如何在C++中检查科学计数法的双精度数

Question

如何在C++中检查科学计数法的双精度数

3

我主要想检查一行文本是否为有效的科学计数法浮点数。

如果一行文本中含有像a这样的字符，或者只有像help或0.a这样很少的字符，则会被视为无效的科学计数法浮点数并被拒绝，但是像1.234E+9或2.468E9这样的值将被存储，因为它是可以接受的。

我已经编写了一些代码来处理这个问题，但是我需要一些帮助来区分科学计数法浮点数和普通文本字符。

char *temp;
int u=0;
int arrayLen = strlen(temp);
for(int i=0; i<len; i++)
{
  if(isalpha(temp[i]))
  {
    if((temp[i] == 'e') && (!isalpha(temp[i-1])))
    {
      break;
    }
    u++;
  }
}

if(u > 0)
{
   temp[0] = 0;
   break;
}

- Rews Prog

3

请使用 std::stod 或其 C 语言等效函数 strtod。 - T.C.

3

你能使用正则表达式吗？这是一个一行就能搞定的操作。 - UniversE

@T.C. 呃 - 或者只是那咳嗽太棒了。太简单了。:-D - UniversE

1

使用std::istringstream和适当的I/O操纵符怎么样？ - πάντα ῥεῖ

如果你得到了"1234.45E2text"，你想要"1234.45E2"还是"0"？ - George Houpis

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1个回答

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- George Houpis · Accepted Answer

正如T.C.建议的那样，使用strtod函数。但要检查返回指针以查看它是否读取了所有内容。

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

double convert_number( const char * num )
{
   char * endptr = 0;
   double retval;
   retval = strtod( num, &endptr );
   return ( !endptr || ( *endptr != '\0' ) ) ? 0 : retval;
}


int main( int argc, char ** argv )
{
    int index;
    for( index = 0; index < argc; ++index )
        printf( "%30s --> %f\n", argv[index], convert_number( argv[index] ) );
    return 0;
}

例子：

./a.out 13.2425 99993.3131.1134  13111.34e313e2  1313e4 1 324.3 "2242e+3"
                       ./a.out --> 0.000000
                       13.2425 --> 13.242500
               99993.3131.1134 --> 0.000000
                13111.34e313e2 --> 0.000000
                        1313e4 --> 13130000.000000
                             1 --> 1.000000
                         324.3 --> 324.300000
                       2242e+3 --> 2242000.000000

--------------------按请求的替代方案-----------------------------

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int convert_number( const char * num, double * retval )
{
   char * endptr = 0;
   *retval = strtod( num, &endptr );
   return ( !endptr || ( *endptr != '\0' ) ) ? 0 : 1;
}


int main( int argc, char ** argv )
{
    int index;
    double dvalue;
    for( index = 0; index < argc; ++index )
        if( convert_number( argv[index], &dvalue ) )
            printf( "%30s --> %f\n", argv[index], dvalue );
        else
            printf( "%30s --> Rejected\n", argv[index], dvalue );
    return 0;
}

结果：

./a.out 13.2425 99993.3131.1134  13111.34e313e2  1313e4 1 324.3 "2242e+3" 0.0
                       ./a.out --> Rejected
                       13.2425 --> 13.242500
               99993.3131.1134 --> Rejected
                13111.34e313e2 --> Rejected
                        1313e4 --> 13130000.000000
                             1 --> 1.000000
                         324.3 --> 324.300000
                       2242e+3 --> 2242000.000000
                           0.0 --> 0.000000

新手版：

int convert_number( const char * num, double * retval )
{
   char * endptr = 0; /* Prepare a pointer for strtod to inform us where it ended extracting the double from the string. */
   *retval = strtod( num, &endptr );  /* Run the extraction of the double from the source string (num) and give us the value so we can store it in the address given by the caller (retain the position in the string where strtod stopped processing). */
   if( endptr == NULL )
       return 0;  /* endptr should never be NULL, but it is a defensive programming to prevent dereferencing null in the next statement. */
   else if( *endptr != '\0 ) /* If we are pointing to a non-null character, then there was an invalid character that strtod did not accept and stopped processing.  So it is not only a double on the line.  So we reject. */
       return 0; /* Not strictly the double we are looking for. */ 
   /* else */
   return 1;  /* Return 1 to the caller to indicate truth (non-zero) that the value in *retval has valid double value to use. */
}