Iterate elements of array using pointer - c++

Question

I try execute code:
char* m[3] = {"123", "456", "789"};
for(char* st=*m; st!=0; st=*(m+1))
{
cout <<st;
}
but it hung up and print: 123456456456...

Answer 1

The loop you wrote is invalid
char* m[3]={"123","456","789"};
for(char* st=*m;st!=0; st=*(m+1))
{
cout <<st;
}
Expression *m has type char * and array m does not contain NULL pointer. So condition st!=0;
is wrong. And the pointer always will point the same element because expression st=*(m+1) gives always the second element of the array
Also take into account that the correct definition of the array will be
const char* m[3] = { "123", "456", "789" };
because it is an array of pointers to string literals and string literals may not be changed.
You could use simply the range based for statement
const char* m[3] = { "123", "456", "789" };
for ( const char* st : m )
{
cout <<st << std::endl;
}
Or you could use iterators in the loop. For example
const char* m[3] = { "123", "456", "789" };
for ( auto it = std::begin( m ); it != std::end( m ); ++it )
{
cout << *it << std::endl;
}
And you could use standard algorithm std::copy
#include <iostream>
#include <algorithm>
#include <iterator>
//,,,
const char* m[3] = { "123", "456", "789" };
std::copy( std::begin( m ), std::end( m ),
std::ostream_iterator<const char *>( std::cout, "\n" ) );

Answer 2

Stop using char* for strings immediately and start using std::string. The same goes for C-style arrays: start using std::array or std::vector. This will save you years of your life and unnecessary pain.
Just look at how beautiful this code is:
std::array<std::string, 3> m = {"123", "456", "789"};
for (auto& s : m)
std::cout << s;
and here's the live example.

Answer 3

Using the code you provided here is one way of doing it:
const char* m[3]={"123","456","789"};
const unsigned int element_count = sizeof(m) / sizeof(m[0]);
//not a nice to have sizeof here, but i used this to keep the code
// consistent with the code provided in the question
for( const char** st = m; st!= (m + element_count) ; st ++ )
{
cout << *st;
}
As you can see to iterate trough an array of char* elements you need to use a pointer to char* and therefore the type is char ** for the iteration variable st.
The code you has was an infinite loop because you assigned st with *(m+1) which is never equal to 0, in fact it's the second element every time.
However there are safer ways of using doing this, by using a std::vector for example:
std::vector<std::string> m = {"123", "456", "789"};
for (auto& st : m)
std::cout << st;
//or:
for(std::vector<std::string>::iterator i = m.begin(),e = m.end();i != e;i++)
std::cout << *i;
//or:
for(std::size_t i = 0;i < m.size() ;i++)
std::cout << m[i];

Answer 4

The problem is that you point a pointer to the same location in each iteration:
char* m[ 3]={ "123","456","789"};
for( char* st = *m; st != 0; st = *( m + 1))
// ^^^^^^^^^ <- this doesn't change,
// st points to *(m+1)
{
std::cout << st;
}
Also you should not use a non-const char * to point to a nonmodifiable string literal:
const char* m[ 3]={ "123","456","789"};
const unsigned int n = sizeof( m) / sizeof( m[ 0]);
for( const char** st = m; st != (m + n) ; st ++ )
{
cout << *st;
}
Even range - based iteration will work:
const char* m[ 3] = { "123", "456", "789" };
for ( const char* st : m )
{
std::cout << st << std::endl;
}
If for some reason you have to stick to simple iteration, this will also work:
const char* m[ 3]={ "123","456","789"};
const char* st;
int r = 0;
for( st = *m; r < 3; st = *( m+r))
{
std::cout << st << std::endl;
r++;
}
Please consider using a standard container, i.e std::vector or std::array. It will help you a lot, mainly in dynamic memory management.
std::vector< std::string> v = { "123", "456", "789"};
for ( auto& s : v)
std::cout << s;

Answer 5

This is the least intrusive solution:
char* m[4] = {"123", "456", "789", NULL};
for(char** st=m; *st!=0; st=(st+1))
{
cout <<*st;
}