I read that the size of array should be known at compile time.
If I am right then below code should not work, but it is compiling without any warnings and is working fine.
Could you please explain me what is happening ?
std::string s = "ABCDEFGHIJ";
int n = s.length();
// declaring character array
char char_array[n + 1];
std::cout << sizeof(char_array) << std::endl;
//Actual result
11
Assuming you're using g++, it's a nonstandard compiler extension. The keyword to search for is "variable length array".
Since you didn't explain what compiler you're using, I'll show an example of how to enable errors for this kind of code with g++ using either the -pedantic flag or explicitly enabling vla warnings / errors.
demo:/tmp$ cat ex.cc
#include <iostream>
#include <string>
int main() {
std::string s = "ABCDEFGHIJ";
char char_array[s.length() + 1];
std::cout << sizeof(char_array) << std::endl;
}
demo:/tmp$ cat ex.cc | g++ -x c++ -
demo:/tmp$ cat ex.cc | g++ -Werror=vla -x c++ -
<stdin>: In function ‘int main()’:
<stdin>:6:33: error: variable length array ‘char_array’ is used [-Werror=vla]
cc1plus: some warnings being treated as errors
demo:/tmp$ cat ex.cc | g++ -pedantic -x c++ -
<stdin>: In function ‘int main()’:
<stdin>:6:33: warning: ISO C++ forbids variable length array ‘char_array’ [-Wvla]
demo:/tmp$ g++ --version | head -n 1
g++ (Debian 8.3.0-6) 8.3.0
