Question

I have been working on a question in c++ but im stuck in a part.
It is a long code but I will only post a small part of it where i am stuck now. Here it is;

char* x = (char*)malloc(sizeof(char));
char y = (char)malloc(sizeof(char));
y = *x;

The problem is when I do it, if X points to a char named “HELLO”, when i print y it only prints ‘H’ letter.

I want to copy the whole word that char* has into a char variable. Also I dont know the sizes of the char’s, because they are given by the user. So the length could be anything. I tried strcpy() but couldnt managed to solve the problem.

Any help appreciated.

**Thanks for all the comments. Now I decided to use std::string according to your comments. And I guess Im never gonna use malloc again. Now I have to go back to code and change everything according to string and check if everything works.

Answer 1

To paraphrase Inigo Montoya, I don’t think this code means what you think it means.

malloc(sizeof(char))

This allocates a certain amount of memory. How much memory is allocated is dictated by malloc‘s parameter. In this case you’re passing sizeof(char) which is, by definition, one byte exactly. Therefore, you are allocating one byte of memory.

If you insist on using malloc (more on this later) then what you should be doing is figuring out how long the string you want to store is, add one more byte for the NULL terminator, and then malloc that much. In the case of the string Hello, world. which is 13 characters, the corresponding call would be:

malloc(14)

Next:

char y = (char)malloc(sizeof(char));

malloc returns a pointer to the memory it allocated for you. y in this case isn’t a pointer, it’s just a char. The two aren’t the same. It compiles and appears to work because you use the bludgeoning tool known as a C-style cast: (char)malloc(...). This tells the compiler, “I know I’m pointing the gun at my foot. Just do what I tell you and don’t complain.” Which it dutifully does. But you’re doing the wrong thing for several reasons:

1. malloc returns a pointer but you’re trying to cast it to a char
2. You only allocated 1 byte but you assumed you were allocating a whole string’s worth of memory
3. y is just a char but you treat it as if it were a whole string.

So if you again insist on using malloc, you need to do something like this:

static const char* HELLO = "Hello, malloc.";

char* x = malloc(strlen(HELLO)+1);
strcpy(x, HELLO);
char* y = malloc(strlen(x)+1);
strcpy(y, x);

But, you shouldn’t be using malloc at all in C++. Instead, you should be using std::string:

std::string x = "Hello, string.";
std::string y = x;

This is better because:

1. strings manage thier own memory. You don’t leak memory like you did in your code.
2. string is type-safer. You violated type safety when you cast the return from malloc to a char. You can do it, but its not a char. That’s bad. string won’t let you shoot yourself like this without great effort.
3. It’s less code. The best code is the code you never write.