Logic in detecting int in C - c

Question

I would appreciate some help with this. I'm trying to create this simple program that repeatedly loops asking for the user to enter in an int. If he enters an int, it exits but if he enters something else or bigger than int (ex.4gs4r33) it will loop again asking to enter an int. This is what I have tried, and it's not working. It says it's an int even if it's not.
#include<stdio.h>
unsigned int findInt();
int main() {
printf("Please input an int.\n");
findInt();
}
unsigned int findInt() {
char input;
long num = 0;
int b = 0;
do {
scanf("%c", &input);
if (isdigit(input)){
num = num*10+input+'0';
b = 1;
}
else if (input == '\n')
b = 1;
else
b = 0;
} while(input != '\n');
if (b == 1)
printf("Great!\n");
else{
printf("Not an int \n");
findInt();
}
return 0;
}

Answer 1

Two possible approaches. One would be to modify your code:
b = 1; // start off with good intentions…
do {
scanf("%c", &input);
if (isdigit(input)){
num = num*10+input -'0'; // *** SUBTRACT '0', don't add it!
}
else if (input != '\n') {
b = 0;
break; // *** break when you find non-digit
}
} while (input != '\n');
Two changes: getting the math right as you compute the integer, and fixing the logic (so you break out of your loop when you find a non digit character)
Second approach:
char buf[100];
char intAsString[100];
fgets(buf, 100, stdin);
sscanf(buf, "%d", num);
sprintf(intAsString, "%d\n", num);;
if(strcmp(buf, intAsString) == 0 ) {
printf("yay - you entered an integer!\n");
}
I'm sure you can figure out how that works.
update a complete code snippet that solves the issue of "loop logic" as well: you call the findInt function once from the top level, and it keeps going until you get the int. Note - in order for this to work properly, I read the entire input at once (rather than one at a time), then pick off the characters one by one using sscanf (and updating the pointer manually). It has a number of advantages - not least of which is that you start with a fresh input every time you call findInt, instead of having the rest of the input buffer that still needs reading (and which was giving rise to "no,no,no,great!" - as you would keep reading the bad input until you got to the newline, and accept that...)
#include<stdio.h>
#include <ctype.h>
unsigned int findInt();
int main() {
findInt();
}
unsigned int findInt() {
char input;
char buf[100];
char *temp;
long num = 0;
int b = 0;
printf("please enter an int:\n");
fgets(buf, 100, stdin);
temp = buf;
do {
sscanf(temp++, "%c", &input);
if (isdigit(input)){
num = num*10+input-'0';
b = 1;
}
else if (input == '\n')
{
b = 1;
break;
}
else {
b = 0;
break;
}
} while(input != '\n');
if (b == 1)
printf("Great! %d is an integer!\n", num);
else{
printf("Not an int \n");
findInt();
}
return 0;
}

Answer 2

In the else branch - i.e. not a digit or a newline - you set b to 0. Now if a digit DOES follow you reset that to 1.
You'll probably want to break or somehow record the permanent failure instead of just continuing.

Answer 3

#include <stdlib.h>
#include <limits.h>
#include <stdbool.h>
void findInt(){
int x;
bool ok;
do{
char buff[32], *endp;
long long num;
ok = true;//start true(OK)
printf("Enter a number: ");
fgets(buff, sizeof(buff), stdin);
//strtoll : C99
x=(int)(num=strtoll(buff, &endp, 0));//0: number literal of C. 10 : decimal number.
if(*endp != '\n'){
if(*endp == '\0'){
printf("Too large!\n");//buffer over
while('\n'!=getchar());
} else {
printf("Character that can't be interpreted as a number has been entered.\n");
printf("%s", buff);
printf("%*s^\n", (int)(endp - buff), "");
}
ok = false;
} else if(num > INT_MAX){
printf("Too large!\n");
ok = false;
} else if(num < INT_MIN){
printf("Too small!\n");
ok = false;
}
}while(!ok);
}
,