C++ Question

[DiscoBall 13]

Given the declaration. int *a; The statement. a = new int [50]; Dynamically allowcates an array of 50 components of the type? int, int*, pointer, address?

[koothkeeper]

You get a pointer to an array of 50 unique integers.

[toxcct]

int* a; declares a as a pointer to an int. a = new int[50] allocates an array of 50 int, and then returns the address of the array in memory into a. then, you can code a[1] to access the 2nd cell on the array. if you wanted an array of pointers to int, you should write this :

int* a;
a = new int*[50];

cheers,

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TOXCCT >>> GEII power
[toxcct][VisualCalc]

[Anonymous]

Not quite. That would be: // 50 new pointers to integers. int **a = new (int *)[50]; // Parentheses might not be necessary. int *b = a[0]; // Assign one element of the array to an int pointer. int c = 0; // New integer on the stack b = &c; // Assign a value to the pointer. a[0] = &c; // Same effect as the above line. *b = 5; // Assign a value to the integer pointed to by b. *(a[0]) = 5; // Same effect as the above line. (Note: I wrote this code in the input box and didn't compile it to double-check if it works, so if I got something wrong please correct me.) Like the other posters said, the code "int *a = new int[50]" allocates an array of 50 integers on the heap and assigns a to the address of the beginning of that array.

[Lost User]

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