Own strncpy() C++ - c++

Question

I am trying to implement my own version of strncpy(), i found a source code from this link.
But I encountered a Unhandled exception at 0x00411ad5 in exercise 2.exe: 0xC0000005: Access violation writing location 0x00417800. everytime the code reaches this code while((x++ < n) && (*dest++ = *source++));
Here is the complete code:
char *strncpy(char * destination, const char * source, size_t n){
char *dest;
dest = destination;
size_t x=0;
while((x++ < n) && (*dest++ = *source++)); //this is where unhandled exception occurs
while(x++ < n){
*dest++ = 0;
}
return dest;
}
int main(){
char *sample = "blue";
char * sample2 = "red";
cout << strncpy(sample, sample2, 5);
getch();
return 0;
}
Please tell me why this occurs and how should I fix it? Thanks!

Answer 1

Your destination is "blue" which is a string literal, that is a constant. As such it is located in a read-only part of memory (and pointed at by local sample variable), thus error when writing.
Try this:
int main(){
char sample[] = "blue";
char * sample2 = "red";
cout << strncpy(sample, sample2, 5);
getch();
return 0;
}
which makes sample an array in local, writeable memory.

Answer 2

You cannot write to a string constant (sample); write to a char array instead:
int main(){
char *sample = "blue";
char buffer[5];
cout << strncpy(buffer, sample, sizeof(buffer));
getch();
return 0;
}

Answer 3

First, it was already explained to you that you can't overwrite a string that is defined like that.
Second, you cant use cout << strncpy if that function returns pointer to the end of the copied string.

Answer 4

There are two main problems with your program
The first one is that function strncpy has to return destination instead of dest
char *strncpy(char * destination, const char * source, size_t n){
char *dest;
dest = destination;
size_t x=0;
while((x++ < n) && (*dest++ = *source++)); //this is where unhandled exception occurs
while(x++ < n){
*dest++ = 0;
}
// return dest;
return destination;
}
The second one is that string literals are immutable. Any attempt to modify a string literal results in undefined behaviour.
Thus main function should be rewritten the following way
int main(){
char sample[] = "blue";
char * sample2 = "red";
cout << strncpy(sample, sample2, sizeof( sample ) );
getch();
return 0;
}
Also it is a bad style of programming to use variable with name x as a count. It is better to use for example i.
I would write the function simpler
char * strncpy( char *destination, const char *source, size_t n )
{
char *dest = destination;
while ( n-- && ( *dest++ = *source++ ) );
while ( n-- ) *dest++ = '\0';
return destination;
}