*p++、*++p、++*p、(*p)++

Arrangement c I found several interesting things when I was using the language pointer , Go directly to the code to explain .

Background conditions

GCC edition 7.4.0
i++： First use i, Re increment
++i： Increase first , Reuse i

Priority table ：

Operators of the same priority , The operation order is determined by the combination direction .

*p++

#include <stdlib.h>
int buf[]={
    1,3,5};
int main()
{
    
    int *p = buf;
    printf("%d\r\n",*p++);
    printf("%d",*p);
    return(0);
}

result ：

analysis

According to the priority table * and ++ At the same priority . Then look at the direction , From right to left . therefore p++, then *.
Look again i++ The nature of , First use i, Re increment . It was printed first when printing for the first time *p yes buf[0]=1, After printing , Re execution *（p++）, here p Point to buf[1]. So the second print out is 3.

*++p

#include <stdlib.h>
int buf[]={
    1,3,5};
int main()
{
    
    int *p = buf;
    
    printf("%d\r\n",*++p);
    printf("%d",*p);
    return(0);
}

result ：

analysis

This is very simple The previous question is the same as the previous one . First ++p then *.
++p It is to calculate first and then execute . So both point to buf[1]=3

++*p

#include <stdlib.h>
int buf[]={
    1,3,5};
int main()
{
    
int *p= buf;
    
    printf("%d\r\n",++*p);
    printf("%d",*p);
    return(0);
}

result ：

analysis

First *p, then ++, because *p=1, then ++（*p）, It is to calculate first and then execute , So the result is 2.

(*p)++

#include <stdlib.h>
int buf[]={
    1,3,5};
int main()
{
    
int *p= buf;
    
    printf("%d\r\n",(*p)++);
    printf("%d",*p);
    return(0);
}

result ：

analysis

First in parentheses *p=1, And then again ++.
The first printing is incremental because it is executed first , So the result is 1
The second printing is over because of the increment, and then 2