在 Unix 和 Windows 上都可以使用的 C 中 64 位整数 (uint64_t) 的 atoi 等价物是什么？

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

提问by JosephH

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

我正在尝试将 64 位整数字符串转换为整数，但我不知道该使用哪个。

回答by cnicutar

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

strtoull如果您拥有它或_strtoui64()与 Visual Studio 一起使用，请使用它。

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 
);

回答by Flexo

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

您已将此问题标记为c++，因此我假设您也可能对 C++ 解决方案感兴趣。您可以使用boost::lexical_cast或者std::istringstream如果您无法使用boost来执行此操作：

#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).

这两种风格都适用于 Windows 和 Linux（以及其他）。

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

在 C++11 中，还有对进行操作std::string的函数，包括std::stoull您可以使用的函数：

#include <string>

int main() {
  const std::string str="594348534879";
  unsigned long long v = std::stoull(str);
}

回答by Dmitri

Something like...

就像是...

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

..then just use atoll(). You may want to change the #ifdef WINDOWSto 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).

..然后只需使用atoll(). 您可能想要将 更改#ifdef WINDOWS为其他内容，只需使用您可以依赖的内容来指示atoll()缺少但atoi64()存在的内容（至少对于您关注的场景）。

回答by Kerrek SB

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

尝试strtoull()，或strtoul()。前者仅在 C99 和 C++11 中存在，但通常广泛可用。

回答by maroun2

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

在现代 C++ 中，我会使用 std::stoll。

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

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.

回答by Don Pedro

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.

当在 strtoll（当然也易于与 std::string 一起使用）和 std::stoll（乍一看似乎更适合 std::string）或 boost::lexical_cast 之类的 C 样式函数之间进行选择时：请注意，后两者将在它们无法解析输入字符串或范围溢出的情况下抛出异常。有时这很有用，有时则没有，这取决于您要实现的目标。

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).

如果您无法控制要解析的字符串（因为它是外部数据），但您想编写健壮的代码（这始终是您的愿望），您总是需要预料到某些恶意攻击者注入的损坏数据或损坏的外部组件。对于损坏的数据 strtoll 不会抛出但需要更明确的代码来检测非法输入数据。std::stoll 和 boost::lexical_cast 会自动检测并发出糟糕的输入信号，但您必须确保在某处捕获异常以避免被终止^(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.

所以选择一个或另一个取决于周围代码的结构，解析结果的需要（有时非法数据被“解析”成0是绝对可以的）要解析的数据来源以及最后但并非最不重要的你个人喜好。可用的功能通常都不优于其他功能。

回答by Hiloliddin Jaloliddinzoda

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":

这里我们将包含 HEX 字符的 String 转换为 uint64_t 十六进制值。字符串的所有单个字符都被转换为一个十六进制整数。例如在基数 10 -> String = "123" 中：

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

• 第一个循环：值为 1
• 第二个循环：值为 1*10 + 2 = 12
• 第三个循环：值为 12*10 + 3 = 123

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

所以就像这个逻辑用于将 HEX 字符的 String 转换为 uint_64hex 值。

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;
  }
}