How to scan only one whole and repeat reading if the user enters non-digital characters?

Question

Here is a minor tyro issue with C code trying to simply prevent the user from entering a character or integer less than 0 or more than 23.

#include <stdio.h> #include <stdlib.h> int main(void) { const char *input; char *iPtr; int count = 0; int rows; printf("Enter an integer: "); scanf("%s", input); rows = strtol(input, &iPtr, 0); while( *iPtr != '\0') // Check if any character has been inserted { printf("Enter an integer between 1 and 23: "); scanf("%s", input); } while(0 < rows && rows < 24) // check if the user input is within the boundaries { printf("Select an integer from 1 to 23: "); scanf("%s", input); } while (count != rows) { /* Do some stuff */ } return 0; } 

I did it halfway, and a slight boost will be appreciated.

Answer 1

Use scanf("%d",&rows) instead of scanf("%s",input)

This allows you to get the direct integer value from stdin without the need for conversion to int.

If the user enters a string containing non-digital characters, you must clear your stdin before the next scanf("%d",&rows) .

your code might look like this:

 #include <stdio.h> #include <stdlib.h> int clean_stdin() { while (getchar()!='\n'); return 1; } int main(void) { int rows =0; char c; do { printf("\nEnter an integer from 1 to 23: "); } while (((scanf("%d%c", &rows, &c)!=2 || c!='\n') && clean_stdin()) || rows<1 || rows>23); return 0; } 

Explanation

1)

 scanf("%d%c", &rows, &c) 

This means that an integer is entered from the user and a non-digital character is next to him.

Example1: If the user enters aaddk and then ENTER , scanf will return 0. Nothing is closed

Example2: If the user enters 45 and then ENTER , scanf will return 2 (2 items are closed). Here %d is capting 45 and %c is capting \n

Example 3: If the user enters 45aaadd and then ENTER , scanf will return 2 (2 items are closed). Here %d is capting 45 and %c is capting a

2)

 (scanf("%d%c", &rows, &c)!=2 || c!='\n') 

In Example 1: this is a TRUE condition because scanf return 0 ( !=2 )

In Example 2: this is a FALSE condition because scanf return 2 and c == '\n'

In Example 3: this is a TRUE condition because scanf return 2 and c == 'a' (!='\n')

3)

 ((scanf("%d%c", &rows, &c)!=2 || c!='\n') && clean_stdin()) 

clean_stdin() always TRUE because the function always returns 1

In example 1: (scanf("%d%c", &rows, &c)!=2 || c!='\n') - TRUE , so the condition after && must be checked to execute clean_stdin() , and that's it condition TRUE

In Example 2: (scanf("%d%c", &rows, &c)!=2 || c!='\n') is FALSE , so the condition after && will not be checked (since its result will be equal to the whole FALSE ), therefore clean_stdin() will not be executed, and the whole condition is FALSE

In example 3: (scanf("%d%c", &rows, &c)!=2 || c!='\n') - TRUE , so the condition after && must be checked to execute clean_stdin() and the whole condition TRUE

So, you may notice that clean_stdin() will only execute if the user enters a string containing a non-digital character.

And this condition ((scanf("%d%c", &rows, &c)!=2 || c!='\n') && clean_stdin()) will return FALSE only if the user enters integer and nothing else

And if the condition ((scanf("%d%c", &rows, &c)!=2 || c!='\n') && clean_stdin()) is FALSE and integer is between and 1 and 23 , then the while will break otherwise, while will continue

Answer 2

 #include <stdio.h> main() { char str[100]; int num; while(1) { printf("Enter a number: "); scanf("%[^0-9]%d",str,&num); printf("You entered the number %d\n",num); } return 0; } 

%[^0-9] in scanf() copies everything that is not between 0 and 9 . Essentially, it clears the input stream without numbers and puts it in str . Well, the length of the non-digital sequence is limited to 100. The next %d selects only integers in the input stream and puts it in num .

Answer 3

You can create a function that reads an integer from 1 to 23 or returns 0 if not-int

eg.

 int getInt() { int n = 0; char buffer[128]; fgets(buffer,sizeof(buffer),stdin); n = atoi(buffer); return ( n > 23 || n < 1 ) ? 0 : n; } 

Answer 4

 char check1[10], check2[10]; int foo; do{ printf(">> "); scanf(" %s", check1); foo = strtol(check1, NULL, 10); // convert the string to decimal number sprintf(check2, "%d", foo); // re-convert "foo" to string for comparison } while (!(strcmp(check1, check2) == 0 && 0 < foo && foo < 24)); // repeat if the input is not number 

If the input is a number, you can use foo as your input.

Answer 5

You will need to repeat your strtol call inside your loops, where you ask the user to try again. In fact, if you do a loop a do { ... } while(...); and not time, you will not get the same duplication twice.

You should also format your code so that you can see where the code is inside the loop, not.

Answer 6

MOHAMED, your answer is great and it really helped me. Here I have posted the code, which I think is a bit simpler:

 #include <stdio.h> int getPositive(void); void clean_input(void); int main(void) { printf("%d\n",getPositive()); return 0; } int getPositive(void) { int number; char buffer; // Holds last character from user input. // (ie '\n' or any other character besides numbers) int flag; // Holds scanf return value do{ flag = scanf("%d%c", &number, &buffer); // Gets input from user // While scanf did not read 2 objects (ie 1 int & 1 char) // or the user inputed a number and then a character (eg. 12te) // ask user to type a valid value while (flag !=2 || buffer!='\n') { clean_input(); printf("%s","You have typed non numeric characters.\n" "Please type an integer\n?"); flag = scanf("%d%c", &number, &buffer); } if(number<0) { printf("%s","You have typed a non positive integer\n" "Please type a positive integer\n?"); } else { // If user typed a non negative value, exit do-while. break; } }while(1); } void clean_input(void) { while (getchar()!='\n'); return; } 

In my case, I want the number to be just positive. If you want your number to be between 1 and 23, you replace number<0 with number<1 || number>23 number<1 || number>23 in the if statement . You will also need to modify printf to print the corresponding message.

Answer 7

The above script created by Yonggoo Noh worked well for me.
I changed it to include the value 0 to return to main () at any time, but I have a problem :
- Sometimes, after returning to main (), when I enter the script again, it asks me for the value ("check1") 2 or more times.
Why is this going to happen?

 char check1[10], check2[10]; int foo; do{ printf(">> "); scanf(" %s", check1); foo = strtol(check1, NULL, 10); // convert the string to decimal number sprintf(check2, "%d", foo); // re-convert "foo" to string for comparison } while (!(strcmp(check1, check2) == 0 && ((0 < foo && foo < 24) || foo==0))); // repeat if the input is not number if (n==0){ main(); }