Basics of strtol?

31,709

Solution 1

The first argument is the string. It has to be passed in as a C string, so if you have a std::string use .c_str() first.

The second argument is optional, and specifies a char * to store a pointer to the character after the end of the number. This is useful when converting a string containing several integers, but if you don't need it, just set this argument to NULL.

The third argument is the radix (base) to convert. strtol can do anything from binary (base 2) to base 36. If you want strtol to pick the base automatically based on prefix, pass in 0.

So, the simplest usage would be

long l = strtol(input.c_str(), NULL, 0);

If you know you are getting decimal numbers:

long l = strtol(input.c_str(), NULL, 10);

strtol returns 0 if there are no convertible characters at the start of the string. If you want to check if strtol succeeded, use the middle argument:

const char *s = input.c_str();
char *t;
long l = strtol(s, &t, 10);
if(s == t) {
    /* strtol failed */
}

If you're using C++11, use stol instead:

long l = stol(input);

Alternately, you can just use a stringstream, which has the advantage of being able to read many items with ease just like cin:

stringstream ss(input);
long l;
ss >> l;

Solution 2

Suppose you're given a string char const * str. Now convert it like this:

#include <cstdlib>
#include <cerrno>

char * e;
errno = 0;

long n = std::strtol(str, &e, 0);

The last argument 0 determines the number base you want to apply; 0 means "auto-detect". Other sensible values are 8, 10 or 16.

Next you need to inspect the end pointer e. This points to the character after the consumed input. Thus if all input was consumed, it points to the null-terminator.

if (*e != '\0') { /* error, die */ }

It's also possible to allow for partial input consumption using e, but that's the sort of stuff that you'll understand when you actually need it.

Lastly, you should check for errors, which can essentially only be overflow errors if the input doesn't fit into the destination type:

if (errno != 0) { /* error, die */ }

In C++, it might be preferable to use std::stol, though you don't get to pick the number base in this case:

#include <string>

try { long n = std::stol(str); }
catch (std::invalid_argument const & e) { /* error */ }
catch (std::out_of_range const & e)     { /* error */ }
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James Thompson
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James Thompson

Updated on August 10, 2022

Comments

  • James Thompson
    James Thompson over 1 year

    I am really confused. I have to be missing something rather simple but nothing I am reading about strtol() is making sense. Can someone spell it out for me in a really basic way, as well as give an example for how I might get something like the following to work?

    string input = getUserInput;
    int numberinput = strtol(input,?,?);