Reading character-by-character - always reaching an invalid character - c

Question

I'm new to C and to the programming world. I've been requested to get a reversed number in base 20 and print out his conversion to decimal.
I know it's not the most efficient code out there, but that's the best I managed to do with my current knowledge.
When I run this program and input a number, I'm always getting the default case...
What am I missing?
Thank you :)
#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
int main() {
char x;
int exponent= 1;
int sum = 0;
int flag = 0;
printf("Enter a reversed number is base 20\n");
scanf(" %c", &x);
while (x != "\n") {
switch (x) {
case '0':sum += 0;
break;
case '1':sum += 1 * exponent;
break;
case '2': sum += 2 * exponent;
break;
case '3':sum += 3 * exponent;
break;
case '4':sum += 4 * exponent;
break;
case '5':sum += 5 * exponent;
break;
case '6':sum += 6 * exponent;
break;
case '7':sum += 7 * exponent;
break;
case '8':sum += 8 * exponent;
break;
case '9':sum += 9 * exponent;
break;
case 'A':
case 'a': sum += 10 * exponent;
break;
case 'B':
case 'b':sum += 11 * exponent;
break;
case 'C':
case 'c' : sum += 12 * exponent;
break;
case 'D':
case 'd': sum += 13 * exponent;
break;
case 'E':
case 'e': sum += 14 * exponent;
break;
case 'F':
case 'f': sum += 15 * exponent;
break;
case 'G':
case 'g': sum += 16 * exponent;
break;
case 'H':
case 'h': sum += 17 * exponent;
break;
case 'I':
case 'i': sum += 18 * exponent;
break;
case 'J':
case 'j': sum += 19 * exponent;
break;
default:flag++;
break;
}
if (flag == 1) {
printf("Error! %c is not a valid digit in base 20", x);
break;
}
else {
exponent *= 20;;
scanf("%c", &x);
}
}
if (flag == 1)
return 0;
else
printf(sum);
return 0;
}

Answer 1

You should really read your compiler's warnings...:
<source>: In function 'main':
<source>:10:14: warning: comparison between pointer and integer
10 | while (x != "\n") {
| ^~
<source>:10:14: warning: comparison with string literal results in unspecified behavior [-Waddress]
<source>:77:16: warning: passing argument 1 of 'printf' makes pointer from integer without a cast [-Wint-conversion]
77 | printf(sum);
| ^~~
| |
| int
In file included from <source>:2:
/usr/include/stdio.h:332:43: note: expected 'const char * restrict' but argument is of type 'int'
332 | extern int printf (const char *__restrict __format, ...);
| ~~~~~~~~~~~~~~~~~~~~~~~^~~~~~~~
You're comparing x to a an address - the address of the string constant "\n" (that is located in some section of your program's memory space). You probably meant to compare x to '\n'.
Also, the printf() function takes a format string before the argument, so it should be printf("%d\n", sum); (to also print a newline character).
PS 1: Always compile your code with more warnings turned on, e.g. gcc -W -Wall (and perhaps other warning flags) for a better chance catch these kinds of typos and minor errors.
PS 2: As #Jabberwocky notes, your program could be made much shorter (with no loss of readability); and, on the other hand, you should always check the return value of scanf() and similar functions which may fail due to user input:
#include <stdio.h>
#include <stdlib.h>
int main() {
char x;
int exponent = 1;
int sum = 0;
int num_scanned;
printf("Enter a reversed number is base 20\n");
num_scanned = scanf(" %c", &x);
if (num_scanned != 1) {
fprintf(stderr, "Invalid input\n");
exit(EXIT_FAILURE);
}
while (x != '\n') {
int digit;
if (x >= '0' && x <= '9') { digit = x-'0'; }
else if (x >= 'a' && x <= 'j') { digit = x-'a'; }
else if (x >= 'A' && x <= 'J') { digit = x-'A'; }
else {
fprintf(stderr, "%c is not a valid digit in base 20.", x);
exit(EXIT_FAILURE);
}
sum += digit * exponent;
exponent *= 20;
num_scanned = scanf("%c", &x);
if (num_scanned != 1) {
fprintf(stderr, "Invalid input\n");
exit(EXIT_FAILURE);
}
}
printf("%d\n",sum);
return EXIT_SUCCESS;
}