[C Language] (* p) ++ and * p ++ and ++ * p open a fog

Environment: win7

IDE: DEV-C ++

Compiler: GCC

 

1. Let's talk about the basics of ++ I and I ++.

The Code is as follows:

#include <stdio.h>//just change simplevoid stop(void){  system("pause"); }int main(void){ int i = 1; printf("i++ = %d\n",i++); printf("i = %d\n",i);  int j = 1; printf("++j = %d\n",++j); printf("j = %d\n",j);  printf("i++ = : %d   ++i = %d\n",i++,++i); printf("i = %d\n",i);  printf("++j = : %d   j++ = %d\n",++j,j++); printf("j = %d\n",j);  stop(); return 0;}

Running result:

 i++ =  i =  ++j =  j =  i++ = :    ++i =  i =  ++j = :    j++ =  j = 

1) only when I ++ encounters a semicolon (;) will it affect the value of I. The output value of I ++ is still 1, and the value of I is also 1.

2) After I passes the semicolon, I = 2 because of auto-incrementing

3) ++ j will affect the j value and ++ j value regardless of the Semicolon ";", so J = 2

4) The value of j is also 2.

5) depending on the compiler and the operating system, the printf computing direction is also different. This is calculated from the right, first ++ I, then I ++, so output 3 and 3.

6) The last I ++ encountered a semicolon, So I = 4

7) from the right, the result printed by j is still 2, but the value obtained by the second operation is 3, so ++ j = 4

 

2. Clarify the priority of * And ++ in * p ++.

 #include <stdio.h>    stop(        system(    main(        i =        *      p = &      printf(           v = *p++      printf(      printf(      printf(      system(        }

Running result:

 -p =  v =  -p =  i = 

From the result, the ++ symbol affects the value of p and does not affect the value of I. It seems that the priority of ++ is higher than the pointer * symbol.

 

Let's look at the example.

 #include <stdio.h>    stop(        system(    main(        i =        *      p = &      printf(           v = ++*      printf(      printf(      printf(      system(        }

Calculation Result:

 -p =  v =  -p =  i = 

How can ++ not affect the value of p?

 

Let's look at the example.

Slightly adjust the code: v = * ++ p;

Calculation Result:

 -p =  v =  -p =  i = 

Ah, it turns out that * And ++ have the same priority level. The same level is calculated from right to left.

 

OK. All problems are solved gradually,

* (P ++)

* (++ P)

(* P) ++

++ (* P) should be okay.

Fog finally opened up...