Returning a char array instead of const char *

Question

How to return a local char array from a function which has the return type as const char*?
Note: I want a way without using std::string

const char * getXXX()
{
  char buf[32];
  sprintf(buf,"XXX%d",i);  // some manipulation
  return buf; // How to return???
}

@Spook , if I use "new" key word inside the function, where should I delete that pointer? — Nadaraj
– Nadaraj, Commented Feb 19, 2014 at 8:49
Can you please elaborate on why you don't want to use std::string? Using the standard library will make your life as a C++ programmer so much easier! — Some programmer dude
– Some programmer dude, Commented Feb 19, 2014 at 8:50
@Nadaraj I've already answered that :) If you allocate memory inside your function, the caller is then responsible for freeing it. Unless, of course, you use std::string - then you can simply return it by value and don't have to worry about freeing allocated memory. — Spook
– Spook, Commented Feb 19, 2014 at 8:52
@Joachim Pileborg .. since i'm using a lot of sprintf statements inside the function, i need a char buf. So, i was looking a solution without assigning the char buf again to a temp string to return the value. — Nadaraj
– Nadaraj, Commented Feb 19, 2014 at 9:09
You do know about std::ostringstream? Using it you can use all the normal C++ stream output formatting functionality, and still get a std::string as output. No fixed-size buffers, no special case handling for strings, type safety, automatic appending to the string. All of those things you can't get by using old C-style sprintf and family. — Some programmer dude
– Some programmer dude, Commented Feb 19, 2014 at 9:11

Banukobhan Nagendram · Accepted Answer · 2014-02-21 14:02:14Z

2

const char * getXXX()
{
  static char buf[32];
  sprintf(buf,"XXX%d",i);
  return buf;
}

This will work

answered Feb 21, 2014 at 14:02

Banukobhan Nagendram

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Comments

Spook · Accepted Answer · 2014-02-19 08:55:04Z

1

As you correctly noticed, you should not return buf (this is a local variable and its memory will be freed after the function returns). The solution is to allocate some memory and return it.

char * returnBuf = new char[32];

memcpy(returnBuf, buf, 32 * sizeof(char));

return returnBuf;

Remember though, that the caller will be responsible for freeing memory returned by getXXX:

char * result = GetXXX();

// Process

delete[] result;

edited Feb 19, 2014 at 8:55

answered Feb 19, 2014 at 8:45

Spook

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