CODEWITHSUNDEEP
c++arrays
tgh
tgh

Reputation: 795

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

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

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

In other words, create a dynamic array with all values initialised to the termination character:

char* c = new char[length]; // how do i amend this?

Upvotes: 76

Views: 141140

Answers (10)

AnT stands with Russia
AnT stands with Russia

Reputation: 320531

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' };

Upvotes: 22

songyuanyao
songyuanyao

Reputation: 172924

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] = {};.

Upvotes: 7

Seva Alekseyev
Seva Alekseyev

Reputation: 61380

No internal means, AFAIK. Use this: memset(c, 0, length);

Upvotes: 1

Steve Jessop
Steve Jessop

Reputation: 279255

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]();]

Upvotes: 36

Fred
Fred

Reputation: 1448

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

Upvotes: 126

almathie
almathie

Reputation: 731

you have to initialize it "by hand" :

char* c = new char[length];
for(int i = 0;i<length;i++)
    c[i]='\0';

Upvotes: 0

pm100
pm100

Reputation: 50190

and the implicit comment by many posters => Dont use arrays, use vectors. All of the benefits of arrays with none of the downsides. PLus you get lots of other goodies

If you dont know STL, read Josuttis The C++ standard library and meyers effective STL

Upvotes: 1

anon
anon

Reputation:

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.

Upvotes: 5

Andreas Bonini
Andreas Bonini

Reputation: 44752

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)

Upvotes: 4

mrkj
mrkj

Reputation: 3101

Maybe use std::fill_n()?

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

Upvotes: 22

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