/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

46 votes180 views3 answers
All data types in C++
C++
0 vote128 views1 answer
Number of nodes of bst cpp
C++
2 votes205 views1 answer
Check gpu usage jetson nano
C++
3 votes135 views2 answers
Convert long int to binary string c++
C++
2 votes175 views3 answers
Online ide c++
C++
1 vote161 views1 answer
Restting a queue stl
C++
0 vote142 views1 answer
Output to console in color c++
C++
6 votes175 views2 answers
What is throw in c++
C++
19 votes142 views2 answers
C++ file .
C++
0 vote201 views1 answer
Rotate an array of n elements to the right by k steps
C++