#include<stdio.h>
int main(void)
void jolly(void);//自定义函数
void deny(void);

int main(void)
  {
    jolly();
      jolly();
      jolly();
      deny();

      return 0;
  }

  void jolly(void)//函数1定义开始
  {
      printf("For he`s a jolly good fellow!\n");
  }
  
  void deny(void)//函数2定义开始
  {
      printf("Which nobody can deny!\n");
  }

//online parser: https://www.bccn.net/run/

#include <stdio.h>

void static inline jolly(void)//函数1定义开始
{
  puts("For he`s a jolly good fellow!");
}
  
void static inline deny(void)//函数2定义开始
{
  puts("Which nobody can deny!\n");
}

int main(int argc, char *argv[])
{
  jolly();
  jolly();
  jolly();
  deny();

  return 0;
}

output sample:
For he`s a jolly good fellow!
For he`s a jolly good fellow!
For he`s a jolly good fellow!
Which nobody can deny!

你输出的结果是我想要的结果，我想问下我编的那个哪里出问题？

line 2 lost of a semicolon(;)