/C++/Sizeof’ on array function parameter ‘arr’ will return size of ‘int*’ [-Wsizeof-array-argument]

Sizeof’ on array function parameter ‘arr’ will return size of ‘int*’ [-Wsizeof-array-argument]

There is no way to determine the length inside the function. However you pass arr,
sizeof(arr) will always return the pointer size. So the best way is to pass the
number of elements as a separate argument.
sizeof only works to find the length of the array if you apply it to the original array.

int arr[5]; //real array. NOT a pointer
sizeof(arr); // :)
However, by the time the array decays into a pointer, sizeof will give the 
size of the pointer and not of the array.

void getArraySize(int arr[]){
sizeof(arr); // will give the pointer size
}
There is some reasoning as to why this would take place. How could we make things
so that a C array also knows its length? A first idea would be not having arrays
decaying into pointers when they are passed to a function and continuing to keep 
the array length in the type system.
When you pass an array to a function it decays to pointer. So the sizeof function
will return the size of int *. This is the warning that your compiler complining
about

Source: stackoverflow.com

Similar codes

1 vote239 views2 answers
Lambda capture as const cpp
C++
4 votes239 views1 answer
How to print a string to console in c++
C++
2 votes241 views1 answer
C++ endl
C++
0 vote253 views1 answer
How to create a c++ templeate
C++
1 vote256 views1 answer
C++ isalphanum
C++
15 votes235 views10 answers
Hello world
C++
0 vote236 views1 answer
CamelCase Problem
C++
5 votes255 views1 answer
Break in c++
C++
0 vote231 views1 answer
Consell de forces polítiques de Catalunya
C++
0 vote237 views1 answer
ImportError: cannot import name 'TFLiteConverter' from 'tensorflow.lite'
C++