...

}

However, if the function f () were not declared constexpr, the compiler would instead interpret the code as if it were actually written as follows:

unsigned f (unsigned n) { return n<2?1:n*f(n-1); }

static unsigned num; // uninitialised!

int main () {

num = f (4);

// ...

}

In other words, num is dynamically initialised -- at runtime.

Local objects are always initialised dynamically, as are objects allocated upon the heap (the free store). Static objects may be statically allocated, however this depends on the complexity of the object's constructor as well as the compiler's ability to perform compile-time computation.