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".

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 

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