strcpy doesn't work on the same size arrays

Question

When I try to assign value of one string to other using strcpy runtime error occurs. Below the code:

int main (int argc, char **argv)
{ 
  char str[5];
  char str2[5];//if set size of str2 equal to 6, no error occurs

  str[0] = 'a';
  str[1] = 'b';
  str[2] = 'c';
  str[3] = 'd';
  str[4] = 'e';

  cout<<sizeof(str)<<endl;
  cout<<str[0]<<endl;
  cout<<str[1]<<endl;
  cout<<str[2]<<endl;
  cout<<str[3]<<endl;
  cout<<str[4]<<endl;

  strcpy(str2,str);

  cout<<sizeof(str2)<<endl;
  cout<<str2[0]<<endl;
  cout<<str2[1]<<endl;
  cout<<str2[2]<<endl;
  cout<<str2[3]<<endl;
  cout<<str2[4]<<endl;

  getch();
  return 0;
}

Error is:

Run-Time Check Failure #2 - Stack around the variable 'str' was corrupted

If I set the size of str2 equal to 6 or more program works well. What is a problem here?

Even with str2[6], str2[7] gives error. Barmar answer is correct. — Pranit Kothari, Sep 05 '13 at 05:51
terminate the strings, or if they aren't strings, use `memcpy` — Karoly Horvath, Sep 05 '13 at 05:51
You are correct. strcpy does NOT work on arrays. What is your question — kfsone, Sep 05 '13 at 06:03

score 7 · Answer 1 · answered Sep 05 '13 at 05:49

7

strcpy operates on zero-terminated strings. Your char arrays don't have terminating zero bytes.

If it's working when you declare the arrays as [6] it's just by accident.

answered Sep 05 '13 at 05:49

Barmar

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score 5 · Answer 2 · edited May 23 '17 at 11:56

5

Function strcpy(); expects nul \0 terminated string. str[] is not nul \0 terminated.

Because you are printing array char by char in your code, you can rectify code as suggested by @Karoly Horvath using memcpy instead of strcpy.

void * memcpy ( void * destination, const void * source, size_t count );

memcpy(str2, str, sizeof(str));

edited May 23 '17 at 11:56

Community

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answered Sep 05 '13 at 05:48

Grijesh Chauhan

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What changements can I do so that program works well, but both str and str have the same size? – Nurlan Sep 05 '13 at 05:51
1

terminate the strings, or if they aren't strings, use `memcpy` – Karoly Horvath Sep 05 '13 at 05:52
@NurlanKenzhebekov sorry before I could write answer my Internet filed. :( read other answer – Grijesh Chauhan Sep 05 '13 at 05:52
@NurlanKenzhebekov note you have undefined behavior in your code – Grijesh Chauhan Sep 05 '13 at 06:12

score 3 · Answer 3 · answered Sep 05 '13 at 05:52

3

It is very risky to use string operations without forming null-terminated strings.

Here, strcpy() expects a null terminated string to be copied to a string which also has to be null terminated.

Therefore you must use:

  char str[6];
  char str2[6];

  str[0] = 'a';
  str[1] = 'b';
  str[2] = 'c';
  str[3] = 'd';
  str[4] = 'e';
  str[5] = '\0';
  strcpy(str2,str);

answered Sep 05 '13 at 05:52

Aakash Anuj

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Or `char str[] = "abcde";`. `str` will have 6 elements, including the terminating `\0`. – Pete Becker Sep 05 '13 at 14:16

strcpy doesn't work on the same size arrays

3 Answers3