Syntax for pointing to a array and acessing its elements

Hello, can somebody give me some help about using pointers to arrays?

What I want to do is to use a array of integers. This is the come I came up with:

int *myArray[10];

myArray = (void*)malloc(sizeof(int) *10);

Is that correct? How can I access the 2nd element of my array? I tried *myArray[2] but it didn't work.

These are different:

Code:

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int *myArray[10];
int myArray[10];

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The first is an array of pointers to integers. The second is an array of integers.

Ah, thank you! Just to make sure I'm making sense of pointers, would this be correct:

Code:

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int *MyArray[10];

for (i = 0; i <= 10; i++)
{
    myArray[0] = (int*)malloc(sizeof(int) );
}

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and to access elements:

Code:

---

*myArray[0] = 1;
 *myArray[1] = 1;

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Do you want a pointer to point at an array of integers?
If so you can do like this:

Code:

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int main()
{
        int array[3] = {1,2,3};
        int *ptr;

        ptr = array;

        printf("%d\n", *ptr);
        printf("%d\n", *ptr+1);

        //or perhaps better, like this
        printf("%d\n", ptr[0]);
        printf("%d\n", ptr[1]);


        return 0;
}

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Quote:

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Originally Posted by papagaio

Ah, thank you! Just to make sure I'm making sense of pointers, would this be correct:

Code:

---

int *MyArray[10];

for (i = 0; i <= 10; i++)
{
    myArray[0] = (int*)malloc(sizeof(int) );
}

---

and to access elements:

Code:

---

*myArray[0] = 1;
 *myArray[1] = 1;

---

---

Close.

Code:

---

int *MyArray[10];

for (i = 0; i <= 10; i++)
{
    myArray[i] = malloc(sizeof(int) );
}

---

Close. But still not right.

Code:

---

int *MyArray[10];

for (i = 0; i <= 9; i++)
{
    myArray[i] = malloc(sizeof(int));
}

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or more commonly:

Code:

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int *MyArray[10];

for (i = 0; i < 10; i++)
{
    myArray[i] = malloc(sizeof(*myArray[i]));
}

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and the 10 may well be replaced by a named constant.