Review-C language Inline compilation-Beginner (2)

Assembly takes out the value in memory

1# include <stdio.h>2 3 intMain ()4 {5         inti = -;6         intRET =0;7 8         int*p = &i;9         //ret = *p;Ten __asm__ ( One                 "Ldr%0, [%1]" A  -:"+r"(ret)//Output -:"R"(p)//input the         ); -printf"Hello world!%d\n", ret); -}

Description: [] equivalent to *p in the *,%1 equivalent to address p, that is [%1] = *p, the contents of the address p out!

Change the value of a variable

1#include <stdio.h>2 3 intMain ()4 {5         inti =6;6         intRET =0;7 8         //i = +;9 __asm__ (Ten                 "mov r0, #100 \ n" One                 "str r0, [%1]\n" A                 //"Ldr r0, [%1]\n" -  -:"+r"(ret)//Output the:"R"(&i) -         ); -  -printf"I%d\n", i); +}

Note: The immediate number 100 is uploaded to R0, then r0 to * (&i), that is, change the value of the variable I address.

Assigning a value to an array

1#include <stdio.h>2 3 intMain ()4 {5         intarr[3] = {0};6 7 __asm__ (8                 "mov r0, #1 \ n"9                 "str r0, [%0, #0]\n"Ten                 "add R0, r0, #1 \ n" One                 "str r0, [%0, #4]\n" A                 "add R0, r0, #1 \ n" -                 "str r0, [%0, #8]\n" -  the://"+r" (arr)//Error -:"R"(arr) -:"R0" -         ); +  -printf"Arr[0]%d\n", arr[0]); +printf"Arr[1]%d\n", arr[1]); Aprintf"Arr[2]%d\n", arr[2]); at}

Description: The input section is an array of the first address of ARR, paying special attention.

About the back shift of the address

1#include <stdio.h>2 3 intMain ()4 {5         intarr[3] = {1,2,3};6         inti =Ten; 7         intp =0;8 9 __asm__ (Ten                 //Ldr%0, [%2]\n]//addr and value not change One                 //Ldr%0, [%2, #4]\n]//addr, Value change A                 //Ldr%0, [%2, #4]!\n]//addr Change, value change -                 "Ldr%0, [%2], #4 \ n" //addr Change, value no change -                 "mov%1,%2\n" the      -:"+r"(i),"+r"(p)//Error -:"R"(arr) -:"R0" +         ); -      +printf"i is:%d\n", i); Aprintf"P is:%x\n", p);  atprintf"arr is:%x\n", arr); -}

Description: The difference between line 11th, line 12, and line 13: for example, suppose the arr address is 0x10

11 Line: Is 0x10 + 4 is equal to the address of the 0x14, arr itself or 0x10, take the value of the 0x14 inside to I, the next arr + 4 or 0x14

12 line: More than an exclamation mark, is now the base or 0x10,+ 4, arr value becomes 0x14, the next time arr + 4 becomes 0x18

13 Line: First assign the value of the 0x10 inside to I, then add the address to the 0x14

　　　　

Review-C language Inline compilation-Beginner (2)