简体   繁体   English

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?

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

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。

Use strtoull if you have it or _strtoui64() with visual studio.使用strtoull如果有)或_strtoui64()与 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 
);

You've tagged this question , so I'm assuming you might be interested in C++ solutions too.您已将此问题标记为 ,因此我假设您也可能对 C++ 解决方案感兴趣。 You can do this using boost::lexical_cast or std::istringstream if boost isn't available to you:如果 boost 对您不可用,您可以使用boost::lexical_caststd::istringstream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::stoull which you can use:在 C++11 中,还有std::string进行操作函数,包括您可以使用的std::stoull

#include <string>

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

Something like...就像是...

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

..then just use atoll() . ..然后只需使用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).您可能希望将#ifdef WINDOWS更改为其他内容,只需使用您可以依赖的内容来指示atoll()丢失但atoi64()存在(至少对于您关注的场景)。

Try strtoull() , or strtoul() .尝试strtoull()strtoul() The former is only in C99 and C++11, but it's usually widely available.前者仅在 C99 和 C++11 中存在,但通常广泛可用。

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.

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.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,后两者将在无法解析输入字符串或范围溢出的情况下抛出异常。 Sometimes this is useful, sometimes not, depends what you're trying to achive.有时这很有用,有时则不然,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.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。 For corrupted data strtoll will not throw but needs more explicit code to detect illegal input data.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 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) . 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.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。 Neither of the functions available is generally superiour to the others.可用的功能通常都不优于其他功能。

Here we convert String consisting of HEX character to uint64_t hex value.这里我们将包含 HEX 字符的 String 转换为 uint64_t 十六进制值。 All individual characters of string is converted to hex integer ony by one.字符串的所有单个字符都被转换为一个十六进制整数。 For example in base 10 -> String = "123":例如在基数 10 -> String = "123" 中:

  • 1st loop : value is 1第一个循环:值为 1
  • 2nd loop : value is 1*10 + 2 = 12第二个循环:值为 1*10 + 2 = 12
  • 3rd loop : value is 12*10 + 3 = 123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;
  }
}

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM