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

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?

Answer 1

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".

Answer 2

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

Answer 3

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