Array of pointers 3

[mattias1975]

Hello!
Whats wrong with this?

char *pointer[10];

strcpy(pointer[0], "hello");
printf("%s", pointer[0]);

Thank you

 

[DaveSinkula]

pointer[0] does not point to memory that you can write to.

 

[smgarth]

you can use a two-dimensional array of course:

const int max = 10, size = 256;
char strings[max][size+1];
strcpy(strings[0], "hello");
printf("%s", strings[0]);

 

[Salem]

pointer[0] = "hello";

This is also valid.

It's usefulness on the other hand depends on what you want to do next.

--

 

[gaoanyu]

It's because that the memory space (where "hello" is stored) that pointer[0] points to is not specified.
try this:

void main()
{
	char *pointer[10];
	pointer[0] = new char [10];
	strcpy(pointer[0], "hello");
	printf("%s", pointer[0]);  
	delete [] pointer[0];
}