Inline Functions in C++

================

non-inline functions

================

 

For non-inline functions, if the prototype does not change, often the files that use the function need not be recompiled.

 

lib.cpp

 

#include <stdio.h>

void show(int i) {
  printf("show %d using style 1\n", i);
}

 

main.cpp

void show(int i);

int main(int argc, const char *argv[]) {
  show(100); 
  return 0;
}

$ g++ -c -fPIC lib.cpp

$ g++ -fPIC -shared -Wl,-soname,libref.so -o libref.so lib.o

$ g++ -c main.cpp

$ g++ -o main main.o -lref -L.

$./main

show 100 using style 1

Change printf("show %d using style 1\n", i); to printf("show %d using style 2\n", i);

$ g++ -c -fPIC lib.cpp

$ g++ -fPIC -shared -Wl,-soname,libref.so -o libref.so lib.o

$ ./main

show 100 using style 2

 

================

inline functions

================

lib.h

#include <stdio.h>

inline void show(int i) {
  printf("show %d using style 1\n", i);
}

 caller1.cpp

#include "lib.h"

void call1() {
  show(100);
}

 caller2.cpp

#include "lib.h"

void call2() {
  show(100);
}

 main.cpp

xtern void call1();
extern void call2();
int main() {
  call2();
  call1();
}

$ g++ -c *.cpp

$ g++ -o main *.o

$ ./main

show 100 using style 1

show 100 using style 1

Now change  printf("show %d using style 1\n", i); to printf("show %d using style 2\n", i);

For the following commands:

$ g++ -c caller1.cpp

$ g++ -o main *.o

$ ./main

The result is

show 100 using style 2

show 100 using style 2

 

For the following commands:

$ g++ -c caller2.cpp

$ g++ -o main *.o

$ ./main

The result is

show 100 using style 1

show 100 using style 1