Is there a way to specify how many characters of a string to print out using printf()?

Is there a way to specify how many characters of a string to print out (similar to decimal places in `int`s)?

printf ("Here are the first 8 chars: %s\n", "A string that is more than 8 chars");

Would like it to print: `Here are the first 8 chars: A string`

printf(....."%.8s")

printf ("Here are the first 8 chars: %.8s\n", "A string that is more than 8 chars");

%8s would specify a minimum width of 8 characters. You want to truncate at 8, so use %.8s.

If you want to always print exactly 8 characters you could use %8.8s

Using `printf` you can do

printf("Here are the first 8 chars: %.8s\n", "A string that is more than 8 chars");

If you're using C++, you can achieve the same result using the STL:

using namespace std; // for clarity
string s("A string that is more than 8 chars");
cout << "Here are the first 8 chars: ";
copy(s.begin(), s.begin() + 8, ostream_iterator<char>(cout));
cout << endl;

Or, less efficiently:

cout << "Here are the first 8 chars: " <<
string(s.begin(), s.begin() + 8) << endl;

Print first four characters:

`printf("%.4s\n", "A string that is more than 8 chars");`

See [this link][1] for more information (check .precision -section)

[1]:

[To see links please register here]

In C++ it is easy.

std::copy(someStr.c_str(), someStr.c_str()+n, std::ostream_iterator<char>(std::cout, ""));

EDIT:
It is also safer to use this with string iterators, so you don't run off the end. I'm not sure what happens with printf and string that are too short, but I'm guess this may be safer.