原来C语言也可以面向对象(一)_c语言代码可以具备面向对象的风格吗

C语言是一门博大精深的语言，我相信没有研读过Linux中代码的人，很少有人体会到吧，废话不多说，简单的写了一个测试demo

test.h

typedef void (* func1_callback)(void);
typedef void (* func2_callback)(int i);
 
typedef struct {
	int size;
	func1_callback func1_cb;
	func2_callback func2_cb;
	
} testCallback;
 
typedef struct {
	int size;
	int (*init)(testCallback *_tcb);
	int (*set)(int i);
	int (*release)(void);
} testInterface;

test.c

#include "test.h"
#include <stdio.h>
 
testCallback tcb;
 
int test_init(testCallback *_tcb) {
	printf("This is test init function.\n");
	tcb = *_tcb;	
	tcb.func1_cb();
	return 0;
}
 
int test_set(int i) {
	printf("This is test set function, argumaent i is : %d\n", i);	
	tcb.func2_cb(i);
	return 0;
}
 
int test_release(void) {
	printf("This is test release function.\n");	
	return 0;
}
 
static const testInterface tInterface = {
	.size = sizeof(testInterface),
	.init = test_init,
	.set = test_set,
	.release = test_release,	
};
 
const testInterface *get_test_interface() {
	return &tInterface;	
}

main.c

#include "test.h"
#include <stdio.h>
void func1_cb() {
	printf("This is func1 cb, has no argument.\n");	
}
 
void func2_cb(int i) {
	printf("This is func2 cb, argument i is : %d\n", i);
}
 
testCallback tcb = {
	sizeof(testCallback),
	func1_cb,
	func2_cb,	
};
 
static const testInterface *ti;
 
int main(void) {
	ti = get_test_interface();
	ti->init(&tcb);
	ti->set(10);
	ti->release();
 
	return 0;
}

这个测试demo很简单，不用解释了吧，这篇先这样了。