string test;
test="3527523";
int a=test[3];
std::cout<<a<<std::endl;
I was expecting output to be number 7, but I got 55 instead. I have searched through different posts, they all used this method without any problems.. I just don't know where the problem is in my case. Thank you for your help!
test[3] returns char, which you're converting to an integer, so it returns the ASCII value of the character. 3 in ASCII is 55.
Try using char instead as the type of a:
char a = test[3];
std::cout << a; // 3
If you wanted the result to be an integer, subtract the character from '0':
std::string test = "3527523";
int a = test[3] - '0';
std::cout << a; // 3
You get the char '7' by the brackets, which has the ASCII code 55 (48+7).
Problem is that you assign test[3] to a variable of type int, thus causing the actual character code (55) to be printed instead of the represented character ('7'). Replace int with char or static_cast before printing.
You don't "cast", you "convert".
You can use sscanf(), atoi() among other functions to convert the value of a string:
int iret = sscanf (test, "%d", &a);
The value of test[3] is the ASCII character '7', which is actually the integer "55".
PS: The beauty of "sscanf()" is that you can easily detect if a parse error occurred by checking for "iret != 1".
Use atoi:
int atoi (const char * str); //Convert string to integer
For example:
string test;
test="3527523";
const char c = test[3];
int a=atoi(&c);
std::cout<<a<<std::endl;
