CODEWITHSUNDEEP
c++stdsizeof
Sara
Sara

Reputation: 129

Functioning of sizeof() operator in C++

I wrote a simple program in order to understand the functioning of the function of the standard c++ library sizeof().

It follows:

const char* array[] = {
                          "1234",
                          "5678"
                          };

    std::cout << sizeof(array) << std::endl;//16
    std::cout << sizeof (array[0]) << std::endl;//8

    std::cout << printf("%lu\n",sizeof (char) );//1
    std::cout << printf("%lu\n",sizeof (int) );//24
    std::cout << printf("%lu\n",sizeof (float) );//24
    std::cout << printf("%lu",sizeof (double) );//281

It is possible to see by the output reported the characters has dimension 1 byte in my OS, as expectable. But I do not understand why the dimension of '''array[0]''' is 8, as it contains 4 charcaters and at least other 2 charcaters for the end sequence "\n" which is contained in a string. Thus, I supposed that the number of bytes occupied by the first element of the array should be 6 and not 8. Moreover, if I increase/decrease the number of charcaters contained in the first element of the array, the its size does not change. Clearly, I am wrong. If somebody can explain me this functioning, I would really appreciate. Thanks,

Upvotes: 0

Views: 505

Answers (2)

Basile Starynkevitch
Basile Starynkevitch

Reputation: 1

I wrote a simple program in order to understand the functioning of the function of the standard c++ library sizeof().

Wrong terminology. Please read n3337 (a C++ standard) and the wikipage on sizeof.

sizeof is a compile-time operator, not a function. If v is some variable, sizeof(v) only depends on the type of v and never on its value (in contrast, for most functions f, the value of f(v) depends upon the value of v).

And a good way to understand something about C++ is to refer to documents like standards or good web pages about it.

If somebody can explain me

Yes. Read a good book about C++. This one is written by the main designer of C++. Try to understand more and better the (difficult) semantics of C++. You could also study the source code of existing open source C++ compilers such as GCC or Clang/LLVM (thus effectively using one of your free software freedoms).

BTW, with a lot of pain you might find C++ implementations with sizeof(int) being 1 (e.g. for some DSP processors). On cheap 32 bits ARM processors (those in cheap mobile phones today, for instance; then you would probably use some cross-compiler) or on some Raspberry Pis (or perhaps some mainframes) you could have sizeof(array[0]) or sizeof(void*) being 4 even in 2019.

Upvotes: 6

Adrian Mole
Adrian Mole

Reputation: 51893

Let's break down the meaning of the somewhat confusing output values you see!

First, the sizeof(array) and sizeof(array[0]) (where your output method is fine). You have delared/defined array as an array of two char* values, each of which is a pointer. The size of a pointer on your system is 8 bytes, so the total size of array is: 8 * 2 = 16. For array[0]: this is a single pointer, so its size is simply 8 bytes.

Does all this make sense so far? If so, then let's look at the second part of your code …

The values for sizeof(char), sizeof(int), sizeof(float) and sizeof(double) are, on your system, in order, 1, 4, 4, and 8. These values are actually being output! However, as you are also outputting the return value of printf(), which is the number of characters it has written, you are getting the extra values, "2", "2", "2" and "1" inserted (in a confusing, and possibly undefined, order), for the four calls (the last one has no newline, so it's only one character; all others are one digit + newline = 2 characters).

Change the second part of your code as follows, to get the correct outputs:

printf("%zu\n", sizeof(char));   //1
printf("%zu\n", sizeof(int));    //4
printf("%zu\n", sizeof(float));  //4
printf("%zu\n", sizeof(double)); //8

Upvotes: 2

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