How to convert hexadecimal to decimal in C? - c

Question

So, with my beginner level of experience in C I've been trying to write a code that converts hexadecimal input to a decimal with an array. I believe you will get it more spesificly by looking at my code but it does not work properly. I keep getting an amount that is much larger than intended.
#include <stdio.h>
int main()
{
int i, k, j, N, power, result;
char array[50];
result = 0;
printf("how many charecters are there in your hexadecimal input?: \n");
scanf("%d", &N);
for(k=0; k<=N-1; k++)
{
printf("What is the %d . value of your hexadecimal input?: \n", k+1);
scanf("%s", &array[k]);
}
for(i=0; i<=N-1; i++)
{
power = 1;
for(j=0; j<=i; j++)
{
power = power *16;
}
if((array[i] >= 0) && (array[i] <= 9))
{
result = result + array[i] * power;
}
else
{
result = result + (array[i] - 'A' + 10) * power;
}
}
printf("your result is %d", result);
return 0;
}

Answer 1

Your code is overly complicated and wrong, but the idea is correct.
You want this:
#include <stdio.h>
#include <string.h>
int main()
{
char array[50];
scanf("%49s", array); // just use a single scanf with %s
// "49s" will avoid buffer overflow of array
int length = strlen(array); // get length of string (and use meaningful
// variable name length instead of N)
int result = 0;
for (int i = 0; i < length; i++)
{
int nibble; // nibble is the hexadécimal "digit" from 0 to 15
if ((array[i] >= '0') && (array[i] <= '9'))
{
nibble = array[i] - '0';
}
else
{
nibble = array[i] - 'A' + 10;
}
result *= 16; // no need to maintain power, just multiply by 16
// on each run
result += nibble; // ... and add the nibble
}
printf("your result is %d", result);
return 0;
}
There is still room for improvement:
you should accept lowercase
there is no test for invalid characters such as 'G', 'H' etc.
you should put the conversion code in a function such asint HexStringToDecimal(const char *hexstring)
and certainly a bunch of other things I forgot.
Exercise for you: convert that code so it converts a decimal string instead of a hexadecimal string. Hint: the code will be simpler.
Side note:
Avoid constructs like this:
for(k=0; k<=N-1; k++)
instead use this exact equivalent which is more readable:
for(k=0; k<N; k++)

Answer 2

just edited your code a little bit,
and also the input is supposed to be from left to right
instead of right to left according to which I think you have coded.
Hence, for C6, first C then 6.
#include <stdio.h>
int main()
{
int i, k, j, N, power, result;
char array[50];
result = 0;
printf("how many charecters are there in your hexadecimal input?: \n");
scanf("%d", &N);
for(k=0; k<=N-1; k++)
{
printf("What is the %d . value of your hexadecimal input?: \n", k+1);
scanf(" %c", &array[k]);
}
for(i=0; i<=N-1; i++)
{
power = 1;
for(j=0; j<=N-2-i; j++)
{
power = power*16;
}
if((array[i] >= '0') && (array[i] <= '9'))
{
result = result + (array[i] - '0') * power;
}
else
{
result = result + (array[i] - 'A' + 10) * power;
}
}
printf("your result is %d", result);
return 0;
}
the takeaways
your input taking is inefficeient. and in that too %c instead of %s in scanf.
if((array[i] >= 0) && (array[i] <= 9)), suppose i=1 and array[1] = 7. Now, you are comparing ascii value of 7 which is 55, instead of 7.
same thing in the statement of if, the ascii values of array[i] are being used so I have subtracted accordingly.