Two methods to judge the storage of large and small end

1. Coercive transformation

Given int Variable of type a, The assignment is 1,1 Of 16 Into the system for ：00 00 00 01.

If the small end stores a Stored as ：01 00 00 00.

Big end storage is ：00 00 00 01.

You can take out a Your address is forced to char* type ,*(char*)&a To judge the value as 0/1.

If it is 0 It is big end storage , yes 1 Store for small end .

The code implementation is as follows ：

int main()
{
	int a = 1;
	//00 00 00 01
	if (*(char*)&a == 1)
		printf(" Small end storage \n");
	else
		printf(" Big end storage \n");
	return 0;
}

2. Using Commons

Pictured ：

We can find out a,b The address of is the same , So for b Changes , Will also change a. because b Occupy four bytes , and a There is only one byte .

  In the case of small end storage c In Chinese, it means 1,

On the contrary, if it is large-end storage, then c In Chinese, it means 0.

The code implementation is as follows ：

typedef union A
{
	char a;
	int b;
}A;
int main()
{
	A a;
	a.b = 1;
	if (a.a == 1)
		printf(" Small end storage \n");
	else
		printf(" Big end storage \n");
}