C++ 相当于 atoi
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equivalent of atoi
提问by Mansuro
Is there a function that could replace atoi in c++. I made some research and didn't find anything to replace it, the only solutions would be using cstdlib or implementing it myself
是否有一个函数可以替换 C++ 中的 atoi。我做了一些研究并没有找到任何可以替代它的方法，唯一的解决方案是使用 cstdlib 或自己实现它
采纳答案by David Rinck
If you don't want to use Boost, C++11 added std::stoifor strings. Similar methods exist for all types.
如果您不想使用 Boost，std::stoi则为字符串添加了C++11 。所有类型都存在类似的方法。
std::string s = "123"
int num = std::stoi(s);
Unlike atoi, if no conversion can be made, an invalid_argumentexception is thrown. Also, if the value is out of range for an int, an out_of_rangeexception is thrown.
与 不同atoi，如果无法进行转换，invalid_argument则会引发异常。此外，如果该值超出 int 的范围，out_of_range则会引发异常。
回答by Armen Tsirunyan
boost::lexical_castis your friend
boost::lexical_cast是你的朋友
#include <string>
#include <boost/lexical_cast.hpp>
int main()
{
std::string s = "123";
try
{
int i = boost::lexical_cast<int>(s); //i == 123
}
catch(const boost::bad_lexical_cast&)
{
//incorrect format
}
}
回答by Armen Tsirunyan
You can use the Boost function boost::lexical_cast<> as follows:
您可以使用 Boost 函数 boost::lexical_cast<> 如下：
char* numericString = "911";
int num = boost::lexical_cast<int>( numericString );
More information can be found here(latest Boost version 1.47). Remember to handle exceptions appropriately.
可以在此处找到更多信息（最新的 Boost 版本 1.47）。记住要适当地处理异常。
回答by Torp
Without boost:stringstream ss(my_string_with_a_number); int my_res; ss >> my_res;
About as annoying as the boost version but without the added dependency. Could possibly waste more ram.
不使用 boost：stringstream ss(my_string_with_a_number); int my_res; ss >> my_res;
与 boost 版本一样烦人，但没有添加依赖项。可能会浪费更多的内存。
回答by Alastair
You don't say why atoiis unsuitable so I am going to guess it has something to do with performance. Anyway, clarification would be helpful.
你没有说为什么atoi不合适，所以我猜测它与性能有关。无论如何，澄清会有所帮助。
Using Boost Spirit.Qi is about an order of magnitude faster than atoi, at least in tests done by Alex Ott.
使用 Boost Spirit.Qi 大约比Alex Ottatoi进行的测试快一个数量级。
I don't have a reference but the last time I tested it, Boost lexical_castwas about an order of magnitude slower than atoi. I think the reason is that it constructs a stringstream, which is quite expensive.
我没有参考资料，但上次我测试它时，Boostlexical_cast比atoi. 我认为原因是它构造了一个stringstream，这是相当昂贵的。
Update: Some more recent tests
更新：一些最近的测试
回答by Mirco De Zorzi
You can use the function stoi();
您可以使用该功能 stoi();
#include <string>
// Need to include the <string> library to use stoi
int main(){
std::string s = "10";
int n = stoi(s);
}
To actually compile this you will have to enable c++11, look up on google how to do it (on code::blocks it's: Settings -> Compiler -> "Have g++ follow C++11 ISO C++ language standard") If you compile from terminal you have to add -std=c++11
要实际编译它，您必须启用 c++11，在 google 上查找如何操作（在 code::blocks 上：设置 -> 编译器 ->“让 g++ 遵循 C++11 ISO C++ 语言标准”）如果从终端编译，则必须添加 -std=c++11
g++ -std=c++11 -o program program.cpp
