Related
Does anybody know of a convenient means of determining if a string value "qualifies" as a floating-point number?
bool IsFloat( string MyString )
{
... etc ...
return ... // true if float; false otherwise
}
If you can't use a Boost library function, you can write your own isFloat function like this.
#include <string>
#include <sstream>
bool isFloat( string myString ) {
std::istringstream iss(myString);
float f;
iss >> noskipws >> f; // noskipws considers leading whitespace invalid
// Check the entire string was consumed and if either failbit or badbit is set
return iss.eof() && !iss.fail();
}
You may like Boost's lexical_cast (see http://www.boost.org/doc/libs/1_37_0/libs/conversion/lexical_cast.htm).
bool isFloat(const std::string &someString)
{
using boost::lexical_cast;
using boost::bad_lexical_cast;
try
{
boost::lexical_cast<float>(someString);
}
catch (bad_lexical_cast &)
{
return false;
}
return true;
}
You can use istream to avoid needing Boost, but frankly, Boost is just too good to leave out.
Inspired by this answer I modified the function to check if a string is a floating point number. It won't require boost & doesn't relies on stringstreams failbit - it's just plain parsing.
static bool isFloatNumber(const std::string& string){
std::string::const_iterator it = string.begin();
bool decimalPoint = false;
int minSize = 0;
if(string.size()>0 && (string[0] == '-' || string[0] == '+')){
it++;
minSize++;
}
while(it != string.end()){
if(*it == '.'){
if(!decimalPoint) decimalPoint = true;
else break;
}else if(!std::isdigit(*it) && ((*it!='f') || it+1 != string.end() || !decimalPoint)){
break;
}
++it;
}
return string.size()>minSize && it == string.end();
}
I.e.
1
2.
3.10000
4.2f
-5.3f
+6.2f
is recognized by this function correctly as float.
1.0.0
2f
2.0f1
Are examples for not-valid floats. If you don't want to recognize floating point numbers in the format X.XXf, just remove the condition:
&& ((*it!='f') || it+1 != string.end() || !decimalPoint)
from line 9.
And if you don't want to recognize numbers without '.' as float (i.e. not '1', only '1.', '1.0', '1.0f'...) then you can change the last line to:
return string.size()>minSize && it == string.end() && decimalPoint;
However: There are plenty good reasons to use either boost's lexical_cast or the solution using stringstreams rather than this 'ugly function'. But It gives me more control over what kind of formats exactly I want to recognize as floating point numbers (i.e. maximum digits after decimal point...).
I recently wrote a function to check whether a string is a number or not. This number can be an Integer or Float.
You can twist my code and add some unit tests.
bool isNumber(string s)
{
std::size_t char_pos(0);
// skip the whilespaces
char_pos = s.find_first_not_of(' ');
if (char_pos == s.size()) return false;
// check the significand
if (s[char_pos] == '+' || s[char_pos] == '-') ++char_pos; // skip the sign if exist
int n_nm, n_pt;
for (n_nm = 0, n_pt = 0; std::isdigit(s[char_pos]) || s[char_pos] == '.'; ++char_pos) {
s[char_pos] == '.' ? ++n_pt : ++n_nm;
}
if (n_pt>1 || n_nm<1) // no more than one point, at least one digit
return false;
// skip the trailing whitespaces
while (s[char_pos] == ' ') {
++ char_pos;
}
return char_pos == s.size(); // must reach the ending 0 of the string
}
void UnitTest() {
double num = std::stod("825FB7FC8CAF4342");
string num_str = std::to_string(num);
// Not number
assert(!isNumber("1a23"));
assert(!isNumber("3.7.1"));
assert(!isNumber("825FB7FC8CAF4342"));
assert(!isNumber(" + 23.24"));
assert(!isNumber(" - 23.24"));
// Is number
assert(isNumber("123"));
assert(isNumber("3.7"));
assert(isNumber("+23.7"));
assert(isNumber(" -423.789"));
assert(isNumber(" -423.789 "));
}
Quick and dirty solution using std::stof:
bool isFloat(const std::string& s) {
try {
std::stof(s);
return true;
} catch(...) {
return false;
}
}
I'd imagine you'd want to run a regex match on the input string. I'd think it may be fairly complicated to test all the edge cases.
This site has some good info on it. If you just want to skip to the end it says:
^[-+]?[0-9]*\.?[0-9]+([eE][-+]?[0-9]+)?$
Which basically makes sense if you understand regex syntax.
[EDIT: Fixed to forbid initial whitespace and trailing nonsense.]
#include <sstream>
bool isFloat(string s) {
istringstream iss(s);
float dummy;
iss >> noskipws >> dummy;
return iss && iss.eof(); // Result converted to bool
}
You could easily turn this into a function templated on a type T instead of float. This is essentially what Boost's lexical_cast does.
I always liked strtof since it lets you specify an end pointer.
bool isFloat(const std::string& str)
{
char* ptr;
strtof(str.c_str(), &ptr);
return (*ptr) == '\0';
}
This works because the end pointer points to the character where the parse started to fail, therefore if it points to a nul-terminator, then the whole string was parsed as a float.
I'm surprised no one mentioned this method in the 10 years this question has been around, I suppose because it is more of a C-Style way of doing it. However, it is still perfectly valid in C++, and more elegant than any stream solutions. Also, it works with "+inf" "-inf" and so on, and ignores leading whitespace.
EDIT
Don't be caught out by empty strings, otherwise the end pointer will be on the nul-termination (and therefore return true). The above code should be:
bool isFloat(const std::string& str)
{
if (str.empty())
return false;
char* ptr;
strtof(str.c_str(), &ptr);
return (*ptr) == '\0';
}
With C++17:
bool isNumeric(std::string_view s)
{
double val;
auto [p, ec] = std::from_chars(s.data(), s.data() + s.size(), val);
return ec == std::errc() && p == s.data() + s.size();
}
Both checks on return are necessary. The first checks that there are no overflow or other errors. The second checks that the entire string was read.
You can use the methods described in How can I convert string to double in C++?, and instead of throwing a conversion_error, return false (indicating the string does not represent a float), and true otherwise.
The main issue with other responses is performance
Often you don't need every corner case, for example maybe nan and -/+ inf, are not as important to cover as having speed. Maybe you don't need to handle 1.0E+03 notation. You just want a fast way to parse strings to numbers.
Here is a simple, pure std::string way, that's not very fast:
size_t npos = word.find_first_not_of ( ".+-0123456789" );
if ( npos == std::string::npos ) {
val = atof ( word.c_str() );
}
This is slow because it is O(k*13), checking each char against 0 thur 9
Here is a faster way:
bool isNum = true;
int st = 0;
while (word.at(st)==32) st++; // leading spaces
ch = word.at(st);
if (ch == 43 || ch==45 ) st++; // check +, -
for (int n = st; n < word.length(); n++) {
char ch = word.at(n);
if ( ch < 48 || ch > 57 || ch != 46 ) {
isNum = false;
break; // not a num, early terminate
}
}
This has the benefit of terminating early if any non-numerical character is found, and it checks by range rather than every number digit (0-9). So the average compares is 3x per char, O(k*3), with early termination.
Notice this technique is very similar to the actual one used in the stdlib 'atof' function:
http://www.beedub.com/Sprite093/src/lib/c/stdlib/atof.c
You could use atof and then have special handling for 0.0, but I don't think that counts as a particularly good solution.
This is a common question on SO. Look at this question for suggestions (that question discusses string->int, but the approaches are the same).
Note: to know if the string can be converted, you basically have to do the conversion to check for things like over/underflow.
What you could do is use an istringstream and return true/false based on the result of the stream operation. Something like this (warning - I haven't even compiled the code, it's a guideline only):
float potential_float_value;
std::istringstream could_be_a_float(MyString)
could_be_a_float >> potential_float_value;
return could_be_a_float.fail() ? false : true;
it depends on the level of trust, you need and where the input data comes from.
If the data comes from a user, you have to be more careful, as compared to imported table data, where you already know that all items are either integers or floats and only thats what you need to differentiate.
For example, one of the fastest versions, would simply check for the presence of "." and "eE" in it. But then, you may want to look if the rest is being all digits. Skip whitespace at the beginning - but not in the middle, check for a single "." "eE" etc.
Thus, the q&d fast hack will probably lead to a more sophisticated regEx-like (either call it or scan it yourself) approach. But then, how do you know, that the result - although looking like a float - can really be represented in your machine (i.e. try 1.2345678901234567890e1234567890). Of course, you can make a regEx with "up-to-N" digits in the mantissa/exponent, but thats machine/OS/compiler or whatever specific, sometimes.
So, in the end, to be sure, you probably have to call for the underlying system's conversion and see what you get (exception, infinity or NAN).
I would be tempted to ignore leading whitespaces as that is what the atof function does also:
The function first discards as many
whitespace characters as necessary
until the first non-whitespace
character is found. Then, starting
from this character, takes as many
characters as possible that are valid
following a syntax resembling that of
floating point literals, and
interprets them as a numerical value.
The rest of the string after the last
valid character is ignored and has no
effect on the behavior of this
function.
So to match this we would:
bool isFloat(string s)
{
istringstream iss(s);
float dummy;
iss >> skipws >> dummy;
return (iss && iss.eof() ); // Result converted to bool
}
int isFloat(char *s){
if(*s == '-' || *s == '+'){
if(!isdigit(*++s)) return 0;
}
if(!isdigit(*s)){return 0;}
while(isdigit(*s)) s++;
if(*s == '.'){
if(!isdigit(*++s)) return 0;
}
while(isdigit(*s)) s++;
if(*s == 'e' || *s == 'E'){
s++;
if(*s == '+' || *s == '-'){
s++;
if(!isdigit(*s)) return 0;
}else if(!isdigit(*s)){
return 0;
}
}
while(isdigit(*s)) s++;
if(*s == '\0') return 1;
return 0;
}
I was looking for something similar, found a much simpler answer than any I've seen (Although is for floats VS. ints, would still require a typecast from string)
bool is_float(float val){
if(val != floor(val)){
return true;
}
else
return false;
}
or:
auto lambda_isFloat = [](float val) {return (val != floor(val)); };
Hope this helps !
ZMazz
I'm new to C++. I'm working on a project where I need to read mostly integers from the user through the console. In order to avoid someone entering non-digit characters I thought about reading the input as a string, checking there are only digits in it, and then converting it to an integer. I created a function since I need to check for integers several times:
bool isanInt(int *y){
string z;
int x;
getline(cin,z);
for (int n=0; n < z.length(); n++) {
if(!((z[n] >= '0' && z[n] <= '9') || z[n] == ' ') ){
cout << "That is not a valid input!" << endl;
return false;
}
}
istringstream convert(z); //converting the string to integer
convert >> x;
*y = x;
return true;
}
When I need the user to input an integer I'll call this function. But for some reason when I make a call tho this function the program doesn't wait for an input, it jumps immediately to the for-loop processing an empty string. Any thoughts? Thanks for your help.
There are many ways to test a string for only numeric characters. One is
bool is_digits(const std::string &str) {
return str.find_first_not_of("0123456789") == std::string::npos;
}
This would work:
#include <algorithm> // for std::all_of
#include <cctype> // for std::isdigit
bool all_digits(const std::string& s)
{
return std::all_of(s.begin(),
s.end(),
[](char c) { return std::isdigit(c); });
}
You can cast the string in a try/catch block so that if the cast fails you it would raise an exception and you can write whatever you want in the console.
For example:
try
{
int myNum = strtoint(myString);
}
catch (std::bad_cast& bc)
{
std::cerr << "Please insert only numbers "<< '\n';
}
Character-classification is a job typically delegated to the ctype facets of a locale. You're going to need a function that takes into account all 9 digits including the thousands separator and the radix point:
bool is_numeric_string(const std::string& str, std::locale loc = std::locale())
{
using ctype = std::ctype<char>;
using numpunct = std::numpunct<char>;
using traits_type = std::string::traits_type;
auto& ct_f = std::use_facet<ctype>(loc);
auto& np_f = std::use_facet<numpunct>(loc);
return std::all_of(str.begin(), str.end(), [&str, &ct_f, &np_f] (char c)
{
return ct_f.is(std::ctype_base::digit, c) || traits_type::eq(c, np_f.thousands_sep())
|| traits_type::eq(c, np_f.decimal_point());
});
}
Note that extra effort can go into making sure the thousands separator is not the first character.
try another way like cin.getline(str,sizeof(str)), and str here is char*. I think ur problem may be cause by other functions before calling this function. Maybe u can examine other parts of ur codes carefully. Breakpoints setting is recommended too.
Always use off-the-shelf functions. Never write alone.
I recommend
std::regex
Enjoy.
Welcome. I have two problems. Firstly - function size bool (const char * pass) to check whether the amount of the chars inside of string is at least 8, but there is something wrong. Its always showing that there is minimum 8 chars, even is string contain only 3 chars.
My job is to create several small functions for checking the correctness of the entered string of chars. Can You help you with this one? If all small functions inside of bool check(...) return true i need to write in the console "STRING IS OKAY".
I will be grateful for any suggestions.
#include <iostream>
#include <cctype>
using namespace std;
//Check the amount of chars
bool size (const char* pass){
if(sizeof(pass) > 7)
return true;
}
//Checks if the ASCII are located between 32 to 126
bool isPrint (const char* pass){
for(int x=0; x <= sizeof(pass); x++){
if(isprint(pass[x]))
return true;
}
}
//Check the amount of numbers
bool isNum (const char* pass){
for(int x=0; x <= sizeof(pass); x++){
if(isdigit(pass[x]))
return true;
}
}
//Check the amount of Upper letters
bool isUpperLetter (const char* pass){
for(int x=0; x <= sizeof(pass); x++){
if(isupper(pass[x]))
return true;
}
}
//Check the amount of lower letters
bool isLowerLetter (const char* pass){
for(int x=0; x <= sizeof(pass); x++){
if(islower(pass[x]))
return true;
}
}
//Check the amount of Punctuation Marks
bool isPunctMark (const char* pass){
for(int x=0; x <= sizeof(pass); x++){
if(ispunct(pass[x])){
return true;
}
}
}
//All small moduls together
bool check (const char* pass){
size(pass);
isPrint(pass);
isNum(pass);
isUpperLetter(pass);
isLowerLetter(pass);
isPunctMark(pass);
}
int main() {
char x;
cout << "Enter the string of characters" << endl;
cin >> x;
const char *password = &x;
check(password);
}
sizeof(pass) returns the size of the pointer. Which is implementation specific, and if your function is always returning true, we can guess that sizeof(char*) is 8, implying you have a 64-bit system.
On many other systems, it will return 4, or may even return 2 or 1, dependant on the architecture.
You probably want to check the length of the string the pointer is pointing to like so:
int len=strlen(pass);
if(len>=8) //Check for >=8 is clearer than >7)
{
return true;
}
You could also iterate over the string and check for null. But why bother when there is a nice std library routine that does the job.
to run all checks, do something like
bool isValid(const char* const pass)
{
if(!isPrint(pass))
{
return false;
}
if (!isNum(pass))
{
return false;
}
//etc
}
You could also have a big long
if(isPrint(pass)&&isNum(pass) .....)
{
return true;
}
but that would be messier, and harder to debug.
sizeof gives you the size of the type of the object passed to it. It is evaluated strictly at compile time. You are passing it a const char *, which is 8 on 64 bit systems.
To get the length of a C-style string, you can use the C function strlen, in the header <cstring>.
That said, I would recommend not doing this. I would recommend moving away from C strings to C++ std::strings, because they are much easier to use correctly.
Now as it is, you are very much using C strings INcorrectly!
int main() {
char x;
cout << "Enter the string of characters" << endl;
cin >> x;
const char *password = &x;
check(password);
}
You read a single char (x), then take its address and treat this as a C string. Now there are two significant problems here.
First, you probably meant to read more than one character.
Second, you will hit undefined behaviour, and your computer may well blow up, because C strings are supposed to be a pointer to a NUL-terminated array of char. Any functions that expect a C string will loop through look for an ending '\0', which x does not have, since it is not even an array in the first place.
So, if you use std::string from the <string> header, you can have much safer code, without all this faffing with pointers, NUL-terminators, and such.
(untested)
// one of your functions for example
bool isUpperLetter (const std::string& s){
for(int x=0; x < s.size(); ++x){ // Use <, not <=. C++ uses 0-indexing.
if(isupper(s[x]))
return true;
}
return false; // you forgot this!
}
int main() {
std::string s;
std::cout << "Enter a string:\n";
std::cin >> s;
isUpperLetter(s);
}
By the way, this won't work if your input string contains spaces, but one thing at a time!
(Next steps if you learn fast: Read up on std::getline and the <algorithm> header. std::count_if looks very relevant.)
And while we're at it, kill bad habits early, and read up on why you should avoid using namespace std; and std::endl.
EDIT
From your comment, you are stuck with the signature bool check(const char*), so I guess you are supposed to be learning about how to work with C strings. Let's for the moment assume your instructors know what they are doing.
Then the normal way to loop through a C string is with a pointer, checking for '\0'. So for instance, to count the number of uppercase letters (and really, you don't write it this way for real code. Or at least, if you tried to on a project I was working on I would strongly suggest you fix it):
int countUppercase (const char* c)
{
if(NULL==c) return 0;
int count = 0;
for ( ; '\0' != *c ; ++c ) // loop while not found the NUL
{
if (isupper(*c))
++count;
}
return count;
}
I still strongly recommend reading into a std::string if you can get away with it. If not, you're next best bet is probably std::istream::getline.
You are checking the sizeof(pass), which is size of const char*. You should iterate over the array and chack for str[i]=='\0' instead.
EDIT:
As suggested, you can also use strlen() function instead.
As there was already mentioned you use sizeof pointer in loops of your functions instead of the actual length of the passed string. Moreover sometimes comments before functions do not correspond what they do or have to do. For example
//Check the amount of numbers
bool isNum (const char* pass){
for(int x=0; x <= sizeof(pass); x++){
if(isdigit(pass[x]))
return true;
}
}
In the comment there is written "check the amount of numbers". I think you should return number of digits in the given string.
So I would rewrite the function the following way
//Check the amount of numbers
size_t DigitCount ( const char* pass )
{
size_t count = 0;
for ( ; *pass; ++pass )
{
if ( isdigit( *pass ) ) ++count;
}
return count;
}
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<<str;
while(ss>>word)
{
//if( )
cout<<word<<endl;
}
}
Another version...
Use strtol, wrapping it inside a simple function to hide its complexity :
inline bool isInteger(const std::string & s)
{
if(s.empty() || ((!isdigit(s[0])) && (s[0] != '-') && (s[0] != '+'))) return false;
char * p;
strtol(s.c_str(), &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.).
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;
}
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<int>(word);
}
catch(boost::bad_lexical_cast& e)
{
std::cout << word << "isn't a number" << std::endl;
}
If you're just checking if word is a number, that's not too hard:
#include <ctype.h>
...
string word;
bool isNumber = true;
for(string::const_iterator k = word.begin(); k != word.end(); ++k)
isNumber &&= isdigit(*k);
Optimize as desired.
Use the all-powerful C stdio/string functions:
int dummy_int;
int scan_value = std::sscanf( some_string.c_str(), "%d", &dummy_int);
if (scan_value == 0)
// does not start with integer
else
// starts with integer
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.
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 & 2 doesn't fit your requirements.
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 &e)
{
//Failure! myString did not contain an integer
}
Since C++11 you can make use of std::all_of and ::isdigit:
#include <algorithm>
#include <cctype>
#include <iostream>
#include <string_view>
int main([[maybe_unused]] int argc, [[maybe_unused]] char *argv[])
{
auto isInt = [](std::string_view str) -> bool {
return std::all_of(str.cbegin(), str.cend(), ::isdigit);
};
for(auto &test : {"abc", "123abc", "123.0", "+123", "-123", "123"}) {
std::cout << "Is '" << test << "' numeric? "
<< (isInt(test) ? "true" : "false") << std::endl;
}
return 0;
}
Check out the result with Godbolt.
template <typename T>
const T to(const string& sval)
{
T val;
stringstream ss;
ss << sval;
ss >> 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<double>("4.3");
or
int i = to<int>("4123");
I have modified paercebal's method to meet my needs:
typedef std::string String;
bool isInt(const String& s, int base){
if(s.empty() || std::isspace(s[0])) return false ;
char * p ;
strtol(s.c_str(), &p, base) ;
return (*p == 0) ;
}
bool isPositiveInt(const String& s, int base){
if(s.empty() || std::isspace(s[0]) || s[0]=='-') return false ;
char * p ;
strtol(s.c_str(), &p, base) ;
return (*p == 0) ;
}
bool isNegativeInt(const String& s, int base){
if(s.empty() || std::isspace(s[0]) || s[0]!='-') return false ;
char * p ;
strtol(s.c_str(), &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 >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.
Does anybody know of a convenient means of determining if a string value "qualifies" as a floating-point number?
bool IsFloat( string MyString )
{
... etc ...
return ... // true if float; false otherwise
}
If you can't use a Boost library function, you can write your own isFloat function like this.
#include <string>
#include <sstream>
bool isFloat( string myString ) {
std::istringstream iss(myString);
float f;
iss >> noskipws >> f; // noskipws considers leading whitespace invalid
// Check the entire string was consumed and if either failbit or badbit is set
return iss.eof() && !iss.fail();
}
You may like Boost's lexical_cast (see http://www.boost.org/doc/libs/1_37_0/libs/conversion/lexical_cast.htm).
bool isFloat(const std::string &someString)
{
using boost::lexical_cast;
using boost::bad_lexical_cast;
try
{
boost::lexical_cast<float>(someString);
}
catch (bad_lexical_cast &)
{
return false;
}
return true;
}
You can use istream to avoid needing Boost, but frankly, Boost is just too good to leave out.
Inspired by this answer I modified the function to check if a string is a floating point number. It won't require boost & doesn't relies on stringstreams failbit - it's just plain parsing.
static bool isFloatNumber(const std::string& string){
std::string::const_iterator it = string.begin();
bool decimalPoint = false;
int minSize = 0;
if(string.size()>0 && (string[0] == '-' || string[0] == '+')){
it++;
minSize++;
}
while(it != string.end()){
if(*it == '.'){
if(!decimalPoint) decimalPoint = true;
else break;
}else if(!std::isdigit(*it) && ((*it!='f') || it+1 != string.end() || !decimalPoint)){
break;
}
++it;
}
return string.size()>minSize && it == string.end();
}
I.e.
1
2.
3.10000
4.2f
-5.3f
+6.2f
is recognized by this function correctly as float.
1.0.0
2f
2.0f1
Are examples for not-valid floats. If you don't want to recognize floating point numbers in the format X.XXf, just remove the condition:
&& ((*it!='f') || it+1 != string.end() || !decimalPoint)
from line 9.
And if you don't want to recognize numbers without '.' as float (i.e. not '1', only '1.', '1.0', '1.0f'...) then you can change the last line to:
return string.size()>minSize && it == string.end() && decimalPoint;
However: There are plenty good reasons to use either boost's lexical_cast or the solution using stringstreams rather than this 'ugly function'. But It gives me more control over what kind of formats exactly I want to recognize as floating point numbers (i.e. maximum digits after decimal point...).
I recently wrote a function to check whether a string is a number or not. This number can be an Integer or Float.
You can twist my code and add some unit tests.
bool isNumber(string s)
{
std::size_t char_pos(0);
// skip the whilespaces
char_pos = s.find_first_not_of(' ');
if (char_pos == s.size()) return false;
// check the significand
if (s[char_pos] == '+' || s[char_pos] == '-') ++char_pos; // skip the sign if exist
int n_nm, n_pt;
for (n_nm = 0, n_pt = 0; std::isdigit(s[char_pos]) || s[char_pos] == '.'; ++char_pos) {
s[char_pos] == '.' ? ++n_pt : ++n_nm;
}
if (n_pt>1 || n_nm<1) // no more than one point, at least one digit
return false;
// skip the trailing whitespaces
while (s[char_pos] == ' ') {
++ char_pos;
}
return char_pos == s.size(); // must reach the ending 0 of the string
}
void UnitTest() {
double num = std::stod("825FB7FC8CAF4342");
string num_str = std::to_string(num);
// Not number
assert(!isNumber("1a23"));
assert(!isNumber("3.7.1"));
assert(!isNumber("825FB7FC8CAF4342"));
assert(!isNumber(" + 23.24"));
assert(!isNumber(" - 23.24"));
// Is number
assert(isNumber("123"));
assert(isNumber("3.7"));
assert(isNumber("+23.7"));
assert(isNumber(" -423.789"));
assert(isNumber(" -423.789 "));
}
Quick and dirty solution using std::stof:
bool isFloat(const std::string& s) {
try {
std::stof(s);
return true;
} catch(...) {
return false;
}
}
I'd imagine you'd want to run a regex match on the input string. I'd think it may be fairly complicated to test all the edge cases.
This site has some good info on it. If you just want to skip to the end it says:
^[-+]?[0-9]*\.?[0-9]+([eE][-+]?[0-9]+)?$
Which basically makes sense if you understand regex syntax.
[EDIT: Fixed to forbid initial whitespace and trailing nonsense.]
#include <sstream>
bool isFloat(string s) {
istringstream iss(s);
float dummy;
iss >> noskipws >> dummy;
return iss && iss.eof(); // Result converted to bool
}
You could easily turn this into a function templated on a type T instead of float. This is essentially what Boost's lexical_cast does.
I always liked strtof since it lets you specify an end pointer.
bool isFloat(const std::string& str)
{
char* ptr;
strtof(str.c_str(), &ptr);
return (*ptr) == '\0';
}
This works because the end pointer points to the character where the parse started to fail, therefore if it points to a nul-terminator, then the whole string was parsed as a float.
I'm surprised no one mentioned this method in the 10 years this question has been around, I suppose because it is more of a C-Style way of doing it. However, it is still perfectly valid in C++, and more elegant than any stream solutions. Also, it works with "+inf" "-inf" and so on, and ignores leading whitespace.
EDIT
Don't be caught out by empty strings, otherwise the end pointer will be on the nul-termination (and therefore return true). The above code should be:
bool isFloat(const std::string& str)
{
if (str.empty())
return false;
char* ptr;
strtof(str.c_str(), &ptr);
return (*ptr) == '\0';
}
With C++17:
bool isNumeric(std::string_view s)
{
double val;
auto [p, ec] = std::from_chars(s.data(), s.data() + s.size(), val);
return ec == std::errc() && p == s.data() + s.size();
}
Both checks on return are necessary. The first checks that there are no overflow or other errors. The second checks that the entire string was read.
You can use the methods described in How can I convert string to double in C++?, and instead of throwing a conversion_error, return false (indicating the string does not represent a float), and true otherwise.
The main issue with other responses is performance
Often you don't need every corner case, for example maybe nan and -/+ inf, are not as important to cover as having speed. Maybe you don't need to handle 1.0E+03 notation. You just want a fast way to parse strings to numbers.
Here is a simple, pure std::string way, that's not very fast:
size_t npos = word.find_first_not_of ( ".+-0123456789" );
if ( npos == std::string::npos ) {
val = atof ( word.c_str() );
}
This is slow because it is O(k*13), checking each char against 0 thur 9
Here is a faster way:
bool isNum = true;
int st = 0;
while (word.at(st)==32) st++; // leading spaces
ch = word.at(st);
if (ch == 43 || ch==45 ) st++; // check +, -
for (int n = st; n < word.length(); n++) {
char ch = word.at(n);
if ( ch < 48 || ch > 57 || ch != 46 ) {
isNum = false;
break; // not a num, early terminate
}
}
This has the benefit of terminating early if any non-numerical character is found, and it checks by range rather than every number digit (0-9). So the average compares is 3x per char, O(k*3), with early termination.
Notice this technique is very similar to the actual one used in the stdlib 'atof' function:
http://www.beedub.com/Sprite093/src/lib/c/stdlib/atof.c
You could use atof and then have special handling for 0.0, but I don't think that counts as a particularly good solution.
This is a common question on SO. Look at this question for suggestions (that question discusses string->int, but the approaches are the same).
Note: to know if the string can be converted, you basically have to do the conversion to check for things like over/underflow.
What you could do is use an istringstream and return true/false based on the result of the stream operation. Something like this (warning - I haven't even compiled the code, it's a guideline only):
float potential_float_value;
std::istringstream could_be_a_float(MyString)
could_be_a_float >> potential_float_value;
return could_be_a_float.fail() ? false : true;
it depends on the level of trust, you need and where the input data comes from.
If the data comes from a user, you have to be more careful, as compared to imported table data, where you already know that all items are either integers or floats and only thats what you need to differentiate.
For example, one of the fastest versions, would simply check for the presence of "." and "eE" in it. But then, you may want to look if the rest is being all digits. Skip whitespace at the beginning - but not in the middle, check for a single "." "eE" etc.
Thus, the q&d fast hack will probably lead to a more sophisticated regEx-like (either call it or scan it yourself) approach. But then, how do you know, that the result - although looking like a float - can really be represented in your machine (i.e. try 1.2345678901234567890e1234567890). Of course, you can make a regEx with "up-to-N" digits in the mantissa/exponent, but thats machine/OS/compiler or whatever specific, sometimes.
So, in the end, to be sure, you probably have to call for the underlying system's conversion and see what you get (exception, infinity or NAN).
I would be tempted to ignore leading whitespaces as that is what the atof function does also:
The function first discards as many
whitespace characters as necessary
until the first non-whitespace
character is found. Then, starting
from this character, takes as many
characters as possible that are valid
following a syntax resembling that of
floating point literals, and
interprets them as a numerical value.
The rest of the string after the last
valid character is ignored and has no
effect on the behavior of this
function.
So to match this we would:
bool isFloat(string s)
{
istringstream iss(s);
float dummy;
iss >> skipws >> dummy;
return (iss && iss.eof() ); // Result converted to bool
}
int isFloat(char *s){
if(*s == '-' || *s == '+'){
if(!isdigit(*++s)) return 0;
}
if(!isdigit(*s)){return 0;}
while(isdigit(*s)) s++;
if(*s == '.'){
if(!isdigit(*++s)) return 0;
}
while(isdigit(*s)) s++;
if(*s == 'e' || *s == 'E'){
s++;
if(*s == '+' || *s == '-'){
s++;
if(!isdigit(*s)) return 0;
}else if(!isdigit(*s)){
return 0;
}
}
while(isdigit(*s)) s++;
if(*s == '\0') return 1;
return 0;
}
I was looking for something similar, found a much simpler answer than any I've seen (Although is for floats VS. ints, would still require a typecast from string)
bool is_float(float val){
if(val != floor(val)){
return true;
}
else
return false;
}
or:
auto lambda_isFloat = [](float val) {return (val != floor(val)); };
Hope this helps !
ZMazz