How to convert each character in a character array to integer?

Question

I have tried the standard methods but still I get error in my answer.

My code:

int main() {   
   int val;
   char str[] = {'1', '45', '0'};

   val = str[1] - '0';

   printf("Int value = %d\n", val);

   return(0);
}

I am getting answer as 5 instead of 45.

How do I solve this issue?

[update from comment:]

I actually need to process an array of strings..suppose I want to convert octal numbers to decimal, and my input has to be in the form of an array of strings. I wish to convert these no.s to decimal : {45,17,100} For that I would, at first be requiring to extract each element and change it to integer. Could you plz suggest what would be the best way to do it?

`char` **is** and integer type. What is your problem, what do you want to accomplish? — too honest for this site, Jul 08 '16 at 06:31
`'45'` is a multi-character constant. See [this answer to another SO question](https://stackoverflow.com/a/3684046/434551) for further info. — R Sahu, Jul 08 '16 at 06:31
Have a look at what the compiler tells you, when compiling the code shown. — alk, Jul 08 '16 at 07:12

alk · Answer 1 · 2016-07-08T07:53:20.490

I actually need to process an array of strings

What you have defined here

  char str[] = {'1', '45', '0'};

is not an array of strings, but exactly one array of char with 3 elements. It is not even a C-"string", as this would require a trailing value of 0. Note that the value of the character '0' isn't 0, but, for example, for the ASCII character set it's 48.

'45' is a multi-byte character literal, which in fact is an integer. The code above tries to initialise the 2nd element of the char-array str (str[1], which is a char) using this very literal.

This does not work.

A char cannot (necessarily) hold the int value of this multi-byte character literal. In this case the initialisation of str[1] overflows, resulting in the unexpected value of 5.

To see the issue try the following code:

#include <limits.h>
#include <stdio.h>


int main(void)
{
  char c_min = CHAR_MIN;
  char c_max = CHAR_MAX;
  unsigned char uc = '45';

  printf("'1'=%d\n", '1');
  printf("'45'=%d\n", '45');
  printf("'0'=%d\n", '0');

  printf("lowest possible value for char=%d\n", c_min);
  printf("highest possible value for char=%d\n", c_max);

  printf("'45' converted to an (unsigned) char=%u\n", uc);

  return 0;
}

The example above shows how the value of 45 gets truncated when being assigned to char.

Depending on the C implementation you use the conversion of '45' to a char might even invoke the infamous Undefined Behaviour. This is not good.

What you seem to be wanting is:

  #define ARRAY_SIZE_MAX 3

  char * str[ARRAY_SIZE_MAX] = {"1", "45", "0"}; /*. Mind the double quotes. */

This defines an array of 3 pointers to char , with each pointing to a C-"string".

score 1 · Answer 2 · edited Jul 08 '16 at 08:34

1

Here, you are getting 5 instead of 45 because st[1] = '5', this is because we have only ASCII value of 0 to 9 integers and 45 have no ASCII value. To store 45 in your string you have to declare multidimensional string. for example:

char st[3][3]={'1', '45', '0'};

Here is the working code:

#include<stdio.h>
int main(){
    int i,ans,j;
    char st[3][3]={{'1'}, {'4','5'},{'0'}};
    for(j=0;j<3;j++){
        for(i=0;st[j][i]>=48 && st[j][i]<=57;i++){
            ans=st[j][i]-'0';
            printf("%d",ans);
        }
        printf("\n");
     }
     return 0;
}

Output is

1
45  
0

edited Jul 08 '16 at 08:34

Draken

3,134
13
34
54

answered Jul 08 '16 at 08:29

Naman Gupta

39
2
12

To make your code fully portable, that is independent from the ASCII character set, you want to change the two Magic Numbers `48` to `'0'` and `57` to `'9'`. – alk Jul 08 '16 at 10:51

score -1 · Answer 3 · edited May 23 '17 at 11:44

It depends, what do you want.

Every character has it's own int value- It's simply ASCII code

If you will iterate over integers and make it print like %c you will get ASCII table (you can see it eg. there.

But if you want to read int values from string / char datatype you will have to parse it- atoi (ascii to integer) function- example there

Btw I dont know how exactly your example works but the problem is you are doing following: int val = '45' - '0'; int value of '0' should be 48, '45' rly I dont know, but '4' is 52int and '5' is 53 int, so something like that..

As i wrote you should to do something like int val = atoi('45') - atoi('0') just for sure maybe better to cast into int as follows int val = (int) (atoi('45') - atoi('0')) - exactly asi in your example

int main() {   
   int val;
   char str[] = {'1', '45', '0'};

   val = (int) (atoi(str[1]) - atoi('0'));

   printf("Int value = %d\n", val);

   return(0);
}

Not sure, but i think that should works, hope that will help

`atoi()` takes either a `char`-array or a `char`-pointer, but does not take a `char`, nor an `int`. `str[1]` is a `char` and `'0'` is `char` literal, which is of type `int`. — alk, Jul 08 '16 at 07:04

How to convert each character in a character array to integer?

3 Answers3

Output is