How do I find the number of bytes used by a pointer?
I have a pointer (uint8_t *myPointer), that I pass a开发者_StackOverflows parameter to a method, and then this method sets a value to this pointer, but I want to know how many bytes are used (pointed at ?) by the myPointer variable.
Thanks in advance.
The size of the pointer: sizeof(myPointer)
(Equal to sizeof(uint8_t*)
)
The size of the pointee: sizeof(*myPointer)
(Equal to sizeof(uint8_t)
)
If you meant that this points to an array, there is no way to know that. A pointer just points, and cares not where the value is from.
To pass an array via a pointer, you'll need to also pass the size:
void foo(uint8_t* pStart, size_t pCount);
uint8_t arr[10] = { /* ... */ };
foo(arr, 10);
You can use a template to make passing an entire array easier:
template <size_t N>
void foo(uint8_t (&pArray)[N])
{
foo(pArray, N); // call other foo, fill in size.
// could also just write your function in there, using N as the size
}
uint8_t arr[10] = { /* ... */ };
foo(arr); // N is deduced
You can't. You must pass the size of the buffer pointed to by the pointer yourself.
You can't. Unless the array is instantiated at compile time. Example int test[] = {1, 2, 3, 4};
In that case, you can use sizeof(test)
to return you the correct size. Otherwise, it's impossible to tell the size of an array without storing a counter of your own.
I assume you mean the memory needed for the pointer only. You can check this at compile time with sizeof(myPointer)
.
精彩评论