Hexadecimal

[alabalap]

I need some help with this program. If you can say me what exactly means those notation: 'a', 'f', and what happens in every major sequence of every function. This program convert a string to int and hexadecimal numbers to int.
Thanks.

  #include<iostream>
#include<cstring>
using namespace std;

int str2int(const char* c)
{
	int x=strlen(c),nr=0,pow=1;
	for (int i=x-1; i>=0; i--,pow*=10)
		nr=nr+(c[i]-'0')*pow;
	return nr;
}
int str2int2(const char* c)
{
	int x=strlen(c),nr=0,pow=1;
	for (int i=x-1; i>=0; i--,pow*=16)
		if (c[i]>='a' && c[i]<='f')
			nr=nr+(c[i]-'a'+10)*pow;
		else
			if (c[i]>='A' && c[i]<='F')
				nr=nr+(c[i]-'A'+10)*pow;
			else
				nr=nr+(c[i]-'0')*pow;
	return nr;
}
int main()
{
	char c[]="1234";
	char c2[]="ffff";
	cout<<str2int(c)<<endl;
	cout<<str2int2(c2);
	return 0;
}

[Uk Marine]

Well as i recall from my IT class, 'A' is base 10 and 'F' is base 15, it goes like this
1
2
3
4
5
6
7
8
9
A : 10
B : 11
C : 12
D : 13
E : 14
F : 15

[alabalap]

yes.this I know
but what exactly means this sequence for example: c[i]>='a' && c[i]<='f'
with what is compared c[i]?

[Chervil]

This c[i]>='a' && c[i]<='f'
is a combination of two conditions: c[i] >= 'a' and c[i] <= 'f'
The && (logical and) operator means that the complete condition is true only if both of the conditions are true, that is, it is testing that c[i] is somewhere in the range 'a' to 'f' inclusive.

[alabalap]

What means " '0' "?

[Chervil]

'A' is the letter A, similarly '0' is the character representing the digit 0, as distinct from the number zero.

[alabalap]

What values have 'a', '0' and 'f'. this is what i don't understand

[Chervil]

http://www.asciitable.com/
'a' is decimal 97
'A' is decimal 65
'0' is decimal 48
etc.

[alabalap]

ok.I understood.
now I want to know what value will return this nr=nr+(c[i]-'0')*pow; what does exactly this sequence?

[Chervil]

Look at each term in this expression: nr + (c[i]-'0') * pow
the part in parentheses is evaluated first, then the multiplication by pow and finally the addition. Lastly, the end result is assigned to nr.

[alabalap]

and, for example, for nr=4, pow=10 and c[i]=3 what value has this?

[Chervil]

Surely you can work this out for yourself? Or at least have a go. If all else fails, you could add a cout << statement to display the result.

[alabalap]

the last question
if 'a'=10 in hexadecimal why is necessary to write nr=nr+(c[i]-'a'+10)*pow;
and not more simply nr=nr+c[i]*pow?

[giblit]

Because 'a' == 97.
97 - 97 == 0
0 + 10 = 10

say you have b
'b' == 98
98 - 98 = 1
1 + 10 = 11

ect...from a - f

[alabalap]

i don't understand.98-98=1?

[Chervil]

if 'a'=10 in hexadecimal
But it doesn't. That's an erroneous starting point.
'a' = 97 in decimal, or 61 in hexadecimal (6* 16 + 1*1 = 97).

The character 'a' or 'A' is used to represent the hexadecimal digit which has a value of 10(decimal).

[alabalap]

thanks.now i understand all
thank you very much

[giblit]

I meant 98 - 97 == 1 it was a typo.