如何在vc ++中验证数字字符串 [英] How to validate a numeric string in vc++

问题描述

大家好，



我是vc环境中的新手。是否有人建议我如何验证数字字符串？



我的要求如下：

a）数字字符串有负数。

b）范围来自 - 1000到1000.



目前我做以下事情：

a）删除前导零。

b）删除特殊符号和字母。

c）仅限制四个字符的字符串。



用户输入任何类型的组合，如：

a）-1-0

b）1-00

c） - 00

等等。



我不明白如何限制上面的那个。

请有人建议我这件事。

Hi everybody,

I am new in vc environment. Is some one suggest me how I validate the numeric string?

My requirements are as follows:
a)numeric string have negative numbers.
b)range is from -1000 to 1000.

Currently I do the following things:
a)remove leading zeros.
b)remove special symbols and alphabets.
c)restrict the string only for four characters.

User enters combinations of any type like :
a)-1-0
b)1-00
c)--00
and so on.

I am not understanding how to restrict the above one.
Please anyone suggest me regarding to this things.

推荐答案

有像atoi，atof，atol这样的内置函数。

通过使用此功能，您不需要关心restictions：给定的字符串将被转换或您回零！



看看例如atoi：

There are built-in functions like atoi, atof, atol and so long.
By using this function you don''t need to care of restictions: the given string will be converted or you get zero back!

Take a look at e.g. atoi:

function
atoi
<cstdlib>

int atoi (const char * str);

Convert string to integer
Parses the C-string str interpreting its content as an integral number, which is returned as a value of type int.

例如：

For example:

#include <stdio.h>      /* printf, fgets */
#include <stdlib.h>     /* atoi */

int main ()
{
  int i;
  char buffer[256];
  printf ("Enter a number: ");
  fgets (buffer, 256, stdin);
  i = atoi (buffer);
  printf ("The value entered is %d. Its double is %d.\n",i,i*2);
  return 0;
}

这些是标准模板解决方案：

These are standard template solutions:

#include <string>
#include <sstream>
#include <stdexcept>

// Checks if a std::string can be converted to a type T.
template <typename T> bool Is(const std::string& s)
{
    T value;
    std::istringstream is(s);
    is >> value;

    if(!(is && (is >> std::ws).eof()))
        return false;

    return true;
}

// Converts a std::string into a type T.
// (A std::invalid_argument exception is thrown if the conversion is not possible).
template <typename t=""> const T To(const std::string& s)
{
    T value;
    std::istringstream is(s);
    is >> value;
 
    if(!(is && (is >> std::ws).eof()))
        throw std::invalid_argument("Function: To<t>. Unallowed conversion.");
 
    return value;
}
</t></typename>

没有限制，所以你应该转换后应用限制。

我担心这不适用于你所指出的奇怪输入。 （与-00一起使用，但不与其他人一起使用）。



我希望这会有所帮助。



祝你好运！

No restrictions are done, so you should apply restrictions after conversion.
I am afraid this does not work with strange inputs as you have pointed out. (Works with "-00" but not with the others).

I hope this sheds some light.

Best regards!

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