when do we need to pass the size of array as a parameter

Question

I am a little bit confused about pass an array in C/C++. I saw some cases in which the signature is like this

void f(int arr[])

some is lik

Answer 1

C and C++ are not the same thing. They have some common subset, though. What you observed here is that the "first" array dimension when passed to a function always results just in a pointer being passed. The "signature" (C doesn't use this term) of a function declared as

void toto(double A[23]);

is always just

void toto(double *A);

That is that the 23 above is somewhat redundant and not used by the compiler. Modern C (aka C99) has an extension here that lets you declare that A always has 23 elements:

void toto(double A[static 23]);

or that the pointer is const qualified

 void toto(double A[const 23]);

If you add other dimension the picture changes, then the array size is used:

void toto(double A[23][7]);

in both C and C++ is

void toto(double (*A)[7]);

that is a pointer to an array of 7 elements. In C++ these array bounds must be an integer constant. In C it can be dynamic.

void toto(size_t n, size_t m, double A[n][m]);

They only thing that you have to watch here is that here n and m come before A in the parameter list. So better you always declare functions with the parameters in that order.

Answer 2

WHen an array is passed in C or C++ only its address is passed. That is why the second case is quite common, where the second parameter is the number of elements in the array. The function has no idea, only by looking at the address of the array, how many elements it is supposed to contain.