arrays - C code to convert hex to int -

i writing code convert hex entry integer equivalent. 10 , b 11 etc. code acts weirdly, in seg. faults @ random locations , including newline character @ times work. trying debug it, can understand doing wrong here. can take , me here ? lot time.

/* fixed working code interested */

         #include <stdio.h>             #include <stdlib.h>               unsigned int hextoint(const char temp[])             {              int i;             int answer = 0;             int dec;             char hexchar[] = "aabbccddeeff" ;               ( i=0; temp[i] != '\0'; i++ )             {                  if ( temp[i] == '\0')                 {                      return ;                         }                  if (temp[i] == '0' || temp[i] == 'x' || temp[i] == 'x' )                 {                            printf("0");                     answer = temp[i];                 }                  // compare each temp[i] contents in hexchar[]                 int j;                 int = temp[i];                 ( j=0; hexchar[j] != '\0'; j++)                 {                     if ( temp[i] == hexchar[j] )                     {                     answer *= 16;                     answer = answer + 10 + (j/2);   //                    printf("%d\n",answer );                     break;                           }                 }              }              return answer;                }               main()             {             char *test[] =              {   "bad",                 "aabbdd"                 "0100",                 "0x1",                 "0xa",                 "0x0c0be",                 "abcdef",                 "123456",                 "0x123456",                 "deadbeef",                  "zog_c"             };              int answer=0;              // calculate number of char's.             int numberofchars;             numberofchars = sizeof test /sizeof test[0];              printf("main():number of chars = %d\n",numberofchars);             int i;             // go through each character , convert hex integers.             ( = 0; i<numberofchars;i++)             {                 // need take first char , go through , convert                                                     it.                 answer = hextoint(test[i]);                 printf("%d\n",answer );              }               } 

let's take look.

unsigned int hextoint(const char temp[]) {     int i;     int answer = 0;     char hexchar[] = "aabbccddeeff" ;      ( i=0; temp[i] != '\0'; i++ )     {         printf("in here");         printf("%c\t",temp[i] );     }      return answer;   } 

this doesn't seem try conversion. should return 0, since answer never assigned other value. normally, you'd like:

for (i=0; input[i] != '\0'; i++) {     answer *= 16;     answer += digit_value(input[i]); } return answer; 

where digit_value (obviously enough) returns value of individual digit. 1 way is:

int digit_value(char input) {      input = tolower(input);     if (input >= '0' && input <= '9')         return input - '0';     if (input >= 'a' && input <= 'f')         return input - 'a' + 10;     return -1; // signal error. } 

then, looking @ main:

main() { 

depending on "implicit int" rule poor practice, @ least imo. it's better specify return type.

// calculate number of char's. int numberofchars; numberofchars = sizeof test /sizeof test[0]; 

this calculates number of strings, not number of chars.

for ( = 0; i<=numberofchars;i++) 

valid subscripts run 0 through number of items - 1, attempts read past end of array (giving undefined behavior).