select all options that applies #include #define NAME_LEN 100struct part{int number;char name [NAME_LEN+1];int on_hand;};struct part build_part(int number, char *name, int on_hand) {struct part p;p.number = number;strcpy(p.name, name);on_hand;p.on_handreturn p;}int main() {}=struct part plIn the code above:=build_part(10, "Memory", 4);a copy of the contents of p is assigned to p1.the location of p is assigned to p1.Op1 will not have a meaningful value, as the contents of p cannot be accessed outside the functionbuild_part.the strcpy call will not work, as name and p.name have different types.

Answers:- The correct options are: the location of p is assigned to p1. p1 will not have a…

#include #define NAME_LEN 100 struct part{ int number; char name [NAME_LEN+1]; int on hand; }; struct part build_part(int number, char *name, int on_hand) { struct part p; p.number = number; strcpy(p.name, name); p.on_hand = on_hand; return p; } int main() { struct part pl = build_part(10, "Memory", 4); } in the code above: a copy of the contents of p is assigned to p1. the location of p is assigned to p1. p1 will not have a meaningful value, as the contents of p cannot be accessed outside the function build_part. the strcpy call will not work, as name and p.name have different types.