Basics of strtol?

Question

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,?,?);

Answer 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 */
}

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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;

Answer 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 */ }

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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 */ }

Answer 3

Quote from C++ reference:

long int strtol ( const char * str, char ** endptr, int base );

Convert string to long integer

Parses the C string str interpreting its content as an integral number of the specified base, which is returned as a long int value. If endptr is not a null pointer, the function also sets the value of endptr to point to the first character after the number.

So try something like

long l = strtol(pointerToStartOfString, NULL, 0)

Answer 4

I always use simply strol(str,0,0) - it returns long value. 0 for radix (last parameter) means to auto-detect it from input string, so both 0x10 as hex and 10 as decimal could be used in input string.