Different results using atoi - c++

Question

Could someone explain why those calls are not returning the same expected result?
unsigned int GetDigit(const string& s, unsigned int pos)
{
// Works as intended
char c = s[pos];
return atoi(&c);
// doesn't give expected results
return atoi(&s[pos]);
return atoi(&static_cast<char>(s[pos]));
return atoi(&char(s[pos]));
}
Remark: I'm not looking for the best way to convert a char to an int.

Answer 1

None of your attempts are correct, including the "works as intended" one (it just happened to work by accident). For starters, atoi() requires a NUL-terminated string, which you are not providing.
How about the following:
unsigned int GetDigit(const string& s, unsigned int pos)
{
return s[pos] - '0';
}
This assumes that you know that s[pos] is a valid decimal digit. If you don't, some error checking is in order.

Answer 2

What you are doing is use a std::string, get one character from its internal representation and feed a pointer to it into atoi, which expects a const char* that points to a NULL-terminated string. A std::string is not guaranteed to store characters so that there is a terminating zero, it's just luck that your C++ implementation seems to do this.
The correct way would be to ask std::string for a zero terminated version of it's contents using s.c_str(), then call atoi using a pointer to it.
Your code contains another problem, you are casting the result of atoi to an unsigned int, while atoi returns a signed int. What if your string is "-123"?

Answer 3

Since int atoi(const char* s) accepts a pointer to a field of characters, your last three uses return a number corresponding to the consecutive digits beginning with &s[pos], e.g. it can give 123 for a string like "123", starting at position 0. Since the data inside a std::string are not required to be null-terminated, the answer can be anything else on some implementation, i.e. undefined behaviour.
Your "working" approach also uses undefined behaviour.
It's different from the other attempts since it copies the value of s[pos]to another location.
It seems to work only as long as the adjacent byte in memory next to character c accidentally happens to be a zero or a non-digit character, which is not guaranteed. So follow the advice given by #aix.
To make it work really you could do the following:
char c[2] = { s[pos], '\0' };
return atoi(c);

Answer 4

if you want to access the data as a C string - use s.c_str(), and then pass it to atoi.
atoi expects a C-style string, std::string is a C++ class with different behavior and characteristics. For starters - it doesn't have to be NULL terminated.

Answer 5

atoi takes pointer to char for it's argument. In the first try when you are using the char c it takes pointer to only one character hence you get the answer you want. However in the other attempts what you get is pointer to a char which has happened to be beginning of a string of chars, therefore I assume what you are getting after atoi in the later attempts is a number converted from the chars in positions pos, pos+1, pos+2 and up to the end of the s string.

Answer 6

If you really want to convert just a single char in the string at the position (as opposed to a substring starting at that position and ending at the end of the string), you can do it these ways:
int GetDigit(const string& s, const size_t& pos) {
return atoi(string(1, s[pos]).c_str());
}
int GetDigit2(const string& s, const size_t& pos) {
const char n[2] = {s[pos], '\0'};
return atoi(n);
}
for example.