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As many would recognise, this is a common interview question.
two differences I can think of are:
1) new calls the constructor and malloc does not.
2) new returns a pointer of the exact type of the object and malloc returns a void *.
anyhing else to add?
I
3) "new" is a C++ operator, i.e. it is a keyword in C++, malloc() is a function call.Quote:
Originally Posted by indrajit_p1
4) Operator new/new[] must be followed somewhere by delete/delete[], malloc is paired with free()
5) Operator new/new[] can be overloaded, malloc() cannot be overloaded.
6) Operator new/new[] throws an exception if there is an error, malloc() returns a NULL pointer if there is an error.
Regards,
Paul McKenzie
7) Initializers can be provided for non-arrays using new. The same cannot be done with malloc() (calloc() can be used for zero initialization).
8) This one isn't really about malloc() but I thought it worth pointing out that memory allocated via malloc() can easily be reallocated using another library routine, realloc(). When using new this isn't directly possible. You have to allocate a different block of adequate size, copy the contents of the old block and finally deallocate the old block.
9) The is "placement new" which allows you so specify WHERE to allocate the memory (useful especially in embedded systems to put an object at the location of a piece of hardware)
10) You can implement different allocation schemes by providing type specific "operator new" implementations.
11) "malloc" and "new" are both spelled and pronounced differently. ;) :wave:
absolutely hillarious. :-DQuote:
Originally Posted by TheCPUWizard
thanks a LOT for the inputs folks.
I
Of course, this should not be used as a favorable argument for using malloc/realloc, but for using std::vector rather than new[].Quote:
Originally Posted by angelorohit
I don't know if "allocate" is the right term, since using placement new does not set aside the memory at that address per se. More accurately, I think, it constructs an object at a pre-allocated (or otherwise valid, if not on the heap) address.Quote:
Originally Posted by TheCPUWizard
(I know you knew that, but I thought I should clarify the wording a bit.)
Hermit, you are 100% correct. Thanks for the clarification. :wave:
I think initializers can also be provided for arrays using new. Consider :Quote:
Originally Posted by angelorohit
If T is a user defined type with a default constructor which initializes the relevant (internal) variables, you are ok. However if T is a basic data type (like int) then initialization will not happen. To remedy this replace the statement above with the following :Code:T *p = new T[num];
The following code illustrates the idea :Code:T *p = new T[num] ();
This yields the following output :Code:#include <iostream>
using namespace std;
const int num = 3;
class a{
private:
int i;
public:
a(){i=10;}
friend ostream& operator<<(ostream& input_stream, const a &arg){
input_stream<<arg.i;
return input_stream;
}
};
template<typename T>
void test0(){
T *p = new T[num];
for(int i=0; i<num; ++i){ cout<<*p<<endl;}
cout<<endl;
}
template<typename T>
void test1(){
T *p = new T[num] ();
for(int i=0; i<num; ++i){ cout<<*p<<endl;}
cout<<endl;
}
int main()
{
//Initializes only user defined types
test0<int>();
test0<a>();
//Initializes user and basic types
test1<int>();
test1<a>();
return 0;
}
As we see test0<T> does not initialize correctly if T=int, whereas test1<T> does initialize correctly if T=int. Both these functions work fine for the user defined type "a".Code:-842150451
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10
10
10
0
0
0
10
10
10
I would say that is an edge case where you can value initialise with array new. You cannot, however, provide some other specific initial value with array new (according to the compiler I have at hand (g++ 4.2.3); my reading of the standard does not make things so clear).Quote:
I think initializers can also be provided for arrays using new.