C++: check if string is a valid integer using "strtol" - c++

Question

I have heard that I should use strtol instead of atoi due to its better error handling. I wanted to test out strtol by seeing if I could use this code to check if a string is an integer:
#include <iostream>
#include <stdlib.h>
using namespace std;
int main()
{
string testString = "ANYTHING";
cout << "testString = " << testString << endl;
int testInt = strtol(testString.c_str(),NULL,0);
cout << "errno = " << errno << endl;
if (errno > 0)
{
cout << "There was an error." << endl;
cout << "testInt = " << testInt << endl;
}
else
{
cout << "Success." << endl;
cout << "testInt = " << testInt << endl;
}
return 0;
}
I replaced ANYTHING with 5 and it worked perfectly:
testString = 5
errno = 0
Success.
testInt = 5
And when I do it with 2147483648, the largest possible int + 1 (2147483648), it returns this:
testString = 2147483648
errno = 34
There was an error.
testInt = 2147483647
Fair enough. But, when i try it with Hello world!, it incorrectly thinks it's a valid int and returns 0:
testString = Hello world!
errno = 0
Success.
testInt = 0
Notes:
I am using Code::Blocks with GNU GCC Compiler on Windows
"Have g++ follow the C++11 ISO C++ language standard [-std=c++11]" is checked in "Compiler Flags".

Answer 1

According with the man page of strtol. You must define your function such as:
bool isNumeric(const std::string& str) {
char *end;
long val = std::strtol(str.c_str(), &end, 10);
if ((errno == ERANGE && (val == LONG_MAX || val == LONG_MIN)) || (errno != 0 && val == 0)) {
// if the converted value would fall out of the range of the result type.
return false;
}
if (end == str) {
// No digits were found.
return false;
}
// check if the string was fully processed.
return *end == '\0';
}
In C++11, I prefer to use std::stol instead of std::strtol, such as:
bool isNumeric(const std::string& str) {
try {
size_t sz;
std::stol(str, &sz);
return sz == str.size();
} catch (const std::invalid_argument&) {
// if no conversion could be performed.
return false;
} catch (const std::out_of_range&) {
// if the converted value would fall out of the range of the result type.
return false;
}
}
std::stol calls std::strtol, but you works directly with std::string and the code is simplified.

Answer 2

strtol stops on the first non digit
but if you read the man page http://man7.org/linux/man-pages/man3/strtol.3.html you can see
If endptr is not NULL, strtol() stores the address of the first
invalid character in *endptr. If there were no digits at all,
strtol() stores the original value of nptr in *endptr (and returns
0). In particular, if *nptr is not '\0' but **endptr is '\0' on
return, the entire string is valid.
ie
string testString = "ANYTHING";
cout << "testString = " << testString << endl;
char *endptr;
int testInt = strtol(testString.c_str(),&endptr,0);
if(**endptr)
cout << "bad input";

Answer 3

Do not use the C++11 way solution with exceptions, because it is slower. Here is a fast C++11 version:
#include <algorithm>
bool is_decimal(const std::string& s)
{
return !s.empty() && std::find_if(s.begin(), s.end(), [](char c){ return !std::isdigit(c); }) == s.end();
}
If you are sure that your strings mostly are not empty, then you can delete !s.empty(). If you are not, keep it because !s.empty() (!(s.length()==0)) is cheaper than if you call find_if (reference) with an empty string.
Edit:
If you have to handle overflow use the exception version above. Only if you can not use exceptions use this:
#include <string>
#include <sstream>
#include <limits>
template <class T>
bool is_decimal_and_fit(const std::string& s)
{
long double decimal = 0;
return (!(std::istringstream(s) >> decimal).fail() && (decimal >= std::numeric_limits<T>::lowest()) && (decimal <= std::numeric_limits<T>::max()));
}