5.8 증가, 감소 연산자 예제

#include <stdio.h>

int main()
{
	//int a = 0;
	//a++; // a = a + 1 or a += 1; 
	//printf("%d\n", a);

	//++a; // a = a + 1 or a += 1; 
	//printf("%d\n", a);

	/*double b = 0;
	b++;
	printf("%f\n", b);

	++b;
	printf("%f\n", b);*/

	//int count = 0;
	//while (count < 10) // ++count or count++
	//{
	//	printf("%d ", count++); // ++count or count++ 
	//	//count;
	//}


	/*int i = 1, j = 1;
	int i_post, pre_j;

	i_post = i++;
	pre_j = ++j;

	printf("%d %d\n", i, j);
	printf("%d %d\n", i_post, pre_j);*/

	/*int i = 3;
	int l = 2 * --i;
	printf("%d %d\n", i, l);

	i = 1;
	l = 2 * i--;
	printf("%d %d\n", i, l); */

	//	/* very high precedence */
	//int x, y, z;
	//x = 3, y = 4;
	//z = (x + y++) * 5; // (x+y)++ or x+(y++) ? 
	//printf("%d %d %d", x, y, z);

	/* ++ and affect modifiable l-vaues */
	//int x = 1, y = 1, z;
	//z = x * y++; // (x) * (y++), not (x*y)++ 
	//z = (x * y) ++; //error
	//z = 3++; //error

	/* Bad practices */
	/*int n = 1;
	printf("%d %d", n, n * n++);
	int x = n / 2 + 5 * (1 + n++);
	int y = n++ + n++;*/

	return 0;
}