How do I check if a C++ std::string starts with a certain string, and convert a substring to an int?

Question

How do I implement the following (Python pseudocode) in C++?

if argv[1].startswith('--foo='):
    foo_value = int(argv[1][len('--foo='):])

(For example, if argv[1] is --foo=98, then foo_value is 98.)

Update: I'm hesitant to look into Boost, since I'm just looking at making a very small change to a simple little command-line tool (I'd rather not have to learn how to link in and use Boost for a minor change).

Answer 1

If you're already using Boost, you can do it with boost string algorithms + boost lexical cast:

#include <boost/algorithm/string/predicate.hpp>
#include <boost/lexical_cast.hpp>

try {    
    if (boost::starts_with(argv[1], "--foo="))
        foo_value = boost::lexical_cast<int>(argv[1]+6);
} catch (boost::bad_lexical_cast) {
    // bad parameter
}

This kind of approach, like many of the other answers provided here is ok for very simple tasks, but in the long run you are usually better off using a command line parsing library. Boost has one (Boost.Program_options), which may make sense if you happen to be using Boost already.

Otherwise a search for "c++ command line parser" will yield a number of options.

Answer 2

std::string s = "tititoto";
if (s.rfind("titi", 0) == 0) {
  // s starts with prefix
}

Who needs anything else? Pure STL!

Answer 3

You would do it like this:

std::string prefix("--foo=");
if (!arg.compare(0, prefix.size(), prefix))
    foo_value = atoi(arg.substr(prefix.size()).c_str());

Looking for a lib such as Boost.ProgramOptions that does this for you is also a good idea.

Answer 4

Just for completeness, I will mention the C way to do it:

If str is your original string, substr is the substring you want to check, then

strncmp(str, substr, strlen(substr))

will return 0 if str starts with substr. The functions strncmp and strlen are in the C header file <string.h>

^{(originally posted by Yaseen Rauf here, markup added)}

For a case-insensitive comparison, use strnicmp instead of strncmp.

This is the C way to do it, for C++ strings you can use the same function like this:

strncmp(str.c_str(), substr.c_str(), substr.size())

Answer 5

Code I use myself:

std::string prefix = "-param=";
std::string argument = argv[1];
if(argument.substr(0, prefix.size()) == prefix) {
    std::string argumentValue = argument.substr(prefix.size());
}

Answer 6

Nobody used the STL algorithm/mismatch function yet. If this returns true, prefix is a prefix of 'toCheck':

std::mismatch(prefix.begin(), prefix.end(), toCheck.begin()).first == prefix.end()

Full example prog:

#include <algorithm>
#include <string>
#include <iostream>

int main(int argc, char** argv) {
    if (argc != 3) {
        std::cerr << "Usage: " << argv[0] << " prefix string" << std::endl
                  << "Will print true if 'prefix' is a prefix of string" << std::endl;
        return -1;
    }
    std::string prefix(argv[1]);
    std::string toCheck(argv[2]);
    if (prefix.length() > toCheck.length()) {
        std::cerr << "Usage: " << argv[0] << " prefix string" << std::endl
                  << "'prefix' is longer than 'string'" <<  std::endl;
        return 2;
    }
    if (std::mismatch(prefix.begin(), prefix.end(), toCheck.begin()).first == prefix.end()) {
        std::cout << '"' << prefix << '"' << " is a prefix of " << '"' << toCheck << '"' << std::endl;
        return 0;
    } else {
        std::cout << '"' << prefix << '"' << " is NOT a prefix of " << '"' << toCheck << '"' << std::endl;
        return 1;
    }
}

---

Edit:

As @James T. Huggett suggests, std::equal is a better fit for the question: Is A a prefix of B? and is slight shorter code:

std::equal(prefix.begin(), prefix.end(), toCheck.begin())

Full example prog:

#include <algorithm>
#include <string>
#include <iostream>

int main(int argc, char **argv) {
  if (argc != 3) {
    std::cerr << "Usage: " << argv[0] << " prefix string" << std::endl
              << "Will print true if 'prefix' is a prefix of string"
              << std::endl;
    return -1;
  }
  std::string prefix(argv[1]);
  std::string toCheck(argv[2]);
  if (prefix.length() > toCheck.length()) {
    std::cerr << "Usage: " << argv[0] << " prefix string" << std::endl
              << "'prefix' is longer than 'string'" << std::endl;
    return 2;
  }
  if (std::equal(prefix.begin(), prefix.end(), toCheck.begin())) {
    std::cout << '"' << prefix << '"' << " is a prefix of " << '"' << toCheck
              << '"' << std::endl;
    return 0;
  } else {
    std::cout << '"' << prefix << '"' << " is NOT a prefix of " << '"'
              << toCheck << '"' << std::endl;
    return 1;
  }
}

Answer 7

Given that both strings — argv[1] and "--foo" — are C strings, @FelixDombek's answer is hands-down the best solution.

Seeing the other answers, however, I thought it worth noting that, if your text is already available as a std::string, then a simple, zero-copy, maximally efficient solution exists that hasn't been mentioned so far:

const char * foo = "--foo";
if (text.rfind(foo, 0) == 0)
    foo_value = text.substr(strlen(foo));

And if foo is already a string:

std::string foo("--foo");
if (text.rfind(foo, 0) == 0)
    foo_value = text.substr(foo.length());

Answer 8

Using STL this could look like:

std::string prefix = "--foo=";
std::string arg = argv[1];
if (prefix.size()<=arg.size() && std::equal(prefix.begin(), prefix.end(), arg.begin())) {
  std::istringstream iss(arg.substr(prefix.size()));
  iss >> foo_value;
}

Answer 9

At the risk of being flamed for using C constructs, I do think this sscanf example is more elegant than most Boost solutions. And you don't have to worry about linkage if you're running anywhere that has a Python interpreter!

#include <stdio.h>
#include <string.h>

int main(int argc, char **argv)
{
    for (int i = 1; i != argc; ++i) {
        int number = 0;
        int size = 0;
        sscanf(argv[i], "--foo=%d%n", &number, &size);
        if (size == strlen(argv[i])) {
            printf("number: %d\n", number);
        }
        else {
            printf("not-a-number\n");
        }
    }
    return 0;
}

Here's some example output that demonstrates the solution handles leading/trailing garbage as correctly as the equivalent Python code, and more correctly than anything using atoi (which will erroneously ignore a non-numeric suffix).

$ ./scan --foo=2 --foo=2d --foo='2 ' ' --foo=2'
number: 2
not-a-number
not-a-number
not-a-number

Answer 10

text.substr(0, start.length()) == start

Answer 11

With C++17 you can use std::basic_string_view & with C++20 std::basic_string::starts_with or std::basic_string_view::starts_with.

The benefit of std::string_view in comparison to std::string - regarding memory management - is that it only holds a pointer to a "string" (contiguous sequence of char-like objects) and knows its size. Example without moving/copying the source strings just to get the integer value:

#include <string_view>
#include <exception>
#include <iostream>

const char * argument = "--foo=42"; // Emulating command argument.
const char * argumentPrefix = "--foo";
int inputValue = 0;

std::string_view argView = argument;
if (argView.starts_with(argumentPrefix))
{
    std::string_view prefixView = argumentPrefix; // Helper for getting the size of argumentPrefix.
    try
    {
        // The underlying data of argView is nul-terminated, therefore we can use data().
        inputValue = std::atoi(argView.substr(prefixView.size() + 1).data());
    }
    catch (std::exception& e)
    {
        std::cerr << e.what();
    }
}

Answer 12

I use std::string::compare wrapped in utility method like below:

static bool startsWith(const string& s, const string& prefix) {
    return s.size() >= prefix.size() && s.compare(0, prefix.size(), prefix) == 0;
}

Answer 13

Why not use gnu getopts? Here's a basic example (without safety checks):

#include <getopt.h>
#include <stdio.h>

int main(int argc, char** argv)
{
  option long_options[] = {
    {"foo", required_argument, 0, 0},
    {0,0,0,0}
  };

  getopt_long(argc, argv, "f:", long_options, 0);

  printf("%s\n", optarg);
}

For the following command:

$ ./a.out --foo=33

You will get

33

Answer 14

In case you need C++11 compatibility and cannot use boost, here is a boost-compatible drop-in with an example of usage:

#include <iostream>
#include <string>

static bool starts_with(const std::string str, const std::string prefix)
{
    return ((prefix.size() <= str.size()) && std::equal(prefix.begin(), prefix.end(), str.begin()));
}

int main(int argc, char* argv[])
{
    bool usage = false;
    unsigned int foos = 0; // default number of foos if no parameter was supplied

    if (argc > 1)
    {
        const std::string fParamPrefix = "-f="; // shorthand for foo
        const std::string fooParamPrefix = "--foo=";

        for (unsigned int i = 1; i < argc; ++i)
        {
            const std::string arg = argv[i];

            try
            {
                if ((arg == "-h") || (arg == "--help"))
                {
                    usage = true;
                } else if (starts_with(arg, fParamPrefix)) {
                    foos = std::stoul(arg.substr(fParamPrefix.size()));
                } else if (starts_with(arg, fooParamPrefix)) {
                    foos = std::stoul(arg.substr(fooParamPrefix.size()));
                }
            } catch (std::exception& e) {
                std::cerr << "Invalid parameter: " << argv[i] << std::endl << std::endl;
                usage = true;
            }
        }
    }

    if (usage)
    {
        std::cerr << "Usage: " << argv[0] << " [OPTION]..." << std::endl;
        std::cerr << "Example program for parameter parsing." << std::endl << std::endl;
        std::cerr << "  -f, --foo=N   use N foos (optional)" << std::endl;
        return 1;
    }

    std::cerr << "number of foos given: " << foos << std::endl;
}

Answer 15

You can also use strstr:

if (strstr(str, substr) == substr) {
    // 'str' starts with 'substr'
}

but I think it's good only for short strings because it has to loop through the whole string when the string doesn't actually start with 'substr'.

Answer 16

Ok why the complicated use of libraries and stuff? C++ String objects overload the [] operator, so you can just compare chars.. Like what I just did, because I want to list all files in a directory and ignore invisible files and the .. and . pseudofiles.

while ((ep = readdir(dp)))
{
    string s(ep->d_name);
    if (!(s[0] == '.')) // Omit invisible files and .. or .
        files.push_back(s);
}

It's that simple..

Answer 17

std::string text = "--foo=98";
std::string start = "--foo=";

if (text.find(start) == 0)
{
    int n = stoi(text.substr(start.length()));
    std::cout << n << std::endl;
}

Answer 18

Since C++11 also std::regex_search can be used, e.g. as follows (returns an empty string on failure):

#include <regex>

std::string startsWith(const std::string &str, const std::string &prefix) {
  std::smatch match;
  std::regex_search(str, match, std::regex("^" + prefix));
  return match.suffix();
}

Answer 19

With C++11 or higher you can use find() and find_first_of()

Example using find to find a single char:

#include <string>
std::string name = "Aaah";
size_t found_index = name.find('a');
if (found_index != std::string::npos) {
    // Found string containing 'a'
}

Example using find to find a full string & starting from position 5:

std::string name = "Aaah";
size_t found_index = name.find('h', 3);
if (found_index != std::string::npos) {
    // Found string containing 'h'
}

Example using the find_first_of() and only the first char, to search at the start only:

std::string name = ".hidden._di.r";
size_t found_index = name.find_first_of('.');
if (found_index == 0) {
    // Found '.' at first position in string
}

• More about find
• More about find_first_of

Good luck!

Answer 20

if(boost::starts_with(string_to_search, string_to_look_for))
    intval = boost::lexical_cast<int>(string_to_search.substr(string_to_look_for.length()));

This is completely untested. The principle is the same as the Python one. Requires Boost.StringAlgo and Boost.LexicalCast.

Check if the string starts with the other string, and then get the substring ('slice') of the first string and convert it using lexical cast.