stoi和stoll在c ++ [英] stoi and stoll in c++

问题描述

我有#include（字符串）在我的声明在程序的顶部，但是当我尝试运行stoi（字符串）或stoll（字符串）我得到以下错误。我正在运行Cygwin g ++ v4.5.3。

I have #include(string) in my declaratives at the top of the program but when I try to run stoi(string) or stoll(string) i get the following error. I am running Cygwin g++ v4.5.3.

Z：\G\CSCE 437> g ++ convert.cpp -o conv
convert.cpp：在函数 void transfer（std :: string *）'：
convert.cpp：103：36：error： stoll'在此范围中声明
convert.cpp：116：35：错误：`stoi'未在此范围内声明

Z:\G\CSCE 437>g++ convert.cpp -o conv convert.cpp: In function void transfer(std::string*)': convert.cpp:103:36: error:stoll' was not declared in this scope convert.cpp:116:35: error: `stoi' was not declared in this scope

    fileTime[numRec] = stoll(result[0]);    //converts string to Long Long
    if(numRec = 0){
       beginningTime = fileTime[0];
    }
    fileTime[numRec] = timeDiff;
    hostName[numRec] = result[1];
    diskNum[numRec] = stoi(result[2]);
    type[numRec] = result[3];
    offset[numRec] = stoi(result[4]);
    fileSize[numRec] = stoi(result[5]);
    responseTime[numRec] = stoi(result[6]);`

一个字符串数组。

推荐答案

这些函数在C ++ 11中是新的，GCC只有在你指定版本的语言使用命令行选项 -std = c ++ 11 （或 -std = c ++ 0x

These functions are new in C++11, and GCC only makes it available if you specify that version of the language using the command-line option -std=c++11 (or -std=c++0x on some older versions; I think you'll need that for version 4.5).

如果你不能使用C ++ 11某些原因，你可以使用字符串流进行转换：

If you can't use C++11 for some reason, you could convert using string streams:

#include <sstream>

template <typename T> from_string(std::string const & s) {
    std::stringstream ss(s);
    T result;
    ss >> result;    // TODO handle errors
    return result;
}

或者，如果你感觉自闭症，C函数在<在< cstring> 中声明的code> strtoll 。

or, if you're feeling masochistic, the C functions in such as strtoll declared in <cstring>.

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