How to pass char array values to another function using Call by value in C++?

[jaiganeshbe]

Hello all,
I have tried to pass character array from main( ) function to getAddress () function using call by reference.But unforunately i have done some modification in that array.So original value was changed.Now I have to use call by value to pass my array from main () to getAddress.

Is it possible to pass array using call by value in C++? or I have to put inside structure after that i have to pass structure to any function.

kindly explain me using sample programs in C++.

Regards
R.Jaiganesh

[Graham]

First of all, it would help if you posted some sample code illustrating your problem.

However, have you tried declaring the argument const?

void func(const char* arg);

that way, the called function can't change the contents of the char array. (If it needs to change them internally, but without reflecting that back to the caler, then you need to revisit your design.)

[jaiganeshbe]

Hello,

In my getAddress(),i have to change my value.I cannot change my design.Could you explain How to do passing char array using call by value in C++.

Regards
R.Jaiganesh

[GNiewerth]

You simply can´t. Arrays decay to pointers when passed to a function, so there´s no way to pass them by value. You can either clone it manually before processing or you can use vector<char> and pass the vector by value.

[PredicateNormative]

I would love to know why you can't pass a char* in to your function. Anyhow, as GNiewerth has said you can use vector<char> and you can use it as follows:

int main()
{
  char* ca = "Hello";
  std::vector<char> cvec(&ca[0],&ca[5]); //copy the char array
  std::cout << &cvec[0] << std::endl; //print the char array stored by the vector.

  return 0;
};

The break down is as follows:

  char* ca = "Hello";

The above creates a char array with 6 elements (5 for Hello and 1 for the null terminator). The next thing to do is copy the char array to a vector<char>...

  std::vector<char> cvec(&ca[0],&ca[5]);

The above uses the vector range initialiser to copy the char array (this only requires one vector allocation and is the most efficient way to copy the char array into the vector). Note that all 6 elements are copied including the null terminator which is important if you are going to use the vector as you would a char array. Next, as an example to prove that it works, print the vector<char> as follows:

  std::cout << &cvec[0] << std::endl;

Basically, &cvec[0] references the first element in the array, and is also incidently the way to pass a vector through a C API. The output stream will stop printing once it hits the null terminator (without the null you would get buffer overrun which is really dangerous, hence why the null is required).

[Graham]

Hello,

In my getAddress(),i have to change my value.I cannot change my design.

Then you've got bigger problems than the one you posted...

Could you explain How to do passing char array using call by value in C++.

As others have said, you can't. Either copy the contents of the string locally (use strdup()) or do the C++ thing and use a std::string, which you can pass by value.

[PredicateNormative]

Well, technically there is a way to do it (if you forgive the pass by reference instead of by value) but its very stupid, since someone is sure to call the function not realising the input char parameter is assumed to be the first element of a char array. Here is example code, but DON'T copy it, it is a stupid idea to the nth degree! Use a vector<char> if you have to pass by value (as already mentioned the example below is not actually pass by value, but pass by reference).

#include <iostream>
#include <vector>

void f(char &c)
{
 char* array = &c;
 std::cout << array << std::endl;
}

int main()
{
  char* cstr = "Hello";
  char& c = cstr[0];
  f(c);

  return 0;
};