C, convert hex number to decimal number without functions - linux

Question

i'm trying to convert hexadecimal number to decimal number. What i've come up so far is:
#include <unistd.h>
#include <stdio.h>
long convert(char *input, short int *status){
int length = 0;
while(input[length])
{
length++;
}
if(length = 0)
{
*status = 0;
return 0;
}
else
{
int index;
int converter;
int result = 0;
int lastNumber = length-1;
int currentNumber;
for(index = 0; index < length; index++){
if(index == 0)
{
converter = 1;
}
else if(index == 1)
{
converter = 16;
}
else{
converter *= 16;
}
if(input[lastNumber] < 45 || input[lastNumber] > 57)
{
*status = 0;
return 0;
}
else if(input[lastNumber] > 45 && input[lastNumber] < 48)
{
*status = 0;
return 0;
}
else{
if(input[lastNumber] == 45)
{
*status = -1;
return result *= -1;
}
currentNumber = input[lastNumber] - 48;
result += currentNumber * converter;
lastNumber--;
}
}
*status = -1;
return result;
}
}
int main(int argc, char **argv)
{
char *input=0;
short int status=0;
long rezult=0;
if(argc!=2)
{
status=0;
}
else
{
input=argv[1];
rezult=convert(input,&status);
}
printf("result: %ld\n", rezult);
printf("status: %d\n", status);
return 0;
}
Somehow i always get resoult 0. Ia am also not allowed to use any other outher functions (except printf). What could be wrong with my code above?

Answer 1

This:
if(dolzina = 0)
{
*status = 0;
return 0;
}
is not merely testing dolzina, it's first setting it to 0. This causes the else clause to run, but with dolzina equal to 0 which is not the expected outcome.
You should just use == to compare, of course.