C++ getenv() wrapper function not setting the value

Question

The cplusplus.com documentation on getenv() states...

The pointer returned points to an internal memory block, whose content or validity may be altered by further calls to getenv

...which I take to mean, "If you want to keep the content, copy it." So, since I need to retrieve several variables, I wrote a couple of little wrapper functions:

#include <iostream>
#include <string.h>

using namespace std;

void getEnv (char *val, const char *var) {
    val = nullptr;
    char *enVar = getenv(var);
    if (enVar != nullptr) {
        val = new char[strlen(enVar) + 1];
        strcpy(val, enVar);
    }
}

void getEnv (int &val,  const char *var) {
    val = -1;
    char *enVar = getenv(var);
    if (enVar != nullptr) {
        val = atoi(enVar);
    }
}

int main() {
    char *textMode = nullptr;
    int  cLen = 0;

    getEnv(cLen, "CONTENT_LENGTH");
    cout << cLen << endl << endl;

    getEnv(textMode, "TEXT_MODE");
    if (textMode == nullptr)
        cout << "Not set.";
    else 
        cout << "[" << textMode << "]<br>\n";

    return 0;
}

The int version works as expected, but I get nothing back from the char version, and I mean nothing: if I don't initialize *textMode at declaration it remains an uninitialized pointer.

It's pointers, right? Right? I know it is. Gotta be pointers. I'll figure them out one of these days, but hey -- at least I got my linked list to work! Yay!

It compiles fine. I've been fiddling with it for the past hour, trying various things, and it compiles every time. — alanlittle, Feb 10 '17 at 21:21
Notice how you take the `int` as a reference but take the `char*` by value. — Kevin, Feb 10 '17 at 21:22
Yes, because I don't need to allocate memory for the `int`. An `int` is an `int`, so I can just pass it by reference, but the `char` has to be a pointer so I can allocate memory for it. — alanlittle, Feb 10 '17 at 21:24
Allocating memory has nothing to do with it. You're modifying `val` in both cases but this only has an outside affect when it's a reference. See my answer. — Kevin, Feb 10 '17 at 21:26
@alanlittle Why do you continue to use pointers? If you want to have a decent wrapper, don't. — LogicStuff, Feb 10 '17 at 21:27
Surely you should be using `std::string` in some shape or form. — Jonathan Leffler, Feb 10 '17 at 21:29
FWIW, your program leaks memory. If you are intending to use this in a large program, you may want to use a different strategy. Use a class called `EnvironmentVariable` whose destructor can take care of deallocating memory. — R Sahu, Feb 10 '17 at 21:29
@LogicStuff I don't know how else to deal with char strings, apart from `std::string`, but I'm trying to do as much as possible with c-strings, as 1) I need to learn, and 2) it's more efficient. — alanlittle, Feb 10 '17 at 21:32
@RSahu Where is it leaking memory? Where is the `EnvironmentVariable` class? — alanlittle, Feb 10 '17 at 21:33
@alanlittle, `getEnv` allocates memory to make a copy of the value returned by `getenv`. I don't see any code that deallocates that memory. You don't have a class called `EnvironmentVariable`. I am suggesting that you should create one and let it take care of deallocating memory. — R Sahu, Feb 10 '17 at 21:35
@RSahu Ah, OK, I see. I guess I thought it deallocated when it went out of scope. I should have known better! :) I'll have to check if I have that problem anywhere else. It amazes me how many little details that are taken care of in higher languages, have to be given direct attention in C++. But that's a good thing -- I'm totally geeking. Thanks for the...pointer ::groan:: — alanlittle, Feb 10 '17 at 21:40
It's not taken care of for you because you didn't ask for it. `std::string` frees the memory it uses in its destructor. You used `char*` which has no destructor. — Kevin, Feb 10 '17 at 21:52
@RSahu Well, I added `delete enVar;` to the functions, and the program crashes at that point. I don't know about the inner workings of `getenv()` I assumed that it allocates memory, and simply returns a pointer to that, but apparently not. Does the function somehow manage the deallocation itself? I can't imagine how that would work. I know that if I write a function which allocates memory and returns a pointer to it, the caller can then `delete` it, so I'm not sure what's going on. — alanlittle, Feb 13 '17 at 14:12

score 1 · Accepted Answer · answered Feb 10 '17 at 21:25

1

Your second function takes val (an int) by reference: void getEnv (int &val, const char *var) and so can modify the variable passed to it as you expect.

Your first function takes val (a char*) by value: void getEnv (char *val, const char *var) so modifying val has no affect on the variable passed to it. A simple solution is to simply take it as a reference as well: void getEnv (char *&val, const char *var)

answered Feb 10 '17 at 21:25

Kevin

6,993
1
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Ahh! I see. I didn't know you could do that -- I just figured that, since it's a pointer, it's already passing an address, so it wouldn't be a problem. I think I see, though -- it's passing the *value* of `textMod`, which is `nullptr`, and the function can't do anything with that. OK, thanks. I'll get it one of these days! – alanlittle Feb 10 '17 at 21:29

score 0 · Answer 2 · answered Feb 13 '17 at 16:10

Follow up to my comments and the OP's response to them.

Here's what I was thinking:

#include <iostream>
#include <string.h>

using namespace std;

// Use a class to encapsulate the data need to be captured
// in an environment variable.    
class EnvironmentVariable
{
   public:

      EnvironmentVariable(char const* name) : name_(name), isSet_(false)
      {
         char *val = getenv(name);
         if ( val != nullptr )
         {
            isSet_ = true;
            this->value_ = val;
         }
      }

      bool isSet() const
      {
         return isSet_;
      }

      void getValue(char const*& val) const
      {
         if ( isSet_ )
         {
            val = this->value_.c_str();
         }
         else
         {
            val = nullptr;
         }
      }

      void getValue(int& val) const
      {
         if ( isSet_ )
         {
            val = stoi(this->value_);
         }
         else
         {
            val = 0; // Find a suitable default value
         }
      }

   private:
      std::string name_;
      std::string value_;
      bool isSet_;
};

int main() {
   char const* textMode = nullptr;
   int  cLen = 0;

   EnvironmentVariable env1("CONTENT_LENGTH");

   env1.getValue(cLen);
   cout << cLen << endl << endl;

   EnvironmentVariable env2("TEXT_MODE");
   env2.getValue(textMode);

   if (textMode == nullptr)
      cout << "Not set.\n";
   else 
      cout << "[" << textMode << "]<br>\n";

   return 0;
}

I like it. A couple of questions: 1) What is this `: name_(name), isSet_(false)` I'm guessing it assigns default values? Why not just do that in the constructor? And `name_` is not used anywhere. 2) You don't `delete val` (see my last comment above). Is it not necessary? — alanlittle, Feb 13 '17 at 18:01
@alanlittle, `name_` is available for use. You can add a member function `std::string getName() const;` to provide access to the name of the environment variable. — R Sahu, Feb 13 '17 at 18:02
@alanlittle, the `: name_(name), isSet_(false)` part is C++ syntax for initializing those member variables. — R Sahu, Feb 13 '17 at 18:03

C++ getenv() wrapper function not setting the value

2 Answers2