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SizeOf question C++
I have a question about this code https://code.sololearn.com/c8fswm6Kkx2g/#cpp. I am wondering what is happening when I use the sizeof operator on the array "arr" it output 24 which is not even in the sequence. As well as when I use the sizeof operator on the "arr[0]" it outputs 4 even though the first number in the sequence is 10. This is very confusing to me if you could help explain this to me that would be wonderful, thank you.
5 Respostas
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It is pretty much obvious that you wanted to initialize a vector of integers using an array of integers by incorporating pointer arithmetic which is a little bit messy in terms of readability but good for compatibility with C++98/03.
For C++11 compliant implementations, you'd simply use iterators for initializing a vector by another container or in simplest form as the way you initialize an ordinary array directly. Consider the following examples.
// initialize vector using assignment operator (C++11)
std::vector<int> vec = {10, 20, 30, 40, 50};
// initialize vector using iterator and another container (C++11)
int arr[] = {10, 20, 30, 40, 50};
std::vector<int> vec(std::begin(arr), std::end(arr));
// initialize vector using pointer arithmetic and another container (Both C++11 and C++98/03)
int arr[] = {10, 20, 30, 40, 50};
std::vector<int> vec(arr, arr + sizeof(arr) / sizeof(arr[0]));
All three of them yield the same result with different methods which is usually dictated by implementation and project restraints.
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Sizeof returns the size of an element in the memory in bytes. The compiler uses 32 bit integers, which are 4 bytes large (32/8=4). The array contains 6 integers: 6*4=24
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Aaron Eberhardt isn‘t the name of the array basically just a pointer? if so why doesn‘t sizeof(a) return the size of pointers on the system but the lenght of the array
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Thank you very much :)
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Max I guess because C++ knows the array size and also that this must be a pointer to an array. Using an usual pointer instead just returns the pointers size, so these pointers must be treated differently