C 或 C 中 64 位整数(uint64_t)的 atoi 等价于什么?可以在 Unix 和 Windows 上运行吗?

发布于 2024-12-05 22:26:40 字数 41 浏览 0 评论 0原文

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一个。

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

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窝囊感情。 2024-12-12 22:26:40

使用 strtoulll 如果你有的话或者< a href="http://msdn.microsoft.com/en-us/library/85zk715d.aspx" rel="noreferrer">_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 
);

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 
);
何以心动 2024-12-12 22:26:40

您已标记此问题 ,所以我假设您可能也对 C++ 解决方案感兴趣。如果您无法使用 boost,您可以使用 boost::lexical_caststd::istringstream 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;
}

这两种样式都适用于 Windows 和 Linux(以及其他)。

在 C++11 中还有 std::string 进行操作的函数,包括 std::stoull 您可以使用:

#include <string>

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

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);
}
梦幻之岛 2024-12-12 22:26:40

类似.....

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

然后只需使用atoll()。您可能想将 #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() 缺失,但 atoi64() 就在那里(至少对于您关心的场景而言)。

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).

等风来 2024-12-12 22:26:40

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

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

假情假意假温柔 2024-12-12 22:26:40

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.

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.
小鸟爱天空丶 2024-12-12 22:26:40

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 等 C 风格函数之间进行选择时:请注意,如果后两者无法解析输入字符串或范围溢出,则会抛出异常。
有时这很有用,有时没有,这取决于您想要实现的目标。

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),那么您始终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 不会抛出异常,但需要更明确的代码来检测非法输入数据。 std::stoll 和 boost::lexical_cast 会自动检测并发出蹩脚输入信号,但您必须确保在某处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(有时将非法数据“解析”为 0 是绝对可以的)、要解析的数据源以及最后但并非最不重要的您个人的需要来选择其中之一偏好。通常,这两种可用功能都不优于其他功能。

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.

红焚 2024-12-12 22:26:40

这里我们将由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仅按一位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
  • 第二个循环:值为 1*10 + 2 = 12
  • 第三个循环:值为 12*10 + 3 = 123

所以像这样的逻辑用于将十六进制字符的字符串转换为 uint_64hex价值。

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;
  }
}

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;
  }
}
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