C++ 检查 cin 输入流产生一个整数
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Checking cin input stream produces an integer
提问by user2766546
I was typing this and it asks the user to input two integers which will then become variables. From there it will carry out simple operations.
我正在输入这个，它要求用户输入两个整数，然后它们将成为变量。从那里它将执行简单的操作。
How do I get the computer to check if what is entered is an integer or not? And if not, ask the user to type an integer in. For example: if someone inputs "a" instead of 2, then it will tell them to reenter a number.
如何让计算机检查输入的内容是否为整数？如果没有，要求用户输入一个整数。例如：如果有人输入“a”而不是 2，那么它会告诉他们重新输入一个数字。
Thanks
谢谢
#include <iostream>
using namespace std;
int main ()
{
int firstvariable;
int secondvariable;
float float1;
float float2;
cout << "Please enter two integers and then press Enter:" << endl;
cin >> firstvariable;
cin >> secondvariable;
cout << "Time for some simple mathematical operations:\n" << endl;
cout << "The sum:\n " << firstvariable << "+" << secondvariable
<<"="<< firstvariable + secondvariable << "\n " << endl;
}
回答by Chemistpp
You can check like this:
你可以这样检查：
int x;
cin >> x;
if (cin.fail()) {
//Not an int.
}
Furthermore, you can continue to get input until you get an int via:
此外，您可以继续获取输入，直到通过以下方式获得 int：
#include <iostream>
int main() {
int x;
std::cin >> x;
while(std::cin.fail()) {
std::cout << "Error" << std::endl;
std::cin.clear();
std::cin.ignore(256,'\n');
std::cin >> x;
}
std::cout << x << std::endl;
return 0;
}
EDIT: To address the comment below regarding input like 10abc, one could modify the loop to accept a string as an input. Then check the string for any character not a number and handle that situation accordingly. One needs not clear/ignore the input stream in that situation. Verifying the string is just numbers, convert the string back to an integer. I mean, this was just off the cuff. There might be a better way. This won't work if you're accepting floats/doubles (would have to add '.' in the search string).
编辑：为了解决下面关于像 10abc 这样的输入的评论，可以修改循环以接受字符串作为输入。然后检查字符串中是否有任何不是数字的字符，并相应地处理这种情况。在这种情况下，不需要清除/忽略输入流。验证字符串只是数字，将字符串转换回整数。我的意思是，这只是袖手旁观。可能有更好的方法。如果您接受浮点数/双精度数（必须在搜索字符串中添加“.”），这将不起作用。
#include <iostream>
#include <string>
int main() {
std::string theInput;
int inputAsInt;
std::getline(std::cin, theInput);
while(std::cin.fail() || std::cin.eof() || theInput.find_first_not_of("0123456789") != std::string::npos) {
std::cout << "Error" << std::endl;
if( theInput.find_first_not_of("0123456789") == std::string::npos) {
std::cin.clear();
std::cin.ignore(256,'\n');
}
std::getline(std::cin, theInput);
}
std::string::size_type st;
inputAsInt = std::stoi(theInput,&st);
std::cout << inputAsInt << std::endl;
return 0;
}
回答by Dúthomhas
Heh, this is an old question that could use a better answer.
呵呵，这是一个老问题，可以使用更好的答案。
User input should be obtained as a stringand then attempt-convertedto the data type you desire. Conveniently, this also allows you to answer questions like “what type of data is my input?”
用户输入应作为字符串获取，然后尝试转换为您想要的数据类型。方便的是，这还允许您回答诸如“我的输入是什么类型的数据？”之类的问题。
Here is a function I use a lot. Other options exist, such as in Boost, but the basic premise is the same: attempt to perform the string→type conversion and observe the success or failure:
这是我经常使用的一个函数。存在其他选项，例如在 Boost 中，但基本前提是相同的：尝试执行字符串→类型转换并观察成功或失败：
template <typename T>
std::optional <T> string_to( const std::string& s )
{
std::istringstream ss( s );
T result;
ss >> result >> std::ws; // attempt the conversion
if (ss.eof()) return result; // success
return {}; // failure
}
Using the optionaltype is just one way. You could also throw an exception or return a default value on failure. Whatever works for your situation.
使用optional类型只是一种方式。您还可以在失败时抛出异常或返回默认值。什么都适合你的情况。
Here is an example of using it:
下面是一个使用它的例子：
int n;
std::cout << "n? ";
{
std::string s;
getline( std::cin, s );
auto x = string_to <int> ( s );
if (!x) return complain();
n = *x;
}
std::cout << "Multiply that by seven to get " << (7 * n) << ".\n";
limitations and type identification
限制和类型识别
In order for this to work, of course, there must exist a method to unambiguously extract your data type from a stream. This is the natural order of things in C++ — that is, business as usual. So no surprises here.
当然，为了使其工作，必须存在一种方法来明确地从流中提取您的数据类型。这是 C++ 中事物的自然顺序——即一切照旧。所以这里没有惊喜。
The next caveat is that some types subsume others. For example, if you are trying to distinguish between intand double, check for intfirst, since anything that converts to an intis also a double.
下一个警告是某些类型包含其他类型。例如，如果您试图区分intand double，请int首先检查，因为转换为 an 的任何内容int也是 a double。
回答by nook
There is a function in c called isdigit(). That will suit you just fine. Example:
c 中有一个函数叫做isdigit(). 那会很适合你。例子：
int var1 = 'h';
int var2 = '2';
if( isdigit(var1) )
{
printf("var1 = |%c| is a digit\n", var1 );
}
else
{
printf("var1 = |%c| is not a digit\n", var1 );
}
if( isdigit(var2) )
{
printf("var2 = |%c| is a digit\n", var2 );
}
else
{
printf("var2 = |%c| is not a digit\n", var2 );
}
From here
从这里
回答by Zac Howland
If istream fails to insert, it will set the fail bit.
如果 istream 插入失败，它将设置失败位。
int i = 0;
std::cin >> i; // type a and press enter
if (std::cin.fail())
{
std::cout << "I failed, try again ..." << std::endl
std::cin.clear(); // reset the failed state
}
You can set this up in a do-while loop to get the correct type (intin this case) propertly inserted.
您可以在 do-while 循环中进行设置，以正确int插入正确的类型（在本例中）。
For more information: http://augustcouncil.com/~tgibson/tutorial/iotips.html#directly
更多信息：http: //augustcouncil.com/~tgibson/tutorial/iotips.html#directly
回答by Zac Howland
You can use the variables name itself to check if a value is an integer. for example:
您可以使用变量名称本身来检查值是否为整数。例如：
#include <iostream>
using namespace std;
int main (){
int firstvariable;
int secondvariable;
float float1;
float float2;
cout << "Please enter two integers and then press Enter:" << endl;
cin >> firstvariable;
cin >> secondvariable;
if(firstvariable && secondvariable){
cout << "Time for some simple mathematical operations:\n" << endl;
cout << "The sum:\n " << firstvariable << "+" << secondvariable
<<"="<< firstvariable + secondvariable << "\n " << endl;
}else{
cout << "\n[ERROR\tINVALID INPUT]\n";
return 1;
}
return 0;
}
回答by Eduardo Floriani
You could use :
你可以使用：
int a = 12;
if (a>0 || a<0){
cout << "Your text"<<endl;
}
I'm pretty sure it works.
我很确定它有效。
