Hello I know it was asked many times but I hadn't found answer to my specific question.
I want to convert only string that contains only decimal numbers:
For example 256 is OK but 256a is not.
Could it be done without checking the string?
Thanks
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2Your subject line says "convert integer to string", but your question kind of seems to imply a string-to-int conversion... what are you trying to do?!EboMike– EboMike2010-10-04 20:50:51 +00:00Commented Oct 4, 2010 at 20:50
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1This post is vague and needs more punctuation. Could you explain what you mean with an example?JoshD– JoshD2010-10-04 20:51:04 +00:00Commented Oct 4, 2010 at 20:51
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4 Answers
The simplest way that makes error checking optional that I can think of is this:
char *endptr;
int x = strtol(str, &endptr, 0);
int error = (*endptr != '\0');
1 Comment
Evan Teran
I'm not a fan of
atoi since it provides no way of telling the difference between a completely invalid string and a string which is valid and converts to 0In C++ way, use stringstream:
#include <iostream>
#include <string>
#include <sstream>
using namespace std;
int main()
{
stringstream sstr;
int a = -1;
sstr << 256 << 'a';
sstr >> a;
if (sstr.failbit)
{
cout << "Either no character was extracted, or the character can't represent a proper value." << endl;
}
if (sstr.badbit)
{
cout << "Error on stream.\n";
}
cout << "Extracted number " << a << endl;
return 0;
}
2 Comments
Jean-François Fabre
but what if the string is not an integer? what will be the value of
a ?
Donotalo
@Jean-FrançoisFabre: You can check stream condition by verifying the state returned by
sstr.rdstate(), or by checking individual state values. I've updated my answer with the latter choice.An other way using c++ style : We check the number of digits to know if the string was valid or not :
#include <iostream>
#include <sstream>
#include <string>
#include <cmath>
int main(int argc,char* argv[]) {
std::string a("256");
std::istringstream buffer(a);
int number;
buffer >> number; // OK conversion is done !
// Let's now check if the string was valid !
// Quick way to compute number of digits
size_t num_of_digits = (size_t)floor( log10( abs( number ) ) ) + 1;
if (num_of_digits!=a.length()) {
std::cout << "Not a valid string !" << std::endl;
}
else {
std::cout << "Valid conversion to " << number << std::endl;
}
}
1 Comment
stark
Fails for "0", "0001", "-5".
In C++17 and above the best way is to use std::from_chars() from the header charconv. Unlike other solutions, this one handles overflows automatically in a neat way. It is also locale-independent, non-allocating, and non-throwing and hence the performance is expected to be better than solutions based on string streams. You may find more information in cppreference.
The following example is taken from cppreference.
#include <cassert>
#include <charconv>
#include <iomanip>
#include <iostream>
#include <string_view>
#include <system_error>
int main()
{
for (std::string_view const str : {"1234", "15 foo", "bar", " 42", "5000000000"})
{
std::cout << "String: " << std::quoted(str) << ". ";
int result{};
auto [ptr, ec] = std::from_chars(str.data(), str.data() + str.size(), result);
if (ec == std::errc())
std::cout << "Result: " << result << ", ptr -> " << std::quoted(ptr) << '\n';
else if (ec == std::errc::invalid_argument)
std::cout << "This is not a number.\n";
else if (ec == std::errc::result_out_of_range)
std::cout << "This number is larger than an int.\n";
}
}
