Integer promotion and size side byte order

With an integer type number 1 For example His binary representation is

000000000000000000000000000000001;

And in computer storage His data is stored in hexadecimal , namely

0x 00 00 00 01

From left to right is the high to low order of the data , And the memory in the computer also corresponds to high memory to low memory , Then there is the concept of byte order at the big and small ends

Big endian byte order ： It means that the high bit of data is stored in the low address , The low order of the data is stored at the high address

Small endian byte order ： The high bit of data is stored at the high address , The lower order of the data is stored at the lower address

So how do we judge what we use vs When storing data, do you use large end byte order or small end byte order , We can create one char The pointer of type points to the number of integer type 1, because char* Only one byte can be accessed from high to low , So we can pass char A pointer to access 1 The first byte of , Let's see what he is 0 still 1, So as to determine the size of its storage end

The code is as follows :

char*p=NULL;// Initialization pointer 
int a=1;
p=&a;//p Point to a One byte in 
if(*p)
printf(" It's the big end ");
else
printf(" It's the small end ");
// have a look p Access after application a Middle specific byte 

So what is integer lift

Let's look at a piece of code

int main(){
char a=-1;
signed char b=-1;
unsigned char c=-1;
printf("%d %d %d",a,b,c);

that a,b,c What are the results of ;

We know , Numbers are stored in memory in the form of complements -1 The original code of is

10000000 00000000 000000000 00000001

In addition to the sign bit, that is, the highest bit, the bit is reversed Get the inverse ;

11111111 11111111 11111111 11111110;

Add one to the inverse code to get the complement

11111111 11111111 11111111 11111111;

However char Type can only store one byte So truncation happens Take the lower byte namely

11111111

Use a Make a calculation or use %d Output a The integer will be promoted to look at the highest bit of a byte ; if 1 Then the preceding three bits are all 1jiayi Bytes of ; The highest bit is 0 Then fill in the front 0

For now 11111111 The highest bit is 1 Therefore, it is

11111111 11111111 11111111 11111111 For the complement

Add one to it to get the source code and output it The result is -1;

Empathy b and a Also, integer promotion will occur ;char Types are signed by default signed char;

and unsigned char It won't be like this , Because it is set to unsigned character type when defining So it thinks that the highest bit to be intercepted is not a sign bit but a number bit Therefore, all zeros are filled in front during integer lifting

The box %d Output c It will be promoted to

00000000 00000000 00000000 11111111

The highest sign bit is zero So it's a positive number The complement of positive numbers is the same So the direct output is 256

What is the significance of integer upgrade