segmentation fault using c++ copy instead of memcpy

Question

I'm using C++ copy algorithm to copy a string literal, (instead of memcpy) but I'm getting segmentation fault I don't know why though. here is the code:

#include <iostream>
#include <cstring>
#include <stdio.h>

using namespace std;

int main(int argc, char *argv[]) {

    // if using copy with regular pointers, there 
    // is no need to get an output iterator, ex:
    char* some_string = "this is a long string\n";
    size_t some_string_len = strlen(some_string) + 1;

    char* str_copy = new char(some_string_len);
    copy( some_string, some_string + some_string_len, str_copy);
    printf("%s", str_copy);

    delete str_copy;
    return 0;
}

It's for learning purposes. I was trying out the copy algorithm with output iterators, and in a book that I'm reading now, describes that I can replace the iterators in the copy call with low level pointers. — Armen Babakanian
– Armen Babakanian, Commented Oct 7, 2013 at 6:39

P0W · Accepted Answer · 2013-10-07 02:48:42Z

6

Fix :

char* str_copy = new char[some_string_len];
                         ^ notice square bracket

Free memory using :

delete [] str_copy;

answered Oct 7, 2013 at 2:48

P0W

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2 Comments

P0W Over a year ago

@ArmenB and before you ask why ? So, If () is used, the item is value-initialized.

WhozCraig Over a year ago

@AmenB that would be the technical description for saying you were allocating a single char and initializing it to the value of some_string_len. I'm surprised your compiler didn't issue a warning about the down-convert between int and char.

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