How to convert int to char*?

Question

My problem is that I don't know how to convert int value to char array char* m_value. I tried to use itoa but it doesn't work. itoa(m_val, m_wartosc, 10); Maybe there is some other function to do this ?

Main.cpp

int main(int argc, char *argv[])
{
    LargeNumber l1;
    LargeNumber l3(172839); //how to convert this int to char*

    return 0;
}

LargeNumber.h

    class LargeNumber{

            public:

                LargeNumber()
                { 
                    m_array = "0"; //zero for no arg.
                }
                LargeNumber(int val):m_val(val)
                {
                    itoa(m_val, m_array, 10);  //doesn't work
                    //sprintf(m_array, "%d", m_val);
                }



                LargeNumber(const LargeNumber& p):m_array(p.m_array)
                { }  //copy constructor


                ~LargeNumber(){
                    delete []m_array;     //for object with new    
                }
               public: //should be private
                int m_val;
                char* m_array;

};

Why do you need to, you already have a constructor that takes an `int`. — Luchian Grigore, May 28 '12 at 14:30
Why do you want to store your number in base-10 as an ASCII string? — Oliver Charlesworth, May 28 '12 at 14:32
I need to have both of them for my program. I have already one which will take `"string"`. But there is a need for one which will take int value. — mathewM, May 28 '12 at 14:33
I know that storeing number as a char* is wrong, but I need to manage with that. — mathewM, May 28 '12 at 14:35
Side note, if default constructor is used the `delete[] m_array` in the destructor will be attempting to deallocate read-only memory. — hmjd, May 28 '12 at 14:36
Is there a reason `char*` is being used instead of `std::string` for `m_array`? — hmjd, May 28 '12 at 14:39
You are following a Detonator pattern from resign patterns (http://www.lsd.ic.unicamp.br/~oliva/fun/prog/resign-patterns) with: 1. deleting memory of a literal after creating with default constructor. 2. Not allocating memory for an array you are trying to use (m_array). 3. Sharing pointer in your copy constructor - it will be then double-deleted. Conclusion - use `std::string` instead of `char *` — Tadeusz Kopec for Ukraine, May 28 '12 at 14:41

score 5 · Answer 1 · edited May 23 '17 at 12:20

5

The simple answer is: don't. For two reasons:

As you can see from all the (wrong) other answers, memory management is tricky and bug-prone.
I can't see how storing your value in base-10, in an ASCII string, could possibly be useful. (Compared to, say, a base-2³² representation.)

But if you really must store it this way, you will need to allocate the relevant amount of memory, use snprintf to convert (itoa is a non-standard function), and remember to free the memory at the correct time(s) (you will have to read and understand about the Rule of Three).

I would strongly recommend using a std::string instead of a raw C-style array, because it will at least deal with its own memory management, and you will then be able to populate it with a std::stringstream.

edited May 23 '17 at 12:20

Community

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answered May 28 '12 at 14:39

Oliver Charlesworth

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Use `snprintf` (which is C90/C++03). And use `std::vector` to get the necessary memory. – James Kanze May 28 '12 at 14:42
@JamesKanze: Yep, I agree with `snprintf`, answer modified. In terms of storage, I've hinted that there's a host of better idiomatic ways of dealing with this; `vector` is just one possibility. – Oliver Charlesworth May 28 '12 at 14:43
Could you write how snprintf function should look in my program ? – mathewM May 28 '12 at 15:48

score 0 · Answer 2 · answered May 28 '12 at 14:35

0

The second argument of itoa() needs to be an array in memory large enough to store the null-terminated string. An example:

int number = 172839;
char buffer[10];
itoa(number,buffer,10);

answered May 28 '12 at 14:35

Adam27X

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So in my case i can't use itoa before my char* m_array doesn't have declared size ? – mathewM May 28 '12 at 14:39
Yes. You'll need to actually allocate memory in that buffer before trying to store the string there. – Adam27X May 28 '12 at 14:44

score 0 · Answer 3 · answered May 28 '12 at 14:36

0

LargeNumber(int val):m_val(val)
        {
            std::stringstream stream;
            stream << val;
            m_array = new char[stream.str().size()];
            strcpy(m_array, stream.str().c_str());
        }

answered May 28 '12 at 14:36

sithereal

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No memory allocated for `m_array`. – hmjd May 28 '12 at 14:37

score 0 · Answer 4 · answered May 28 '12 at 14:37

0

You have to first allocate the array with

m_array = new char[20]

in constructor before calling iota. the iota doesnt allocate memory.

answered May 28 '12 at 14:37

Arun Kumar M

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How to convert int to char*?

4 Answers4