Modifying pointer value in function

Question

Can anybody point out why does the following code need char** pointer in the modify function. If i just pass char* and modify the value once the function call returns k has garbage value. Can somebody justify this?

char* call()
{
    return "fg";
}
void modify(char** i)
{
    *i = call();
}

int main()
{
    char* k= new char[3];
    modify(k);
}

How did this compile though ? modify takes a parameter of type char** and not char*. Also, use std::string instead. — Mahesh
– Mahesh, Commented Apr 9, 2012 at 17:38
Yes, please post code that actually compiles; otherwise it can be difficult to tell what you're asking about. — Keith Thompson
– Keith Thompson, Commented Apr 9, 2012 at 17:41

Oliver Charlesworth · Accepted Answer · 2012-04-09 17:44:30Z

5

When you pass something into a function, you pass it by value. This means that the function operates on a copy of that thing.

This applies to pointers too. If you pass a char *, then a copy of that pointer gets made; the original pointer is not modified. If you want to modify the original pointer itself, then you need to pass its address, via a char ** argument.

Notes:

1. It's also worth pointing out that your code contains a memory leak. You dynamically-allocate some memory, and then lose the pointer to it, which means that you can never delete it.

2. In C++, you should generally avoid passing raw pointers around like this, because it causes pain and confusion. You should look into smart pointers.

edited Apr 9, 2012 at 17:44

answered Apr 9, 2012 at 17:37

Oliver Charlesworth

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3 Comments

Ricibob Over a year ago

Worth noting that in C++ it generally gives more readable code to pass a reference to a pointer instead of a pointer to a pointer - but in C thats not an option.

Robert Mason Over a year ago

More readable code yes. The declaration can be fun though. Let's throw in a template...

user1177586 Over a year ago

What if i modify the "modify" function program to take a reference to a pointer and pass the address while calling..Will there be a memory leak still?

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