May be i am getting a trouble with my question, I tried to know how can my code run like this...What is my trouble, please help me here is my code:
string arr[3][4];
arr[1][0] = "34234";
printf("%s",arr[1][0]);
But my output is %#$ ( something like this). Please help me, thanks you very much.
It's because the printf function knows nothing about the C++ std::string class. It's there because it's part of the C standard library.
To get a C-style string that you can use in e.g. printf you have to use the c_str method:
printf("%s", arr[1][0].c_str());
But what you really should do is to learn how to use the native C++ stream output:
std::cout << arr[1][0];
PS.
Instead of old C-style arrays, you should also look into the C++ standard containers, like std::vector and std::array.
printf is a C function and know nothing about types such as std::string. You can either use the type safe std::cout, from the <iostream> header:
std::cout << arr[1][0];
or, if you really need to call printf (and give up on type safety), call std::string::c_str(), which returns a const char* to the null-terminated string held by an std::string. printf will understand this.
printf("%s",arr[1][0].c_str());
use std::cout instead of printf with std::string, because C function printf has no view of std::string.
#include <iostream>
std::cout << arr[1][0];
To make your code work, you need to get C char array by calling std::string::c_str function
printf("%s",arr[1][0].c_str());
string?? user-defined??,printf("%s",arr[1])std::cout << arr[1][0] << std::endl;