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。
strtoull_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。如果您无法使用增强功能，您可以使用
boost::lexical_caststd::istringstream#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;
}
两种风格都适用于 Windows 和 Linux（以及其他）。
在 C++11 中还有 对
std::stringstd::stoull#include <string>
int main() {
const std::string str="594348534879";
unsigned long long v = std::stoull(str);
}
类似...
#ifdef WINDOWS
#define atoll(S) _atoi64(S)
#endif
..然后只需使用
atoll()#ifdef WINDOWSatoll()atoi64()尝试
strtoull()strtoul()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 等 C 风格函数之间进行选择时：请注意，后两者如果无法解析输入字符串或范围溢出，将会抛出异常。 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（这始终应该是您的愿望），您总是需要预料到某些恶意攻击者会注入损坏的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部分来选择其中之一个人喜好。一般来说，这两种可用功能都不优于其他功能。
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":
因此，这个逻辑用于将十六进制字符的字符串转换为 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;
}
}