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C++ Pointer to an Array

rahul ummadisetti

7y1k

output :

array :0x7fffffffde20

&array+1 :0x7fffffffde38

*(&array+1) :0x7fffffffde38

(*(&array+1)-array) :6

*(&array+1) and &array+1 refer to the same memory location.

compiler throwing an exception while trying to get the size of the array using a pointer as ((&array+1) - array)

what is the difference between ((&array+1)-array) and (*(&array+1)-array)

2 Replies

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Laxmidhar Sahoo7y ago
&array+1 : This will provide you the address of second index
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Amit Kumar7y ago
Hi,Use below code.
#include int main()
{
using T = int[5] ; // type T is 'array of 5  int'
std::cout << sizeof(T) << '\n' ; // size of 'array of 5 int'
T* ptr = nullptr ; // type T* is 'pointer to array of 5 int'
std::cout << sizeof(ptr) << '\n' ; // size of 'pointer to array of 5   int'
std::cout << sizeof(*ptr) << '\n' ; // size of 'array of 5 int'
}
output:
For more detail follow the below link.http://www.cplusplus.com/forum/beginner/221598/
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