How do I check if a C string is an int

Question

When I use getline  I would input a bunch of strings or numbers  but I only want the while loop to output the  word  if it is not a number  So is there any way to check if  word  is a number or not  I know I could use atoi   for C-strings but how about for strings of the string class   int main        stringstream ss  stringstream  in   stringstream  out     string word    string str    getline cin str     ss lt  lt str    while ss gt  gt word                if               cout lt  lt word lt  lt endl

User · Answer

The accepted answer will give a false positive if the input is a number plus text, because "stol" will convert the firsts digits and ignore the rest.

I like the following version the most, since it's a nice one-liner that doesn't need to define a function and you can just copy and paste wherever you need it.

#include <string>

...

std::string s;

bool has_only_digits = (s.find_first_not_of( "0123456789" ) == std::string::npos);

EDIT: if you like this implementation but you do want to use it as a function, then this should do:

bool has_only_digits(const string s){
  return s.find_first_not_of( "0123456789" ) == string::npos;
}

User · Answer

Since C  11 you can make use of std  all of and   isdigit    include  lt algorithm gt   include  lt cctype gt   include  lt iostream gt   include  lt string view gt   int main   maybe unused   int argc    maybe unused   char  argv          auto isInt      std  string view str  - gt  bool           return std  all of str cbegin    str cend      isdigit               for auto  amp test     abc    123abc    123 0     123    -123    123              std  cout  lt  lt   Is     lt  lt  test  lt  lt     numeric                  lt  lt   isInt test     true     false    lt  lt  std  endl             return 0      Check out the result with Godbolt

User · Answer

You can use boost  lexical cast  as suggested  but if you have any prior knowledge about the strings  i e  that if a string contains an integer literal it won t have any leading space  or that integers are never written with exponents   then rolling your own function should be both more efficient  and not particularly difficult

User · Answer

Ok  the way I see it you have 3 options   1  If you simply wish to check whether the number is an integer  and don t care about converting it  but simply wish to keep it as a string and don t care about potential overflows  checking whether it matches a regex for an integer would be ideal here   2  You can use boost  lexical cast and then catch a potential boost  bad lexical cast exception to see if the conversion failed  This would work well if you can use boost and if failing the conversion is an exceptional condition   3  Roll your own function similar to lexical cast that checks the conversion and returns true false depending on whether it s successful or not  This would work in case 1  amp  2 doesn t fit your requirements

User · Answer

template  lt typename T gt  const T to const string amp  sval            T val          stringstream ss          ss  lt  lt  sval          ss  gt  gt  val          if ss fail                    throw runtime error  string typeid T  name       type wanted      sval           return val      And then you can use it like that   double d   to lt double gt   4 3      or  int i   to lt int gt   4123

User · Answer

You might try boost  lexical cast  It throws an bad lexical cast exception if it fails    In your case   int number  try     number   boost  lexical cast lt int gt  word     catch boost  bad lexical cast amp  e      std  cout  lt  lt  word  lt  lt   isn t a number   lt  lt  std  endl

User · Answer

Another version     Use strtol  wrapping it inside a simple function to hide its complexity    inline bool isInteger const std  string  amp  s       if s empty         isdigit s 0     amp  amp   s 0      -    amp  amp   s 0            return false      char   p     strtol s c str     amp p  10       return   p    0       Why strtol    As far as I love C    sometimes the C API is the best answer as far as I am concerned    using exceptions is overkill for a test that is authorized to fail the temporary stream object creation by the lexical cast is overkill and over-inefficient when the C standard library has a little known dedicated function that does the job    How does it work    strtol seems quite raw at first glance  so an explanation will make the code simpler to read    strtol will parse the string  stopping at the first character that cannot be considered part of an integer  If you provide p  as I did above   it sets p right at this first non-integer character   My reasoning is that if p is not set to the end of the string  the 0 character   then there is a non-integer character in the string s  meaning s is not a correct integer    The first tests are there to eliminate corner cases  leading spaces  empty string  etc     This function should be  of course  customized to your needs  are leading spaces an error  etc     Sources    See the description of strtol at  http   en cppreference com w cpp string byte strtol   See  too  the description of strtol s sister functions  strtod  strtoul  etc

User · Answer

If you re just checking if word is a number  that s not too hard    include  lt ctype h gt         string word  bool isNumber   true  for string  const iterator k   word begin    k    word end      k      isNumber  amp  amp   isdigit  k     Optimize as desired

User · Answer

I have modified paercebal s method to meet my needs   typedef std  string String       bool isInt const String amp  s  int base      if s empty      std  isspace s 0    return false      char   p      strtol s c str     amp p  base       return   p    0        bool isPositiveInt const String amp  s  int base      if s empty      std  isspace s 0      s 0    -   return false      char   p      strtol s c str     amp p  base       return   p    0        bool isNegativeInt const String amp  s  int base      if s empty      std  isspace s 0      s 0    -   return false      char   p      strtol s c str     amp p  base       return   p    0          Note    You can check for various bases  binary  oct  hex and others  Make sure you don t pass 1  negative value or value  gt 36 as base  If you pass 0 as the base  it will auto detect the base i e for a string starting with 0x will be treated as hex and string starting with 0 will be treated as oct  The characters are case-insensitive  Any white space in string will make it return false

User · Answer

Use the all-powerful C stdio string functions   int dummy int  int scan value   std  sscanf  some string c str      d    amp dummy int    if  scan value    0         does not start with integer else        starts with integer

User · Answer

Here is another solution   try      void  std  stoi myString     cast to void to ignore the return value        Success  myString contained an integer    catch  const std  logic error  amp e           Failure  myString did not contain an integer

[c++] How do I check if a C++ string is an int?

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