converting char to int

[Unregistered]

Hi everyone! I'm new to C, and I need to convert to a char into an integer. The char is a part of an array, which seems to be the problem. Apparently atoi() has to accept an array of chars, but that won't do for my project. I thought of using a switch statement, but that seems kind of "unprofessional". Anybody have any ideas?
Thank you again!

[PutoAmo]

I am not sure whether this is what you need ...

#include <stdio.h>

int main (void)
{
 char *str = "123456789";
 int num;

 while (*str)
 {
  sscanf (str, "%1d", &num);

  printf ("%d\n", num);

  str++;
 }

 return 0;
}

[Prelude]

>I'm new to C, and I need to convert to a char into an integer
This may or may not be what you need

char a = '5';
int d = a - '0';
/* d now contains the decimal equivalent of 5 */

-Prelude

[Barjor]

Just do be really picky and difficult

char a = '5';
int d = a - '0';
/* d now contains the decimal equivalent of 5 */

Will only work on the ascii char set. The only reason I thought about that is because I read about it in K&R last night. I don't know about anything not using the ascii char set.

[Prelude]

>Just do be really picky and difficult
I like it that way

>Will only work on the ascii char set.
Please excuse the following pedantry.

/* K&R pp.43 */
As we discussed in Chapter 1, the expression

s[i] - '0'

gives the numeric equivalent of the character stored in s[i], because the values of '0', '1', etc., form a contiguous sequence.

Another example of char to int conversion is the function lower, which maps a single character to lower case for the ASCII character set.

<snip code>

This works for ASCII because the corresponding upper and lower case letters are a fixed distance apart as numeric values and each alphabet is contiguous -- there is nothing but letters between A and Z. This latter observation is not true of the EBCDIC character set, however, so this code would convert more than just letters in EBCDIC.

/* K&R pp.22-23 */
This particular program relies on the properties of the character representation of the digits. For example, the test

if (c >= '0' && c <= '9') ...

determines whether the character in c is a digit. If it is, the numeric value of that digit is

c - '0'

This works only if '0', '1', ..., '9' have consecutive increasing values. Fortunately, this is true for all character sets.

So your statement that the code will only work on ASCII is not explicitly stated by K&R for a number conversion from char to int, in fact it could be interpreted that they even state that it will work on all character sets. I've personally never tried it on a machine that uses EBCDIC, but I have no reason to believe that it will not work since EBCDIC defines 0 through 9 as

hex  character
---  ---------
F0     0
F1     1
F2     2
F3     3
F4     4
F5     5
F6     6
F7     7
F8     8
F9     9

-Prelude

[Barjor]

MM Actually your right. I was typing faster then my brain was thinking(imagine that*smile*) Do you know if EBCDIC is really used? Some of my programs are using conversations that are strictly based on the ascii chart but if this is bad practice I would like to know. I know that all the standard functions are supposed to be "safe" on that issue.

[Prelude]

>Do you know if EBCDIC is really used?
Yes it is, mostly in IBM mainframes, so it's usually safe to assume ASCII if the program is going to remain on microcomputers. However, it's best to write portable code whenever you can on the off chance that some goofball decided to use a weird character set.

-Prelude

[Unregistered]

thank you all for helping me. I have a question though, about putoAmo's code: what is 'str' refer to? Is it a pointer to a string literal? Do you have to allocate memory for it? Thank you again.

[C_Coder]

what is 'str' refer to? Is it a pointer to a string literal?

Yup, the string actually looks like 123456789\0

Do you have to allocate memory for it

Nope, memory is allocated when it is declared like any other variable

[quzah]

Originally posted by Unregistered
thank you all for helping me. I have a question though, about putoAmo's code: what is 'str' refer to? Is it a pointer to a string literal? Do you have to allocate memory for it? Thank you again.

Yep, it's a pointer to a string literal. When using string literals, you never have to allocate space for them. The compiler does it.

Quzah.

[Unregistered]

hmmmm...
So how would I access the string pointed to by str, say if I wanted to print it? I wrote this but it just prints null:

  const char s1[]="hello,my name is sam.";
  const char delimiters[]="., ";
  char *token;
  char *cp;
  strcpy(s1,cp);
  token=strtok(cp,delimiters);
  printf("%s",token);

I tried using &token and *token, but it gave me runtime error.

[quzah]

Originally posted by Unregistered
hmmmm...
So how would I access the string pointed to by str, say if I wanted to print it? I wrote this but it just prints null:

  const char s1[]="hello,my name is sam.";
  const char delimiters[]="., ";
  char *token;
  char *cp;
  strcpy(s1,cp);
  token=strtok(cp,delimiters);
  printf("%s",token);

I tried using &token and *token, but it gave me runtime error.

Of course it gives you errors. You're using strcpy completely wrong. Additionally, you just have pointers that have no space allocated for them, so even if you were trying to copy something some place, 'cp' has not allocated space.

Quzah.

[lliero]

ammm


why not try

int atoi(const char *str)


e.g

#include<stdio.h>
#include<stdlib.h>

main(){
char num1[80],num2[80];

printf("enter first number");
gets(num1);
printf("enter second");
gets (num2);
printf("the sum is %d", atoi(num1) + atoi(num2));
return 0;
}


i hope this will work....