C Programming: how to get normal value of int without using value - 48?

Question

So i have this loop the check user input:

int ch, num;

while ((ch = getchar()) != '\n')
{
    if (isdigit(ch))
    {
        num = ch - 48;   
    }
}

So according this table:

Character   Decimal Value
0           48
1           49
2           50
3           51
4           52
5           53
6           54
7           55
8           56
9           57

I am using this way to get my number: num = ch - 48; And if for example i want to multiple my number by 10 ?

Answer 1

how to get normal value of int without using value - 48?

Like this:

num = ch - '0';

Read more in How to convert char to integer in C?

---

In both cases, you then just do:

num *= 10;

since num is of type int.

---

Warning: Others suggest int atoi (const char * str); to get the number, which expects as an input a string, not a character! So if you pass a character to that function (thus a non-null-terminated string, you will invoke Undefined Behavior).

Answer 2

You should always use the idiomatic

num = ch - '0';

The reason being that this works for any encoding supported by C: C mandates that the digits 0 to 9 appear contiguously and in ascending order. If you hardcode the value of '0' to 48 say, you are not, strictly speaking, writing portable C.

Note that the expression ch - '0' is an int type in C due to the rules of argument promotion ('0' is an int type in C). You are therefore free to multiply this by 10 in the same way as you would apply arithmetic operations to any int type.

Answer 3

Here's the best way to read input from user till the new line. I'm posting one example you can go through it and implement your code as you need.

This example will read for user input in integer untill newLine(\n). Make sure you are not giving space at last.

Sample Code:

#include <stdio.h>
int main(void) {
    int i=0,size,arr[10000];
    char temp; 
    do{
        scanf("%d%c", &arr[i], &temp); 
        i++; 
        } while(temp!= '\n');

    size=i; 

    for(i=0;i<size;i++){ 
        printf("%d ",arr[i]); 
    } 
return 0;
}

Answer 4

You can use atoi() function here.You are getting ch as input from user. You can convert it into string as follows-

char number[2] ="";
number[0]=ch;
number[1]='\0'

num = atoi(number);

This will solve your problem.