Question

I want to know how inline function call is replaced by the inline code.
i read in some book saying compiler may treat the inline function as normal function.

can any own explain how inline function works.

Answer 1

From the C++ FAQ:

When the compiler inline-expands a function call, the function’s code
gets inserted into the caller’s code stream (conceptually similar to
what happens with a #define macro). This can, depending on a zillion
other things, improve performance, because the optimizer can
procedurally integrate the called code — optimize the called code into
the caller.

There are several ways to designate that a function is inline, some of
which involve the inline keyword, others do not. No matter how you
designate a function as inline, it is a request that the compiler is
allowed to ignore: it might inline-expand some, all, or none of the
calls to an inline function. (Don’t get discouraged if that seems
hopelessly vague. The flexibility of the above is actually a huge
advantage: it lets the compiler treat large functions differently from
small ones, plus it lets the compiler generate code that is easy to
debug if you select the right compiler options.)

In the simplest case, the inline function is dropped into its call site as if you had copy-and-pasted it there. Thus for,

inline int madd( int a, int b, int c ) 
{
   return a * b + c;
}


void foo( int data[3] )
{
   int result = madd( data[0], data[1], data[2] );
   printf("%d\n", result); // note to pedants: this is simpler than a cout stream, so there
}

the compiler could turn it into

void foo( int data[3] )
{
   int result = data[0] * data[1] + data[2] ; // madd is replaced inline
   printf("%d\n", result); 
}