How to retrieve numeric value from character data type in c program

Question

I want to retrieve a numeric value from a character type variable.

char x = '10';
int y = 10;
printf("x + y = %d",(int)x+y);

I expect 20 as the result.

Answer 1

printf("x + y = %d",(int)x+y);

Casting char to int doesn't convert it to a number, but rather gives you the integer value corresponding to the character that it's representing (typically ASCII or something compatible, but the C standard doesn't specify that). Here you're using a multi character literal (normally anything enclosed in single quotes should be just one character), to make matters worse; this is implementation defined as well. If you want to store more than a single digit, you will need a string:

const char* x = "10".

You can then convert it to an integer using atoi or strtol. atoi is easier to use, but provides no way of checking whether or not the string contains a valid number.

Example:

const char* x = "10"; int y = 10;
printf("x + y = %d", atoi(x) + y);

Answer 2

You can do:

char x[] = "10"; int y = 10;
printf("x + y = %d",atoi(x)+y);

Answer 3

If you have a C library, use atoi(char *x), secondly characters cannot hold 2 characters, for that you need an array or pointer. Here's the code that you may need:

int main(){
        char *c = "10";
        int i = atoi(c);
        int j = 10;
        printf("i + j=%d",(i + j));
}

Answer 4

Use Below Code..........

#include<conio.h>
#include<stdio.h>
void main()
{
    char x=10;
    int y=10;
    clrscr();

    printf("%d",(x+y));
    getch();
}

Answer 5

Only for 0 to 9..

if (c >= '0' && c <= '9') {
    int v = c - '0';
    // safely use v
}