在C语言中，64位整数(uint64_t)在Unix和Windows上都有效

如果您有strtoull，请使用_strtoui64()。

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 
);

您已经将这个问题标记为c++，所以我假设您可能也对c++解决方案感兴趣。如果boost对您不可用，您可以使用boost::lexical_cast或std::istringstream来完成此操作:

#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;
}

这两种样式都可以在Windows和Linux(以及其他)上使用。

在c++ 11中也有对std::string操作的函数，包括std::stoull，你可以使用:

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

像是…

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

. .那就用atoll()吧。您可能想要将#ifdef WINDOWS更改为其他东西，只需使用您可以依赖的东西来指示atoll()缺失但atoi64()存在(至少对于您所关心的场景)。

尝试strtoull()或strtoul()。前者仅在C99和c++ 11中出现，但通常广泛使用。

在现代c++中，我会使用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.

在strtoll(当然也很容易与std::string一起使用)和std::stoll(乍一看似乎更适合std::string)或boost::lexical_cast这样的c风格函数之间进行选择时，请注意，后两个函数在无法解析输入字符串或范围溢出时将抛出异常。有时这是有用的，有时不是，这取决于你想要达到什么目的。

如果你不能控制要解析的字符串(因为它是外部数据)，但你想编写健壮的代码(这总是应该是你的愿望)，你总是需要预料到一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，但需要更显式的代码来检测非法输入数据。std::stoll和boost::lexical_cast可以自动检测和发出错误输入的信号，但是您必须确保在某个地方捕获异常以避免被终止^(TM)。

因此，根据周围代码的结构、解析结果的需要(有时非法数据被"解析"为0是绝对可以的)、要解析的数据的来源以及最后但并非最不重要的个人偏好来选择一个或另一个。这两种功能通常都不优于其他功能。

这里我们将十六进制字符组成的字符串转换为uint64_t十六进制值。字符串的所有单个字符只被一个转换为十六进制整数。例如以10为基数-> String = "123":

• 第一个循环:值为1
• 第二次循环:值为1*10 + 2 = 12
• 第三次循环:值为12*10 + 3 = 123

所以像这样的逻辑被用来转换十六进制字符的字符串到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;
  }
}