将字符串中保存的数字中的每个数字转换为 int 数组 [英] Convert each digit from number saved in string to array of int

问题描述

我正在 DFA 上编写这个项目，我想将保存为字符串的整数的每个数字保存并转换为 int 数组.这是负责此操作的函数的代码:

I'm writing this project on DFA and i want to save and covert each digit of an integer saved as a string to an int array.This is the code from the function responsible for that:

int l=final_states.size();
int* temp_final;
temp_final=new int [l];
for(int i=0;i<l;i++)
{
    temp_final[i]=atoi(final_states.at(i).c_str());
}

<小时>

这给了我以下错误:request for member 'c_str' in '((DFA*)this)->DFA::final_states.std::basic_string<_CharT, _Traits, _Alloc>::at<char, std::char_traits, std::allocator>(((std::basic_string<char>::size_type)i))'，属于非类类型'char'|.

所以如果你能告诉我如何进行这种转换和保存工作，那就太好了.

So if you could tell me how to make this conversion and saving work , that would be great.

推荐答案

atoi() 函数需要一个 const char*，你不能调用 .c_str() 与 .at(i) 的结果实际上是一个 char& 值.

The atoi() function wants a const char*, you cannot call .c_str() with the result of .at(i) which is actually a char& value.

只需将您的分配行修改为

Just fix your assignment line to

 temp_final[i] = int(final_states[i]) - int('0');

虽然你也可以检查你是否真的有一个数字 那里，然后将其放入结果数组中:

Though you also might check if you really have a digit there, before putting it into your result array:

 if(std::isdigit(final_states[i])) {
     temp_final[i] = int(final_states[i]) - int('0');
 }
 else {
     // Skip, or throw error ...
 }

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