What is wrong with my 'append' algorithm or code? - c

Question

2 strings are given, second word will be append to first one and 3rd variable will store this. For example;
char *str1 = "abc";
char *str2 = "def";
char *str3 = "abcdef"; //should be
Here is my code, I get runtime error:
#include <stdio.h>
#include <malloc.h>
void append(char *str1, char *str2, char *str3, int size1, int size2)
{
int i=0;
str3 = (char*) malloc(size1+size2+1);
str3 = str1;
while (str2[i] != '\0') {
str3[i+size1] = str2[i];
i++;
}
str3[size1+size2] = '\0';
}
int main()
{
char *str1 = "abc";
char *str2 = "def";
char *str3;
append(str1, str2, str3, 3, 3);
return 0;
}

Answer 1

str3 = (char*) malloc(size1+size2+1);
str3 = str1;
Here's your problem. Doing this replaces the pointer to the correct amount of space from malloc to the pointer where str1 is contained. Keeping with your loop design, change this to:
str3 = malloc(size1+size2+1);
for (int j = 0; str1[j] != '\0'; j++)
str3[j] = str1[j];
Also, see this question/answer about casting the result of malloc in C:
Do I cast the result of malloc?

Answer 2

There is another issue with the code. You pass pointer by value. So any malloc inside a function will do only local changes. After function ends your pointer will still point to the old value. You should pass a pointer to pointer if you want to change it. See an example:
#include <stdio.h>
char *c = "Second";
void assign(char *s) { s = c; }
int main()
{
char *str = "First";
assign(str);
printf("String after assign: %s\n", str);
return 0;
}
After running the program you will see 'First' in you console. The correct code is:
#include <stdio.h>
char *c = "Second";
void assign(char **s) { *s = c; }
int main()
{
char *str = "First";
assign(&str);
printf("String after assign: %s\n", str);
return 0;
}

Answer 3

#include <stdio.h>
#include <stdlib.h> //to standard
#include <string.h>
char *append(const char *str1, const char *str2, int size1, int size2){
//parameter char *str3 is local variable.
//It is not possible to change the pointer of the original.
//str3 = str1;//<<-- memory leak
//str3[i+size1] = str2[i];//<<-- write to after str1(can't write!)
char *str3 = (char*) malloc(size1+size2+1);
memcpy(str3, str1, size1);//copy to alloc'd memory.
memcpy(str3 + size1, str2, size2);//copy to after str1
str3[size1+size2] = '\0';
return str3;
}
int main(){
char *str1 = "abc";
char *str2 = "def";
char *str3;
str3 = append(str1, str2, 3, 3);
printf("%s\n", str3);
return 0;
}