Convert input string into a char array

Question

My problem is that, I have a string that I get from keyboard and want to save it in to a char array. After I get the array I want to turn it in to a number or something. I tried many thing but it's doesn't work. This is my best solution so far: ``` string input_string; char char_string[20]; cout << "type in some input text:$" << endl; cin >> input_string; strcpy(char_string, input_string.c_str()); for (int i = 0; i < 20 ; i++) { switch(char_string[i]) { case 'a' : cout << "a" << endl; break; case 'b' : cout << "b" << endl; break; case 'c' : cout << "c" << endl; break; case 'd' : cout << "d" << endl; break; case 'e' : cout << "e" << endl; break; case 'f' : cout << "f" << endl; break; ... ``` but when I run this code I get something thy i don't expect. If the input is `random random` I get this: ``` r a n d o m contain uavialba char contain uavialba char contain uavialba char contain uavialba char contain uavialba char contain uavialba char contain uavialba char contain uavialba char co…

Accepted Answer

The problem is here ``` for (int i = 0; i < 20 ; i++) ``` You always output 20 characters. You should stop outputting when you get to the end of the string. ``` for (int i = 0; char_string[i] != '\0' ; i++) ``` C style strings have a `'\0'` character at the end, you should test for this. Seems the other error is that you read a word instead of a line. Instead of this ``` cin >> input_string; ``` try this ``` getline(cin, input_string); ``` `getline` reads a whole line, `>>` reads a single word. BTW there's an easier way to write the code inside the for loop. You don't need switch, just do this instead ``` cout << char_string[i] << endl; ```

Convert input string into a char array

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