将字符串转换为int C ++ [英] Convert string to int C++

问题描述

只是有一个快速问题。我已经看过互联网了很多，我找到了一些解决方案，但没有一个工作了。看看将字符串转换为int，我不是指ASCII码。

对于快速下移，我们在方程式中作为字符串传递。我们要把它分解，正确格式化并求解线性方程。现在，在说，我不能将一个字符串转换为一个int。

我知道字符串将是格式（-5）或（25）等等，所以它绝对是一个int。但是我们如何从字符串中提取呢？

一种方式是我正在运行一个for循环通过字符串，检查一个数字，提取所有的数字然后看看是否有一个前导' - '，如果有，将int乘以-1。

看起来有点复杂一个小问题。任何想法？

解决方案

在C ++ 11中有一些不错的新转换函数 std ::

$ b

$

code> atoi（str.c_str（））

您可以使用

  std :: stoi（str）
  

其中 str 是您的号码 std :: string 。

有所有版本的数字：
long stol（string）， float stof ， double stod（string），...
请参阅 http://en.cppreference.com/w/cpp/string/basic_string/stol

Just have a quick question. I've looked around the internet quite a bit and I've found a few solutions but none of them have worked yet. Looking at converting a string to an int and I don't mean ASCII codes.

For a quick run-down, we are passed in an equation as a string. We are to break it down, format it correctly and solve the linear equations. Now, in saying that, I'm not able to convert a string to an int.

I know that the string will be in either the format (-5) or (25) etc. so it's definitely an int. But how do we extract that from a string?

One way I was thinking is running a for/while loop through the string, check for a digit, extract all the digits after that and then look to see if there was a leading '-', if there is, multiply the int by -1.

It seems a bit over complicated for such a small problem though. Any ideas?

解决方案

In C++11 there are some nice new convert functions from std::string to a number type.

So instead of

atoi( str.c_str() )

you can use

std::stoi( str )

where str is your number as std::string.

There are version for all flavours of numbers: long stol(string), float stof(string), double stod(string),... see http://en.cppreference.com/w/cpp/string/basic_string/stol

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