c++中atoi()函数的使用
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use of atoi() function in c++
提问by shubz
when I pass a stringvariable in the below code, g++ gives an error:
当我string在下面的代码中传递一个变量时，g++ 给出了一个错误：
cannot convert ‘std::__cxx11::string {aka std::__cxx11::basic_string}' to ‘const char*' for argument ‘1' to ‘int atoi(const char*)'
无法将参数 '1' 的 'std::__cxx11::string {aka std::__cxx11::basic_string}' 转换为 'const char*' 到 'int atoi(const char*)'
My code is:
我的代码是：
#include<iostream>
#include<stdlib.h>
using namespace std;
int main()
{
string a = "10";
int b = atoi(a);
cout<<b<<"\n";
return 0;
}
But if I change the code to :
但是如果我将代码更改为：
#include<iostream>
#include<stdlib.h>
using namespace std;
int main()
{
char a[3] = "10";
int b = atoi(a);
cout<<b<<"\n";
return 0;
}
It works completely fine.
它工作得很好。
Please explain why stringdoesn't work. Is there any difference between string aand char a[]?
请解释为什么string不起作用。有什么区别string a和char a[]？
回答by Trevor Hickey
atoiis an older function carried over from C.
atoi是从 C 继承而来的旧函数。
C did not have std::string, it relied on null-terminated char arrays instead. std::stringhas a c_str()method that returns a null-terminated char*pointer to the string data.
C 没有std::string，它依赖于以空字符结尾的字符数组。 std::string有一个c_str()方法返回一个char*指向字符串数据的空终止指针。
int b = atoi(a.c_str());
In C++11, there is an alternative std::stoi()function that takes a std::stringas an argument:
在 C++11 中，有一个std::stoi()将 astd::string作为参数的替代函数：
#include <iostream>
#include <string>
int main()
{
std::string a = "10";
int b = std::stoi(a);
std::cout << b << "\n";
return 0;
}
回答by Ed Heal
You need to pass a C style string.
您需要传递一个 C 风格的字符串。
I.e use c_str()
即使用 c_str()
Change
改变
int b = atoi(a);
to
到
int b = atoi(a.c_str());
PS:
PS：
This would be better - get the compiler to work out the length:
这会更好 - 让编译器计算出长度：
char a[] = "10";
回答by Remy Lebeau
atoi()expects a null-terminated char*as input. A stringcannot be passed as-is where a char*is expected, thus the compiler error. On the other hand, a char[]can decay into a char*, which is why using a char[]works.
atoi()期望以空字符结尾char*作为输入。Astring不能按原样传递char*，因此会出现编译器错误。另一方面， achar[]可以衰减为 a char*，这就是使用 achar[]作品的原因。
When using a string, call its c_str()method when you need a null-terminated char*pointer to its character data:
使用 a 时string，c_str()当您需要char*指向其字符数据的空终止指针时调用其方法：
int b = atoi(a.c_str());
回答by C-H-M
There is a difference between them. For each one there are different functions:
它们之间是有区别的。每一种都有不同的功能：
String
细绳
As said before, for stringthere is the stoifunction:
如前所述，因为string有stoi函数：
string s("20");
cout << stoi(s) * 2; // output: 40
char*
字符*
In the past, atoiused to handle char*conversion.
过去，atoi用于处理char*转换。
However, now atoiis replaced by strtolwhich gets 3 parameters:
但是， nowatoi被替换为strtolwhich gets 3 个参数：
char*of the characters to parse tolong,char**that returns pointer to after the parsed string,intfor the base the number should be parsed from (2, 10, 16, or whatever).
char*要解析为的字符long，char**返回指向解析后的字符串的指针，int对于基数，应该从（2、10、16 或其他）解析数字。
char c[]="20";
char* end;
cout << strtol(c, &end, 16); // output: 32
There are whole bunch of functions like strtollike strtof, strtod, or strtollwhich converts to float, double, long, and long long.
有喜欢的功能一大堆strtol像strtof，strtod或者strtoll其皈依float，double，long，和long long。
The main advantages of those new functions are mostly error-handling, and multi-base support.
这些新功能的主要优点主要是错误处理和多基支持。
The main disadvantage is that there isn't a function that converts to int, only to long(besides other types).
主要的缺点是没有一个函数可以转换为int，只能转换为long（除了其他类型）。
For more details, see https://stackoverflow.com/a/22866001/12893141
回答by stackerjoe
according to the documentation of atoi(), the function expects a "pointer to the null-terminated byte string to be interpreted"which basically is a C-style string. std::stringis string type in C++ but it have a method c_str()that can return a C-string which you can pass to atoi().
根据 的文档atoi()，该函数需要一个“指向要解释的空终止字节字符串的指针”，它基本上是一个 C 风格的字符串。std::string是 C++ 中的字符串类型，但它有一个方法c_str()可以返回一个 C 字符串，您可以将它传递给atoi().
string a = "10";
int b = atoi(a.c_str());
But if you still want to pass std::stringand your compiler supports C++ 11, then you can use stoi()
但是如果你仍然想通过std::string并且你的编译器支持 C++ 11，那么你可以使用stoi()
string a = "10";
int b = stoi(a);
