Passing string variable by reference - c

Question

I've this piece of code :
#include <stdio.h>
#include <string.h>
void b(char *in, char ** out);
int main()
{
char a[] = {'w','o','r','l','d','\0'};
char *d = nullptr;
b(a, &d);
printf("hello %s\n", d);
return 0;
}
void b(char *in, char ** out)
{
char tmp[10];
for(int i=0;i<6;i++)
tmp[i]=in[i];
*out=tmp;
printf("%s\n", *out);
}
I except to get theses printf :
world
hello world
But I get these :
world
hello
Why the d variable isn't fullfilled ? :(
Thanks for any clue about that !

Answer 1

Inside of b(), you are setting the char* referred by out to point at a local char[] array, which is NOT being null-terminated (thus breaking "%s" in printf()), but also WILL go out of scope when b() exits, thus the caller (ie main()) ends up with a dangling char* pointer to invalid memory.
You tagged your question as c++. C++ is not C. You should use std::string instead of char[] in this situation, eg:
#include <iostream>
#include <string>
void b(const std::string &in, std::string &out);
int main()
{
std::string a = "world";
std::string d;
b(a, d);
std::cout << "hello " << d << "\n";
return 0;
}
void b(const std::string &in, std::string &out)
{
std::string tmp = in.substr(0, 6);
out = tmp;
std::cout << out << "\n";
}
Otherwise, if you really want to use char*, in C or C++, you will need to use dynamic memory allocation, eg:
#include <stdio.h> // or <cstdio> in C++
#include <string.h> // or <cstring> in C++
void b(char *in, char ** out);
int main()
{
char a[] = "world";
char *d = nullptr;
b(a, &d);
printf("hello %s\n", d);
delete[] d; // or free() in C
return 0;
}
void b(char *in, char ** out)
{
char *tmp = new char[7]; // or malloc() in C
for(int i = 0; i < 6; ++i) {
tmp[i] = in[i];
}
tmp[6] = '\0';
*out = tmp;
printf("%s\n", *out);
}