Type cast from char array to int in HEX to DEC funct

[send]

#include <iostream>
#include <string.h>
#include <math.h>
using namespace std;
int hexdec (char * hex) {
int k=strlen(hex);
int i;
int res=0;
for (i=2;i<k;i++) {if (hex[i]=='F') res+=15*pow(16,k-i-1);
else if (hex[i]=='E') res+=14*pow(16,k-i-1);
else if (hex[i]=='D') res+=13*pow(16,k-i-1);
else if (hex[i]=='C') res+=12*pow(16,k-i-1);
else if (hex[i]=='B') res+=11*pow(16,k-i-1);
else if (hex[i]=='A') res+=10*pow(16,k-i-1);
else if (hex[i]>=0 || hex[i]<=9) res+=hex[i]*pow(16,k-i-1);// this string is not workable here how type cast
}
return res;}

int main()
{
cout << hexdec("0x1F")<< endl;
return 0;
}

[Zhuge]

Note that the characters between 0 and 9 are actually '0'..'9', and not 0..9 as you have in your code; the latter are the ASCII characters represented by codes 0 to 9.

As for your actual question, you could make a function that converts those characters for you, or use the ASCII code math trick and subtract the code for '0'.

[send]

res=((int)hex[i])--why it doesnt works or another type-cast.
Or simply implicit typecast. How would you resolve this question?

[send]

It is works very well with numbers such as 0xFFFF, so the issue is decimal digit chars conversion to decimal itself?