C ++错误将字符串转换为双精度 [英] C++ error converting a string to a double

问题描述

我试图将字符串转换为双精度。代码很简单。

I am trying to convert a string to a double. The code is very simple.

            double first, second;
            first=atof(str_quan.c_str());
            second=atof(extra[i+1].c_str());
            cout<<first<<" "<<second<<endl;
            quantity=first/second;

当试图转换额外的，编译器抛出这个宝石的智慧在我：

when trying to convert extra, the compiler throws this gem of wisdom at me:

错误：请求成员c_str in
extra.std :: basic_string< _CharT，
_Traits，_Alloc> :: operator [与_CharT = char，_Traits = std :: char_traits，_Alloc =
std :: allocator（i + 1））），这是非类
类型char

error: request for member c_str in extra.std::basic_string<_CharT, _Traits, _Alloc>::operator[] with _CharT = char, _Traits = std::char_traits, _Alloc = std::allocator(i + 1))), which is of non-class type char

我不知道这是什么意思。如果我cout extra [i + 1]，我得到3.如果我把额外的作为一个字符串，程序试图首先（2）51（ascii为3）。

I have no idea what that means. if I cout extra[i+1], I get 3. If I leave extra as a string, the program tries to div first (2) by 51 (ascii for 3). What the heck is going on?

推荐答案

这听起来像 extra a std :: string ，因此 extra [i + 1] 返回 code>，这是非类类型。

It sounds like extra is a std::string, so extra[i+1] returns a char, which is of non-class type.

听起来你正在尝试解析字符串 extra 从 i + 1 开始。您可以使用：

It sounds like you are trying to parse the string extra starting from the i+1th position. You can do this using:

second = atof(extra.substr(i + 1).c_str());

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