C++ declare an array based on a non-constant variable?

Question

void method(string a) {
  int n = a.size();
  int array[n];
}

The above code can compile correctly using gcc. How can the size of the array come from a non-constant variable? Does the compiler automatically translate the int array[n] to int* array = new int[n] ?

Answer 1

How can the size of the array come from a non-constant variable?

Currently, because that compiler has a non-standard extension which allows you to use C's variable length arrays in C++ programs.

Does the compiler automatically translate the int array[n] to int* array = new int[n] ?

That's an implementation detail. I believe GCC places it on the stack, like normal automatic variables. It may or may not use dynamic allocation if the size is too large for the stack; I don't know myself.

Answer 2

dynamic allocation. The new keyword will do this with a pointer and some allocation.

int * ptr;
int n = a.size();
ptr = new int[n];

Answer 3

根据这个编译器允许在C ++中，只要C90 / 99此表达式。