数据存储例题

1.判断大小端

#include <sdtio.h>
int check_sys()
{
	int i = 1;
	if (*(char*)&i == 1)
	{
		return 1;
	}
	else
	{
		return 0;
	}
}
int main()
{
	int ret = check_sys();
	if (ret = 1)
	{
		printf("小端\n");
	}
	else
	{
		printf("大端\n");
	}

	return 0;
}

对函数化简得：

#include <stdio.h>
int check_sys()
{
	int i = 1;
	return *(char*)&i;
}
int main()
{
	int ret = check_sys();
	if (ret = 1)
	{
		printf("小端\n");
	}
	else
	{
		printf("大端\n");
	}

	return 0;
}

总结：定义一个int类型，用char*强制转换，取出它低字节序的一个字节的内容，判断是否为1或0，若为1则证明低地址处存放的是低字节序的内容，判断为小端，若为0，则相反；

2.判断如下代码a的结果

int main()
{

	char a = -128;
	printf("%u", a);
	return 0;
}

a=4294967168，为解释如下

int main()
{
	//char -128~127
	char a = -128;

	//10000000000000000000000010000000
	//11111111111111111111111101111111
	//11111111111111111111111110000000 - 截断
	//10000000 - a
	//11111111111111111111111110000000 - 提升
	//
	printf("%u\n", a);
	printf("%d\n", a);
	//11111111111111111111111110000000
	//10000000000000000000000001111111
	//10000000000000000000000010000000
	//-128

	//
	//
	//%u - 打印无符号整数
	//
	return 0;
}

3.判断如下代码i的结果

int main()
{
	int i = -20;
	unsigned int j = 10;
	printf("%d\n", i + j);

	return 0;
}

i=-10，解释如下

 int main()
{
	int i = -20;
	//10000000000000000000000000010100
	//11111111111111111111111111101011
	//11111111111111111111111111101100 - -20的补码
	//
	unsigned int j = 10;
	//00000000000000000000000000001010
	printf("%d\n", i + j);
	//11111111111111111111111111101100
	//00000000000000000000000000001010
	//11111111111111111111111111110110 - 补码
	//10000000000000000000000000001001
	//10000000000000000000000000001010 -> -10
	//
	return 0;
}