Allocating struct with variable length array member

Question

I know I can do new char[n] to create an array of n chars. This works even when n is not a compile time constant.

But lets say I wanted a size variable followed by n chars:

My first attempt at this is the following:

struct Test
{
  std::size_t size;
  char a[];
};

However it seems new Test[n] doesn't do what I expect, and instead allocates n sizes.

I've also found that sizeof(std::string) is 4 at ideone, so it seems it can allocate both the size and the char array in one block.

Is there a way I can achieve what I described (presumably what std::string already does)?

Answer 1

While you can do this (and it was often used in C as a workaround of sorts) it's not recommended to do so. However, if that's really what you want to do... here's a way to do it with most compilers (including those that don't play nicely with C99 enhancements).

#define TEST_SIZE(x) (sizeof(Test) + (sizeof(char) * ((x) - 1)))

typedef struct tagTest
{
    size_t size;
    char a[1];
} Test;

int elements = 10; // or however many elements you want
Test *myTest = (Test *)malloc(TEST_SIZE(elements));

The C specifications prior to C99 do not allow a zero-length array within a structure. To work around this, an array with a single element is created, and one less than the requested element count is added to the size of the actual structure (the size and the first element) to create the intended size.