c - Why struct shallow copy does not work? -

i'm testing shallow copy struct code:

#include "stdio.h" #include "conio.h"  int main() {      struct str {         char * name;         int value;     };      struct str str_1 = {"go", 255};     struct str str_2;      str_2 = str_1;      str_1.name = "back";      printf("%s\n",str_1.name);     printf("%s\n",str_2.name);      printf("\n");     system("pause");     return 0; } 

i expected result should be:

back  

but was:

back go 

edit: expected because shallow copy, str_1.name , str_2.name should point same place.

edit: , dynamic allocation, got expected:

#include <stdio.h> #include <conio.h> #include <string.h>  int main() {      struct str {         char * name;         int value;     };      struct str str_1;      struct str str_2;      str_1.name = (char*) malloc(5);     strcpy(str_1.name,"go");      str_2 = str_1;      strcpy(str_1.name,"back");      printf("%s\n",str_1.name);     printf("%s\n",str_2.name);      printf("\n");     system("pause");     return 0; } 

the result is:

back  

what misunderstand here ?

take piece of paper , draw think happening @ each step slowly, , should become clear.

let's draw struct str so:

--------------------- | const char * name | --------------------- | int value         | --------------------- 

and let's denote string @ address 0xabcdef10 so:

--0xabcdef10--------------------- |  "string\0"                   | --------------------------------- 

so, when initialise str_1, need memory location hold "go", lets call 0xaaaaaa00

--0xaaaaaa00--------------------- |  "go\0"                       | --------------------------------- 

then initialise str_1 pointer that:

struct str str_1 = {"go", 255};  --str_1--------------------------- | const char * name = 0xaaaaaa00 | ---------------------------------- | int value = 255                | ---------------------------------- 

now take shallow copy of str_1, , call str_2:

--str_2--------------------------- | const char * name = 0xaaaaaa00 | ---------------------------------- | int value = 255                | ---------------------------------- 

next execute str_1.name = "back";. before, first want create new string. lets put @ 0xbbbbbb00:

--0xbbbbbb00--------------------- |  "back\0"                     | --------------------------------- 

then assign str_1.name, str_1 looks like:

--str_1--------------------------- | const char * name = 0xbbbbbb00 | ---------------------------------- | int value = 255                | ---------------------------------- 

note haven't changed str_2.

so, when @ our final "memory", see:

--0xaaaaaa00--------------------- |  "go\0"                       | ---------------------------------   ....  --0xbbbbbb00--------------------- |  "back\0"                     | ---------------------------------   .... --str_1--------------------------- | const char * name = 0xbbbbbb00 | ---------------------------------- | int value = 255                | --str_2--------------------------- | const char * name = 0xaaaaaa00 | ---------------------------------- | int value = 255                | ---------------------------------- 

so str_1 points @ new string, , str_2 points @ old string.

in case describe dynamic, never update pointers in struct, go through same exercise draw out happens memory in case.