如何检查科学记数法是否加倍或不C++

Question

我基本上想检查线是否是有效的科学替身。

所以基本上，如果一行有一个像a这样的字符，或者只有几个字符，比如help或0.a那么它就会被认为是不科学的替身并被拒绝，但是像1.234E+9或2.468E9这样的东西会被存储，因为它是一个可接受的值

。

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......区分科学替身和仅部分字符

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;
}

Answer 1

正如 T.C. 建议的那样，使用 strtod。 但是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 : retval;
}

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

例：

./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

--------------------根据请求----------------------------备选

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 : 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 --> %fn", argv[index], dvalue );
        else
            printf( "%30s --> Rejectedn", 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 != ' ) /* 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. */
}