sizeof(oVel) returns the size of a pointer

[ghost]

https://github.com/jwatte/EPIC/blob/master/Extrapolator.cpp#L126

oVel is an argument and passed as pointer to ReadPosition(), so sizeof(oVel) returns the size of a pointer.

Fix

Replace sizeof(oVel) with sizeof(Type) * Count

memset(oVel, 0, sizeof(Type) * Count);

[jwatte]

#include <stdio.h>

template<int Count, typename Type>
struct bar {
void func(Type x[Count]);
};

template<int Count, typename Type> void bar<Count, Type>::func(Type x[Count]) {
printf("%d\n", (int)sizeof(x));
}

int main() {
int x[10];
bar<10, int> b;
b.func(x);
return 0;
}

foo.cpp:9:29: warning: ‘sizeof’ on array function parameter ‘x’ will return size of ‘int*’

Never a dull moment in C++ world.

Thanks for the report!

[ghost]

Yes, GCC and Clang do an amazing job. I just noticed it, because I compiled a program with Clang. VS2017 didn't complain about that (didn't try with VS2019 yet).

Thanks