C ++：sizeof用于数组长度

Question

Let's say I have a macro called LengthOf(array) :

sizeof array / sizeof array[0]

When I make a new array of size 23, shouldn't I get 23 back for LengthOf ?

WCHAR* str = new WCHAR[23];
str[22] = '\0';
size_t len = LengthOf(str); // len == 4

Why does len == 4 ?

UPDATE : I made a typo, it's a WCHAR* , not a WCHAR** .

Answer 1

Because str here is a pointer to a pointer, not an array.

This is one of the fine differences between pointers and arrays: in this case, your pointer is on the stack, pointing to the array of 23 characters that has been allocated elsewhere (presumably the heap).

Answer 2

WCHAR** str = new WCHAR[23];

First of all, this shouldn't even compile -- it tries to assign a pointer to WCHAR to a pointer to pointer to WCHAR . The compiler should reject the code based on this mismatch.

Second, one of the known shortcomings of the sizeof(array)/sizeof(array[0]) macro is that it can and will fail completely when applied to a pointer instead of a real array. In C++, you can use a template to get code like this rejected:

#include <iostream>

template <class T, size_t N>
size_t size(T (&x)[N]) { 
    return N;
}

int main() { 
    int a[4];
    int *b;

    b = ::new int[20];

    std::cout << size(a);      // compiles and prints '4'
//    std::cout << size(b);    // uncomment this, and the code won't compile.
    return 0;
}

Answer 3

As others have pointed out, the macro fails to work properly if a pointer is passed to it instead of an actual array. Unfortunately, because pointers and arrays evaluate similarly in most expressions, the compiler isn't able to let you know there's a problem unless you make you macro somewhat more complex.

For a C++ version of the macro that's typesafe (will generate an error if you pass a pointer rather than an array type), see:

• Compile time sizeof_array without using a macro

It wouldn't exactly 'fix' your problem, but it would let you know that you're doing something wrong.

For a macro that works in C and is somewhat safer (many pointers will diagnose as an error, but some will pass through without error - including yours, unfortunately):

• Is there a standard function in C that would return the length of an array?

Of course, using the power of #ifdef __cplusplus you can have both in a general purpose header and have the compiler select the safer one for C++ builds and the C-compatible one when C++ isn't in effect.

Answer 4

The problem is that the sizeof operator checks the size of it's argument. The argument passed in your sample code is WCHAR* . So, the sizeof(WCHAR*) is 4. If you had an array, such as WCHAR foo[23] , and took sizeof(foo) , the type passed is WCHAR[23] , essentially, and would yield sizeof(WCHAR) * 23 . Effectively at compile type WCHAR* and WCHAR[23] are different types, and while you and I can see that the result of new WCHAR[23] is functionally equivalent to WCHAR[23] , in actuality, the return type is WCHAR* , with absolutely no size information.

As a corellary, since sizeof(new WCHAR[23]) equals 4 on your platform, you're obviously dealing with an architecture where a pointer is 4 bytes. If you built this on an x64 platform, you'd find that sizeof(new WCHAR[23]) will return 8.

Answer 5

You wrote:

WCHAR* str = new WCHAR[23];

if 23 is meant to be a static value, (not variable in the entire life of your program) it's better use #define or const than just hardcoding 23.

#define STR_LENGTH 23
WCHAR* str = new WCHAR[STR_LENGTH];
size_t len = (size_t) STR_LENGTH;

or C++ version

const int STR_LENGTH = 23;
WCHAR* str = new WCHAR[STR_LENGTH];
size_t len = static_cast<size_t>(STR_LENGTH);