Question

how to change a char[] in a call by reference to a function?

Asked by: star90

Hi,
I need to pass a char aa[30]="this is a test";
 to a function. I want the function to change aa. so I want to pass it as a reference to the function.

if  I write
string f1(char* data_in)
{
str_cpy(data_in,"change");
return("ggg");
}

and call the function with
main
{
f1(&data_in[0]);
cout<<aa;
}

the value of  aa remains the same.
 How can I do it correctly?
Thanks.
 

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Asked On
2009-06-25 at 16:42:32ID24523784
Topics

C++ Programming Language

,

C Programming Language

Participating Experts
2
Points
125
Comments
9

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Answers

 

by: star90Posted on 2009-06-25 at 17:23:03ID: 24717193

I also want to ask how do I write the function f1 that it would return a char answer[20]  instead of a string?

 

by: evilrixPosted on 2009-06-25 at 23:58:10ID: 24718589

I don't see aa defined anywhere in your simple example case. Maybe you mean to do something like this?

#include <iostream>

void f1(char* data_in)
{
   strcpy(data_in,"change");
}

int main()
{
   char data_in[20];
   f1(data_in);
   std:: cout << data_in << std::endl;
}

Better still, using string...

#include <iostream>
#include <string>

void f1(std::string & data_in)
{
   data_in = "change";
}

int main()
{
   std::string data_in;
   f1(data_in);
   std:: cout << data_in << std::endl;
}

>> I also want to ask how do I write the function f1 that it would return a char answer[20]  instead of a string?

An array can't be returned alone since it has no copy-constructor but you can encapsulate it in a struct

#include <iostream>

struct data_in_
{
   char d[20];
};

data_in_ f1(data_in_ & data_in)
{
   strcpy(data_in.d,"change");
   return data_in;
}

int main()
{
   data_in_ data_in1;
   data_in_ data_in2 = f1(data_in1);

   std:: cout << data_in2.d << std::endl;
}

 

by: Let_Me_BePosted on 2009-06-26 at 00:00:53ID: 24718596

You don't need reference. Once you have a pointer a to variable (and arrays are always passed as a address of the first item) you have read-write access to the variable.

You can't return a static array, but you can return a pointer. The problem is that any local variable your create in the function will stop to exist once the function returns, so you need to allocate memory on the heap (using malloc [C] or new [C++]). I wouldn't recommend this, because then you need to deallocate the memory once its not needed.

#include <string.h>
#include <stdlib.h>
#include <stdio.h>
 
#define DATA "aaargh"
 
char * do_something_with_c_string(char* c_string, int max_length)
{
    if (max_length > strlen(DATA))
      strcpy(c_string,DATA);
 
    char* result = (char*)malloc(strlen(DATA)+1);
    strcpy(result,DATA);
    return result;
}
 
int main()
{
  char * s1, s2[40];
  s1 = do_something_with_c_string(s2,40);
  printf("s1: \"%s\" s2: \"%s\"\n",s1,s2);
  free(s1);
}

                                              
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by: evilrixPosted on 2009-06-26 at 00:04:36ID: 24718615

>> so you need to allocate memory on the heap (using malloc [C] or new [C++])
Nah... you can just wrap it up in a struct and return it by copy. NRVO will optimize the copying away :)

 

by: star90Posted on 2009-06-26 at 07:49:41ID: 24721503

Hi evilrix
In your example (which works and solves my first question) I have a problem:
Since I am passing a char data_in[20];  and "receiving it as char* and not as char *[20]
wouldn't there be a problem of memory corruption inside the function?, because the compiler might only give enough space for one char because you declared a pointer to char* and not to char *[20]?
Thanks.
void f1(char* data_in)
{
   strcpy(data_in,"change");
}

 

by: evilrixPosted on 2009-06-26 at 07:54:49ID: 24721581

>> Since I am passing a char data_in[20];  and "receiving it as char* and not as char *[20] wouldn't there be a problem of memory corruption inside the function?

No, because when you pass an array to a function that takes a pointer of the same type it decays to the pointer type... in essence you can think of it as automatically passing the address of the first item in the array but because an array is a contiguous blob of memory you can still treat it with array semantics. What you can't do; however, is call sizeof on the pointer to get the size of the original array (because it's not an array its the address of the first element).

More reading on arrays and pointers...

Array is not pointer
http://www.cplusplus.com/forum/articles/10/

The difference between pointers and arrays
http://www.cplusplus.com/forum/articles/9/

Exceptions where arrays are not treated as a pointer
http://www.cplusplus.com/forum/articles/8/

 

by: evilrixPosted on 2009-06-26 at 07:55:29ID: 24721590

BTW char *[20] is an array of 20 pointers not a array of chars.

 

by: star90Posted on 2009-06-26 at 08:05:42ID: 24721687

Thanks for the answer .
 I made a mistake. meant *char[20] but your answer solved my problem and Ill also read the links you sent.

 

by: evilrixPosted on 2009-06-26 at 09:44:20ID: 24722682

>> I made a mistake. meant *char[20]
That would be invalid syntax :)

Let me know, after reading the links, if anything is still unclear... I'm happy to try and explain more.

20120131-EE-VQP-002

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