The value assignment between structs of the same type is not necessarily valid.

The value assignment between structs of the same type is not necessarily valid.

The value assignment between structs of the same type is not necessarily valid.

 

Debugging for this issue for a long time today...

 

Previously, I saw a technique about struct usage. This technique can avoid memory fragmentation and ensure that the memory of the struct should not be scattered as much as possible.

Struct struct_name

{

Element_type varible;

...;

Element_type pointer [0];

}

This document describes how to use this function.

 

 

 

For common struct,

 

struct num{int x;int y;};

This type of struct can be assigned operator = Value

 

But!

The special usage described above does not work.

 

#include 
 
  #include 
  
   int main(){struct node{int a;char string[0];};struct num{int x;int y;};struct num m = {1,3};struct num n = {2,4};/***ATTENTION! m = n;*/m = n;printf(m.x:%d m.y:%d n.x:%d n.y:%d,m.x,m.y,n.x,n.y);struct node* p_string_1 = (struct node*)malloc(sizeof(100)+sizeof(int));struct node* p_string_2 = (struct node*)malloc(sizeof(100)+sizeof(int));char hello[] = {Hello world};char panic[] = {Don't panic}; int tmp = 0;for(tmp = 0;tmp < sizeof(hello)/sizeof(hello[0]); tmp++){p_string_1->string[tmp] = hello[tmp];}for(tmp = 0;tmp < sizeof(panic)/sizeof(panic[0]); tmp++){p_string_2->string[tmp] = panic[tmp];}printf(1:%s2:%s,p_string_1->string,p_string_2->string);/***ATTENTION! HERE IS AN EVEIL!*/*p_string_1 = *p_string_2;printf(1:%s2:%s,p_string_1->string,p_string_2->string);free(p_string_1);free(p_string_2);return 0;}
  
 

 

 

Guess what the printed result is?