我试图将64位整数字符串转换为整数，但我不知道要使用哪一个。
答案 0 :(得分：34)
如果有，请使用strtoull或使用visual studio _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
);
答案 1 :(得分：18)
您已标记此问题c++，因此我假设您可能也对C ++解决方案感兴趣。如果您无法获得提升，则可以使用boost::lexical_cast或std::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;
}
这两种样式都适用于Windows和Linux（以及其他）。
在C ++ 11中还有functions that operate on std::string，包括你可以使用的std::stoull：
#include <string>
int main() {
const std::string str="594348534879";
unsigned long long v = std::stoull(str);
}
答案 2 :(得分：7)
像...一样的东西。
#ifdef WINDOWS
#define atoll(S) _atoi64(S)
#endif
..然后只使用atoll()。您可能希望将#ifdef WINDOWS更改为其他内容，只需使用您可以信赖的内容来表明atoll()缺失但atoi64()已存在（至少对于您所使用的方案）关心）。
答案 3 :(得分：4)
尝试strtoull()或strtoul()。前者仅在C99和C ++ 11中，但它通常可以广泛使用。
答案 4 :(得分：1)
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.
答案 5 :(得分：0)
在C样式的函数（例如strtoll（当然也容易与std :: string一起使用）和std :: stoll（乍一看似乎更适合std :: string）之间进行选择时，可以使用boost或c： ：lexical_cast：请注意，如果后两者无法解析输入字符串或范围溢出，则后者将引发异常。 有时候这很有用，有时却没有，取决于您要达到的目标。
如果您无法控制要解析的字符串（因为它是外部数据），但是您想要编写健壮的代码（始终应该是您的期望），则始终需要期待某些恶意攻击者注入的损坏数据或外部损坏的代码组件。对于损坏的数据，strtoll不会抛出，但是需要更明确的代码来检测非法输入数据。 std :: stoll和boost :: lexical_cast会自动检测并提示糟糕的输入，但您必须确保在某处捕获异常，以免被终止（TM）。
因此，根据周围代码的结构选择一个或另一个，解析结果的需求（有时将非法数据“解析”为0是绝对可以的）要解析的数据源，但最后却没有至少是您的个人喜好。可用的功能通常都不比其他功能优越。
答案 6 :(得分：0)
在这里，我们将由十六进制字符组成的String转换为uint64_t十六进制值。字符串的所有单个字符都将转换为十六进制整数一。例如以10为底-> String =“ 123”：
因此，这种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;
}
}