What is atoi equivalent for 64bit integer(uint64_t) in C or C++ that works on both Unix and Windows?
Asked Answered
A

7

37

I'm trying to convert 64bit integer string to integer, but I don't know which one to use.

Azotic answered 21/9, 2011 at 16:0 Comment(3)
@KerrekSB - yes you're right, although the two are exceptionally similar. Related, opposite problem: #7348551Floss
Related, see Convert hexadecimal string with leading “0x” to signed short in C++?Simmons
Possible duplicate of So, we have int32_t, int16_t, uint64_t, etc.. But where are the atoi32, atoi16, atoui64, etc...?Ignaz
F
43

Use strtoull if you have it or _strtoui64() with visual studio.

unsigned long long strtoull(const char *restrict str,
       char **restrict endptr, int base);


/* I am sure MS had a good reason not to name it "strtoull" or
 * "_strtoull" at least.
 */
unsigned __int64 _strtoui64(
   const char *nptr,
   char **endptr,
   int base 
);
Fondafondant answered 21/9, 2011 at 16:3 Comment(2)
Technically sizeof(unsigned long long) is greater than or equal to sizeof(uint64_t), but they often are equal, so that's why strtoull works.Supraorbital
Note: The function _strtoui64() require header file stdlib.hMajors
F
21

You've tagged this question , so I'm assuming you might be interested in C++ solutions too. You can do this using boost::lexical_cast or std::istringstream if boost isn't available to you:

#include <boost/lexical_cast.hpp>
#include <sstream>
#include <iostream>
#include <cstdint>
#include <string>

int main() {
  uint64_t test;
  test = boost::lexical_cast<uint64_t>("594348534879");

  // or
  std::istringstream ss("48543954385");
  if (!(ss >> test))
    std::cout << "failed" << std::endl;
}

Both styles work on Windows and Linux (and others).

In C++11 there's also functions that operate on std::string, including std::stoull which you can use:

#include <string>

int main() {
  const std::string str="594348534879";
  unsigned long long v = std::stoull(str);
}
Floss answered 21/9, 2011 at 16:32 Comment(0)
L
9

Something like...

#ifdef WINDOWS
  #define atoll(S) _atoi64(S)
#endif

..then just use atoll(). You may want to change the #ifdef WINDOWS to something else, just use something that you can rely on to indicate that atoll() is missing but atoi64() is there (at least for the scenarios you're concerned about).

Lindeberg answered 21/9, 2011 at 17:8 Comment(0)
L
5

Try strtoull(), or strtoul(). The former is only in C99 and C++11, but it's usually widely available.

Leflore answered 21/9, 2011 at 16:2 Comment(0)
T
2

In modern c++ I would use std::stoll.

http://en.cppreference.com/w/cpp/string/basic_string/stol

std::stoi, std::stol, std::stoll
  C++  Strings library std::basic_string 
Defined in header <string>
int       stoi( const std::string& str, std::size_t* pos = 0, int base = 10 );
int       stoi( const std::wstring& str, std::size_t* pos = 0, int base = 10 );
(1) (since C++11)
long      stol( const std::string& str, std::size_t* pos = 0, int base = 10 );
long      stol( const std::wstring& str, std::size_t* pos = 0, int base = 10 );
(2) (since C++11)
long long stoll( const std::string& str, std::size_t* pos = 0, int base = 10 );
long long stoll( const std::wstring& str, std::size_t* pos = 0, int base = 10 );
(3) (since C++11)
Interprets a signed integer value in the string str.
1) calls std::strtol(str.c_str(), &ptr, base) or std::wcstol(str.c_str(), &ptr, base)
2) calls std::strtol(str.c_str(), &ptr, base) or std::wcstol(str.c_str(), &ptr, base)
3) calls std::strtoll(str.c_str(), &ptr, base) or std::wcstoll(str.c_str(), &ptr, base)
Discards any whitespace characters (as identified by calling isspace()) until the first non-whitespace character is found, then takes as many characters as possible to form a valid base-n (where n=base) integer number representation and converts them to an integer value. The valid integer value consists of the following parts:
(optional) plus or minus sign
(optional) prefix (0) indicating octal base (applies only when the base is 8 or ​0​)
(optional) prefix (0x or 0X) indicating hexadecimal base (applies only when the base is 16 or ​0​)
a sequence of digits
The set of valid values for base is {0,2,3,...,36}. The set of valid digits for base-2 integers is {0,1}, for base-3 integers is {0,1,2}, and so on. For bases larger than 10, valid digits include alphabetic characters, starting from Aa for base-11 integer, to Zz for base-36 integer. The case of the characters is ignored.
Additional numeric formats may be accepted by the currently installed C locale.
If the value of base is ​0​, the numeric base is auto-detected: if the prefix is 0, the base is octal, if the prefix is 0x or 0X, the base is hexadecimal, otherwise the base is decimal.
If the minus sign was part of the input sequence, the numeric value calculated from the sequence of digits is negated as if by unary minus in the result type.
If pos is not a null pointer, then a pointer ptr - internal to the conversion functions - will receive the address of the first unconverted character in str.c_str(), and the index of that character will be calculated and stored in *pos, giving the number of characters that were processed by the conversion.
Parameters
str -   the string to convert
pos -   address of an integer to store the number of characters processed
base    -   the number base
Return value
The string converted to the specified signed integer type.
Exceptions
std::invalid_argument if no conversion could be performed
std::out_of_range if the converted value would fall out of the range of the result type or if the underlying function (std::strtol or std::strtoll) sets errno to ERANGE.
Tedmund answered 26/10, 2016 at 10:32 Comment(1)
Whilst this may theoretically answer the question, it would be preferable to include the essential parts of the answer here, and provide the link for reference.Trajan
K
0

When choosing between C-style functions like strtoll (which are of course easy to use with std::string as well) and std::stoll (which at first glance appears better suited for std::string) or boost::lexical_cast: Be aware that the two latter will throw exceptions in case they cannot parse the input string or the range overflows. Sometimes this is useful, sometimes not, depends what you're trying to achive.

If you are not in control of the string to parse (as it's external data) but you want to write robust code (which always should be your desire) you always need to expect corrupted data injected by some malicious attacker or broken outside components. For corrupted data strtoll will not throw but needs more explicit code to detect illegal input data. std::stoll and boost::lexical_cast do auto detect and signal crappy input but you must make sure to catch exceptions somewhere to avoid being terminated(TM).

So choose one or the other depending on the structure of the surrounding code, the needs of the parsed results (sometimes illegal data being "parsed" into a 0 is absolutely OK) the source of the data to parse and last but not least your personal preferences. Neither of the functions available is generally superiour to the others.

Kandykane answered 13/7, 2018 at 8:40 Comment(0)
O
-1

Here we convert String consisting of HEX character to uint64_t hex value. All individual characters of string is converted to hex integer ony by one. For example in base 10 -> String = "123":

  • 1st loop : value is 1
  • 2nd loop : value is 1*10 + 2 = 12
  • 3rd loop : value is 12*10 + 3 = 123

So like this logic is used to convert String of HEX character to uint_64hex value.

uint64_t stringToUint_64(String value) {
  int stringLenght = value.length();

  uint64_t uint64Value = 0x0;
  for(int i = 0; i<=stringLenght-1; i++) {
    char charValue = value.charAt(i);

    uint64Value = 0x10 * uint64Value;
    uint64Value += stringToHexInt(charValue);
  }

  return uint64Value;
}

int stringToHexInt(char value) {
  switch(value) {
    case '0':
      return 0;
      break;
    case '1':
      return 0x1;
      break;
    case '2':
      return 0x2;
      break;
    case '3':
      return 0x3;
      break;
    case '4':
      return 0x4;
      break;
    case '5':
      return 0x5;
      break;
    case '6':
      return 0x6;
      break;
    case '7':
      return 0x7;
      break;
    case '8':
      return 0x8;
      break;
    case '9':
      return 0x9;
      break;
    case 'A':
    case 'a':
      return 0xA;
      break;
    case 'B':
    case 'b':
      return 0xB;
      break;
    case 'C':
    case 'c':
      return 0xC;
      break;
    case 'D':
    case 'd':
      return 0xD;
      break;
    case 'E':
    case 'e':
      return 0xE;
      break;
    case 'F':
    case 'f':
      return 0xF;
      break;
  }
}
Osculum answered 25/2, 2020 at 9:26 Comment(0)

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