Memcpy Function crashing in C - c

Question

I am using memcpy in my program and used it as its syntax. But the function crashes. I read some post here and tried to initialize both my char arrays but I am not getting the problem here . Can anyone please look in it?
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(int argc, char *argv[]) {
int i;
char *str_result;
char *str_dst = NULL;
str_result = (char *) malloc(100);
str_dst = (char *) malloc(100);
str_result = "Action done";
size_t len = strlen(str_result);
printf("string length is = %d\n", len);
memcpy(str_dst, str_result, len);
str_dst [len] = '\0';
for (i = 0; i < len; i++) {
printf("%s\n", str_dst[i]);
}
free(str_dst);
free(str_result);
return 0;
}

Answer 1

The error is here:
free (str_result);
After the line str_result = "Action done";, str_result no longer points to the 100-byte block of memory you reserved, but to the string (const char*) "Action done". You can't free this pointer.

Answer 2

str_result = "Action done";
After this str_result does not point to the block which was allocated by malloc . Thus freeing memory not allocated by malloc or similar can cause error in program (one of my favourite statement).
Use strcpy also suggested in comments -
strcpy(str_result,"Action done");
And while printing str_dst -
for (i = 0; i < len; i++) {
printf("%s\n", str_dst[i]); // passing char where char * is expected
}
You don't need a loop to print str_dst .Simply this will work -
printf("%s",str_dst);
Corrected code (Demo)-
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(int argc, char *argv[]) {
int i;
char *str_result;
char *str_dst = NULL;
str_result = malloc(100);
str_dst = malloc(100);
strcpy(str_result, "Action done"); // copy string at location where str_result points to
size_t len = strlen(str_result);
printf("string length is = %zu\n", len);
memcpy(str_dst, str_result, len);
str_dst[len]='\0'; // manually add null character, memcpy won't add it
printf("%s\n", str_dst);
free(str_dst);
free(str_result);
return 0;
}

Answer 3

Below your corrected code with memcpy
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(int argc, char *argv[])
{
int i;
char *str_result;
char *str_dst = NULL;
str_result = (char *) malloc (100);
str_dst = (char *) malloc (100);
strcpy (str_result, "Action done");
size_t len = strlen (str_result);
printf ("string length is = %zu\n", len);
memcpy(str_dst, str_result, len+1);
for (i =0;i<len;i++)
{
printf( "%c\n", str_dst[i]);
}
free (str_dst);
free (str_result);
return 0;
}
Error in your code:
There is a typo in str_result declaration: , instead of ;
You must use strcpy to copy a string to a char *: strcpy (str_result, "Action done"); not str_result = "Action done";
%c is the format to print chars, not %s.
If you want to define len as size_t type you must use %zu format for printf.

Answer 4

You have the following bugs:
str_result = (char *) malloc(100);
Casting the result of malloc is pointless.
str_result = "Action done";
This is not how you assign strings in C. You must use strcpy(). Right now, you create a memory leak. Which is why free(str_result); later causes your program to crash, str_result is no longer pointing at the allocated memory.
printf("string length is = %d\n", len);
The correct format specifier for size_t is %zu.
printf("%s\n", str_dst[i]);
str_dst is a string, not an array of strings. It makes no sense to call it from a loop unless you intend to use %c to print it character by character.