Various results using atoi

Question

Can someone explain why these calls do not return 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])); } 

Note : I'm not looking for a better way to convert char to int .

Answer 1

None of your attempts are correct, including "working as intended" (it just happened by chance). To start atoi() , a null-terminated string is required that you do not provide.

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 do not, some error checking is in order.

Answer 2

What you do is use std::string , get one character from its internal representation and pass a pointer to it in atoi , which expects const char* point to a string with NULL termination. A std::string not guaranteed to preserve characters, so there is a zero end, just the luck is that your C ++ implementation seems to be doing this.

The correct way would be to query std::string for the null version of its contents using s.c_str() , and then call atoi with a pointer to it.

The code contains another problem , you return the result of atoi to unsigned int , and atoi returns the signed int . What if your line is "-123" ?

Answer 3

Since int atoi(const char* s) takes a pointer to a character field, your last three uses return a number corresponding to consecutive digits starting with & s [pos], for example. it can give 123 for a string of type "123" , starting at position 0. Since the data inside a std::string not required to complete with a zero value, the answer may be something else in some implementation, that is, undefined.

Your “working” approach also uses undefined behavior. It differs from other attempts because it copies the value of s[pos] to another location. It seems to work only until the next byte in memory next to the character c accidentally turns out to be a null or non-digital character, which is not guaranteed. Therefore, follow the recommendations of @aix.

To make it work, you can do the following:

 char c[2] = { s[pos], '\0' }; return atoi(c); 

Answer 4

if you want to access the data as a string C - use s.c_str() , and then pass it to atoi .

atoi expects a C-style std::string , std::string is a C ++ class with various behaviors and characteristics. For starters - this should not be NULL terminated.

Answer 5

atoi takes a char pointer for this argument. In the first attempt, when you use char c , it takes a pointer to only one character, so you get the answer you want. However, in other attempts, you get a pointer to char , which, as it turned out, is the beginning of the string char s, so I assume that you get after atoi in subsequent attempts - this is the number converted from the characters at pos , pos+1 , pos+2 and to the end of line s .

Answer 6

If you really want to convert only one char to a string at a position (as opposed to a substring starting at that position and ending at the end of the line), you can do this in the following 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); } 

eg.