c - char *str and malloc'ing the memory for str, still getting SEGFAULT -

i have written small c program reverse string. in though, declare str as
char *str
, then
str = (char*)malloc(20); str = "this test";
however, wont segfault if use
char str[20] = "this test"

#include <stdio.h>  #include <stdlib.h>  #include <string.h>   void swap(char *a, char *b) {    char temp;    temp = *a;   *a = *b;    *b = temp;  }   char* reverse(char *str) {    int len = strlen(str);    printf("len = %d \t %s\n", len, str);    int = 0;     if (len == 0)      return null;    (i=0; i<len/2; i++)    {      swap((str+i), (str+len-1-i));         printf("%s\n", str);    }    return str;  }   int main(void) {    char *str;    str = (char *)malloc(20);    str = "this test";     printf("%s\n", str);    reverse(str);    printf("%s\n", str);    return 0;  }  

my understanding segfault, if declare,
char *str="this test" becuase constant string.
so, thought, when, malloc, str allocated heap , heap memory accessible both functions. but, still segfault error.

when do

str = (char*)malloc(20); str = "this test"; 

you reassign pointer str point somewhere else right after allocation.

in fact make point string literal, array of read-only characters. attempting modify string literal leads undefined behavior.

the simple solution use arrays instead. or copy memory allocate using strcpy.