char to int

Question

can anyone share your knowledge to me for this,
i'm declaring a decimal numbers by using the "char" data type. and also the output must also be the same. instead of characters, it will still return decimal values.
ex:
Input: 1234 (but the data type used was char)
Output: 1234

Answer 1

Is char type capable for accepting 1234 as a value? isn't it limited to 255? it's only one byte (8 bits) data, are you sure, or am I misunderstanding the question?

Answer 2

@Kirk Schafer, thanks, that's an interesting point, certainly a char combination will accept the value because together they form 16 bit data which can take 1234 value. I didn't know the formula for combining byte value as you explained, it was great!.

Another interesting point, so I tested the fact you pointed out, and write a text file with 'aa', and view the content in hex editor I see 61 61, which, combined as 16 bit value makes up 0x6161, the hexadecimal representation of 24,929 just like your example.

I really appreciate your enlightenment, thank you once again :)

Answer 3

@~ swim ~, that's exactly true, however the question did not tag Java, it tagged C instead, and in C/C++ a char is one byte, I was questioning the fact of how a single byte (char) can hold a value as large as 1234.

In Java a char is defined as two bytes because Java is Unicode compliant, in Unicode standard a single character indeed consists of two bytes so that non latin characters can be handled correctly.

Thank you for the explanation mate, there's always something new to learn everyday, I really appreciate it :)

Answer 4

@~ swim ~,
I'll keep that wchar_t in mind, thanks for the advice mate ;-)

Answer 5

@~ swim ~,
Thanks for more info on the Unicode string usage, actually I haven't got that far, but I'm gonna be keeping those in my library for later use.

Big thanks mate :)

Answer 6

@~ swim ~,
Now that's overwhelming, okay, lemme make a note on that...

Thanks for #brainstorm :-D

Answer 7

@lpang Multibyte characters would get this done, but not one.

cout << int('a')<< endl;
97
cout << 'aa'<< endl;
24929
cout << 97*256 + 97 << endl;
24929

You can infer from this that the limit is 256 values, since the next byte up is * 256 to record it properly.

Answer 8

(quietly)
Multibyte character sets (MBCS, UTF-8, 16, 32 ...) can be longer than 2 bytes per represented char.

gcc on SoloLearn (64-bit server I think) is configured in 32-bit mode.

cout << 'abcd'; // is valid here, 32 bits

Each 'slot' is a power of 256, in the same fashion as all bases:
a*256³ + b*256² + c*256¹ + d (*256° or 1)

[sharp turn ahead]

You can use the same formula for IP addresses:
127.0.0.1 = 127*256³ + 0 + 0 + 1 = 2130706433

$ ping 2130706433
PING 2130706433 (127.0.0.1) 56(84) bytes of data.
64 bytes from 127.0.0.1: icmp_seq=1 ttl=64 time=0.194 ms
64 bytes from 127.0.0.1: icmp_seq=2 ttl=64 time=0.227 ms

Answer 9

Oh damn, I meant 0x30. You are right, I'll edit it!
That's what you get for not double checking :P

Answer 10

thank you for your shared ideas!

Answer 11

You can convert a single digit like this:

char one_char = '1';
int one_int = one_char - 0x30;

However it is better to use C's atoi function that converts char*'s to ints.

#include <stdlib.h>
char* string = "1234";
int number = atoi(string);

Answer 12

@Shindlabua, isn't it rather '1' - 48 as 48 == '0' ?

Answer 13

How about converting char to int?

Answer 14

char c = '5', * s = "1234";

Getting the ascii value of a char :
(int)c;

Getting the value of the figure :
(int)(c - '0');

Translating a string into an integer :
atoi(s);

Answer 15

Hahaha no problem, it can happen to anyone ;)