Question

inline int factorial(int n) {     if(!n) return 1;     else return n*factorial(n-1); } 

As I was reading this, found that the above code would lead to ‘infinite compilation’ if not handled by compiler correctly.

How does the compiler decide whether to inline a function or not ?

Answer 1

First, the inline specification on a function is just a hint. The compiler can (and often does) completely ignore the presence or absence of an inline qualifier. With that said, a compiler can inline a recursive function, much as it can unroll an infinite loop. It simply has to place a limit on the level to which it will ‘unroll’ the function.

An optimizing compiler might turn this code:

inline int factorial(int n) {     if (n <= 1)     {         return 1;     }     else     {         return n * factorial(n - 1);     } }  int f(int x) {     return factorial(x); } 

into this code:

int factorial(int n) {     if (n <= 1)     {         return 1;     }     else     {         return n * factorial(n - 1);     } }  int f(int x) {     if (x <= 1)     {         return 1;     }     else     {         int x2 = x - 1;         if (x2 <= 1)         {             return x * 1;         }         else         {             int x3 = x2 - 1;             if (x3 <= 1)             {                 return x * x2 * 1;             }             else             {                 return x * x2 * x3 * factorial(x3 - 1);             }         }     } } 

In this case, we’ve basically inlined the function 3 times. Some compilers do perform this optimization. I recall MSVC++ having a setting to tune the level of inlining that would be performed on recursive functions (up to 20, I believe).