c - 为什么我的编译器不想编译这段代码？

Question

昨天我遇到了这个有问题的代码，我一直想理解并纠正它。到目前为止我已经做了一些研究并纠正了它，但我想知道是否还有其他方法来纠正代码？

# include < stdio .h >
# include < stdlib .h >
int * sub ( int * x , int * y) { //return type should be pointer but is address//
int result = y - x ; //integer becomes pointer//
return &result;
}
int main ( void ) {
int x = 1;
int y = 5;
int * result = sub (&x , &y ); //function has addresses as parameters but not pointers//
printf ("% d\n" , * result );
return EXIT_SUCCESS ;
}

我会简单地删除所有指针和地址:

# include < stdio .h >
# include < stdlib .h >
int sub ( int x , int y) {
int result = y - x ;
return result ;
}
int main ( void ) {
int x = 1;
int y = 5;
int result = sub (x , y );
printf ("% d\n" , result );
return EXIT_SUCCESS ;
}

Answer 1

只需删除导入语句及其周围的空格:

#include <stdio.h>
#include <stdlib.h>

int sub(int x, int y)
{
  int result = y - x;
  return result;
}

int main(void)
{
  int x = 1;
  int y = 5;
  int result = sub(x, y);
  printf("% d\n", result);
  return EXIT_SUCCESS;
}