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write htoi()

(8 posts)

1. 

   rishabh
   guest
   Posted 1 year ago #
   
   

   Given a string that contains hexadecimal number, write a function htoi() that returns the equivalent integer calue

2. 

   Venki
   Moderator
   Posted 1 year ago #
   
   

   // Here is one function
   int getValue(char const* hexString, int const radix)
   {
       int         loopIndex;      // Loop iterator
       int         signFlag = 1;   // Keeps track of sign
       signed int  retValue = 0;   // Final value
       int         currentDigit;   // Current digit under process
       int         len = strlen(hexString);    // Lenght of string
   
       // Check for sign
       if(hexString[0] == '-')
       {
           // Advance the base
           hexString++;
           // Remove '-' from count
           len--;
           // Sing is -ve
           signFlag = -1;
       }
   
       // Iterate from left to right
       // Accumulate the weight of each digit upon occurance of next digit
       for(loopIndex = 0; loopIndex < len; loopIndex++)
       {
           currentDigit = toupper(hexString[loopIndex]) - ASCII_ZERO;
   
           // Is the digit between A through F?
           if(currentDigit > 9)
           {
               // Adjust numerical value
               currentDigit -= (ASCII_A - '9');
           }
   
           // Based on Horner's principle
           retValue = retValue * radix + currentDigit;
       }
   
       // Flip the sign, if required
       retValue *= signFlag;
   
       // This is the final value
       return retValue;
   }
   

   call the function in the following way,

   getValue("1011", 2); // Binary string to integer
   getValue("12345", 10); // Decimal string to integer
   getValue("ABCDE", 16); // Hexadecimal string to integer

   Error handling is omitted.

   Could someone please write test cases for the above program?

3. 

   abc
   guest
   Posted 1 year ago #
   
   

   It fails for getValue("D80", 16).
   It returns 3200, hex for which is C80

4. 

   abc
   guest
   Posted 1 year ago #
   
   

   @Venki:

   I changed the way you calculate currentDigit to the following and then the code works:

    char ch = toupper(hexString[loopIndex]);
         if (ch >= '0' && ch <= '9')
           currentDigit = ch - '0';
         else if (ch >= 'A' && ch <= 'F')
           currentDigit = ch - 'A' + 10;
   

5. 

   Venki
   Moderator
   Posted 1 year ago #
   
   

   Thanks @abc.

6. 

   Sambasiva
   Member
   Posted 1 year ago #
   
   

   #include <stdio.h>
   
   #define hex(c) (( c >= '0' && c <= '9') ? c - '0' :  \
                            10 + ((c >= 'A' && c <= 'F') ? c - 'A' : c - 'a'))
   
   int htoi(const char *s) {
   
       char c;
       int res = 0;
   
       int sign  = (*s == '-') ? s++, -1 : 1;
   
       while((c = *s++)) {
           res = res * 16 +  hex(c);
       }
   
       return sign * res;
   }
   
   int main() {
   
       printf("%d\n", htoi("C80"));
   }
   

7. 

   wgpshashank
   
   Posted 1 year ago #
   
   

   Here is Sweet , Simple & Excellent Version of Program

   #include<stdio.h>
   #include<string.h>
   #include<stdlib.h>
   void htoi(char *);
   int main()
   {
     char *hex="0xFFFF";
     htoi(hex);
     return 0;
   }
   
   void htoi(char *hexstr)
   {
   char c;
   int val=0,i;
   
   int len=strlen(hexstr);
   for(i=0;i<len;i++)
   {
   c=*(hexstr+2); //pointer will points to F in case of 0xFFFF
   
   if(c>='a' && c<='f')
   {
   val=val*16+(c-'a'+10);
   }
   
   else if(c>='A' && c<='F')
   {
   val=val*16+(c-'A'+10);
   }
   
   else
   {
   
   if(c>='0' && c<='9')
   {
   val=val*16+(c-'0');
   }
   
   }
   printf("Val: %d \t ",val);
   hexstr++;
   
   }
   
   printf("\n Val: %d ",val);
   }

   Please Let me know if AnyThing wrong or for any test case its failing..???

8. 

   Ravi Biradar
   guest
   Posted 3 months ago #
   
   

   "Hey your code fails if user enters an invalid number so look at this following code, it holds good "

   #include<stdlib.h>
   int htoi(const char *ptr){
       char c;
       int i,len,value=0;
       if(ptr[0]=='0'&&(ptr[1]=='x'||ptr[1]=='X'))
       {
           //c=*(ptr+2);
           ptr=ptr+2;
           len=strlen(ptr);
           printf("%s\t%d\n",ptr,len);
   
           for(i=0;i<len;i++){
               c=ptr[i];
               if(c>='0'&&c<='9')
                  value=value*16+(c-'0');
               else if(c>='a'&&c<='f')
                    value=value*16+(10+(c-'a'));
               else if(c>='A'&&c<='F')
                    value=value*16+(10+(c-'A'));
               else{
                     printf("invalid number\n");
                     return -1;
                     }
           }
           return value;
       }
       else{
            printf("invalid number\n");
                     return -1;
       }
   }
   void main(){
        char str[10];
        int value=0;
        printf("enter the hex value\n");
        scanf("%s",str);
        if((value=htoi(str))!=-1)
        printf("%u\n",value);
        system("pause");
   }

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