【C语言】数据储存小练习

int main()
{
	//unsigned char 0~255    unsigned char最大能存的数是255及11111111

	unsigned char a = 200;

	//unsigned char a   11001000      无符号型整形提升添加0
	//unsigned char a 运算时整形提升为00000000000000000000000011001000
	unsigned char b = 100;

	//unsigned char b    01100100     无符号型整形提升添加0
	//unsigned char b 运算时整形提升为00000000000000000000000001100100
	

	unsigned char c = 0;
	//a和b整形提升 无符号型整形提升添加0
	//00000000000000000000000011001000
	//00000000000000000000000001100100
	// 相加算出没有被截断的c的值
	//a+b的值 00000000000000000000000100101100

	c = a + b;
	//unsigned char c = 00101100  被截断了
	//00000000000000000000000000101100  整形提升时c的值
	//
	printf("%d %d", a + b, c);
	//              300   44
	//a+b的值 00000000000000000000000100101100           == 300
	//00000000000000000000000000101100  整形提升时c的值  == 44
	return 0;
}

int main()
{
	char killer = 0;
	for (killer = 'A'; killer <= 'D'; killer++)
	{
		if ((killer != 'A') + (killer == 'C') + (killer == 'D') + (killer != 'D') == 3)
		{
			printf("%cn", killer);
		}
	}

	return 0;
}

int main()
{
	int a = 0;
	int b = 0;
	int c = 0;
	int d = 0;
	int e = 0;
	for (a = 1; a <= 5; a++)
	{
		for (b = 1; b <= 5; b++)
		{
			for (c = 1; c <= 5; c++)
			{
				for (d = 1; d <= 5; d++)
				{
					for (e = 1; e <= 5; e++)
					{
						if (((b == 2) + (a == 3) == 1)
							&& ((b==2) + (e==4) == 1)
							&& ((c==1) + (d==2) == 1)
							&& ((c==5) + (d==3) == 1)
							&& ((e==4) + (a==1) == 1)
							)
						{
						    //abcdef中名次不能出现重复
							if(a*b*c*d*e == 120)
								printf("a=%d b=%d c=%d d=%d e=%dn", a, b, c, d, e);
						}
					}
				}
			}
		}
	}

	return 0;
}