scanf c wrong input - c

Question

Good afternoon, my question is conceptual. How can I make it generate a "fancy" error when the user incorrectly enters some data that does not correspond to the scanf() function? So as to only allow integers to be entered in the example below (not characters or array of characters or an inappropriate data).
For example:
#include <stdio.h>
int a;
printf("Enter a number\n");
scanf("%d", &a); //the user is supposed to enter a number
printf("Your number is %d ", a);
//but if the user enters something inappropriate, like a character, the program leads to
//undetermined behavior (which as I understand it interprets said character according to its
//value in the ASCII code).
From already thank you very much

Answer 1

In order to determine whether scanf was able to successfully convert the input to an integer, you should check the return value of scanf:
#include <stdio.h>
#include <stdlib.h>
int main( void )
{
int num;
printf( "Enter a number: " );
if ( scanf( "%d", &num ) != 1 )
{
printf( "Failed to convert input!\n" );
exit( EXIT_FAILURE );
}
printf( "Conversion successful! The number is %d.\n", num );
}
However, using scanf for line-based user input is generally not recommended, because scanf does not behave in an intuitive manner when dealing with that kind of input. For example, scanf will generally not consume an entire line of input at once. Instead, it will generally only consume the input that matches the argument, but will leave the rest of the line on the input stream, including the newline character.
Leaving the newline character on the input stream can already cause a lot of trouble. For example, see this question.
Also, if the user enters for example 6abc, then scanf will successfully match the 6 and report success, but leave abc on the input stream, so that the next call to scanf will probably immediately fail.
For this reason, it is generally better to always read one line of input at a time, using the function fgets. After successfully reading one line of input as a string, you can use the function strtol to attempt to convert the string to an integer:
#include <stdio.h>
#include <stdlib.h>
int main( void )
{
char line[200], *p;
int num;
//prompt user for input
printf( "Enter a number: " );
//attempt to read one line of input
if ( fgets( line, sizeof line, stdin ) == NULL )
{
printf( "Input failure!\n" );
exit( EXIT_FAILURE );
}
//attempt to convert string to integer
num = strtol( line, &p, 10 );
if ( p == line )
{
printf( "Unable to convert to integer!\n" );
exit( EXIT_FAILURE );
}
//print result
printf( "Conversion successful! The number is %d.\n", num );
}
However, this code has the following issues:
It does not check whether the input line was too long to fit into the buffer.
It does not check whether the converted number is representable as an int, for example whether the number is too large to be stored in an int.
It will accept 6abc as valid input for the number 6. This is not as bad as scanf, because scanf will leave abc on the input stream, whereas fgets will not. However, it would probably still be better to reject the input instead of accepting it.
All of these issues can be solved by doing the following:
Issue #1 can be solved by checking
whether the input buffer contains a newline character, or
whether end-of-file has been reached, which can be treated as equivalent to a newline character, because it also indicates the end of the line.
Issue #2 can be solved by checking whether the function strtol set errno to the value of the macro constant ERANGE, to determine whether the converted value is representable as a long. In order to determine whether this value is also representable as an int, the value returned by strtol should be compared against INT_MIN and INT_MAX.
Issue #3 can be solved by checking all remaining characters on the line. Since strtol accepts leading whitespace characters, it would probably also be appropriate to accept trailing whitespace characters. However, if the input contains any other trailing characters, the input should probably be rejected.
Here is an improved version of the code, which solves all of the issues mentioned above and also puts everything into a function named get_int_from_user. This function will automatically reprompt the user for input, until the input is valid.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
#include <limits.h>
#include <errno.h>
int get_int_from_user( const char *prompt )
{
//loop forever until user enters a valid number
for (;;)
{
char buffer[1024], *p;
long l;
//prompt user for input
fputs( prompt, stdout );
//get one line of input from input stream
if ( fgets( buffer, sizeof buffer, stdin ) == NULL )
{
fprintf( stderr, "Unrecoverable input error!\n" );
exit( EXIT_FAILURE );
}
//make sure that entire line was read in (i.e. that
//the buffer was not too small)
if ( strchr( buffer, '\n' ) == NULL && !feof( stdin ) )
{
int c;
printf( "Line input was too long!\n" );
//discard remainder of line
do
{
c = getchar();
if ( c == EOF )
{
fprintf( stderr, "Unrecoverable error reading from input!\n" );
exit( EXIT_FAILURE );
}
} while ( c != '\n' );
continue;
}
//attempt to convert string to number
errno = 0;
l = strtol( buffer, &p, 10 );
if ( p == buffer )
{
printf( "Error converting string to number!\n" );
continue;
}
//make sure that number is representable as an "int"
if ( errno == ERANGE || l < INT_MIN || l > INT_MAX )
{
printf( "Number out of range error!\n" );
continue;
}
//make sure that remainder of line contains only whitespace,
//so that input such as "6abc" gets rejected
for ( ; *p != '\0'; p++ )
{
if ( !isspace( (unsigned char)*p ) )
{
printf( "Unexpected input encountered!\n" );
//cannot use `continue` here, because that would go to
//the next iteration of the innermost loop, but we
//want to go to the next iteration of the outer loop
goto continue_outer_loop;
}
}
return l;
continue_outer_loop:
continue;
}
}
int main( void )
{
int number;
number = get_int_from_user( "Enter a number: " );
printf( "Input was valid.\n" );
printf( "The number is: %d\n", number );
return 0;
}
This program has the following behavior:
Enter a number: abc
Error converting string to number!
Enter a number: 6000000000
Number out of range error!
Enter a number: 6 7 8
Unexpected input encountered!
Enter a number: 6abc
Unexpected input encountered!
Enter a number: 6
Input was valid.
The number is: 6

Answer 2

How to get verified user input of a specific type
#1 Get user input as a string
char s[100];
if (!fgets( s, sizeof(s), stdin )) *s = '\0';
char * p = strptok( s, "\r\n" );
if (!p) complain_and_quit();
*p = '\0';
...
Alternately:
#define __STDC_WANT_LIB_EXT2__ 1
#include <stdio.h>
 
char * s = NULL;
size_t n = 0;
if (getline( &s, &n, stdin ) < 0)
{
free( s );
complain_and_quit();
}
...
free( s );
#2 Get rid of any trailing whitespace
This could easily be put in a trim() function, but here we’ll spell it out:
Can’t believe I forgot this step. Sorry. 😳
p = strchr( s, '\0' );
while (p-- != s) if (!isspace( *p )) break;
p[1] = '\0';
#3 Try to convert that string to the type of thing you want.
char * p;
int user_input = strtol( s, &p, 10 );
if (*p)
{
// Input was not JUST an integer.
// It could be something like "123 xyz", or "not-an-integer".
// Look at the value of p to figure out where the conversion went wrong.
complain();
}
do_something_with_an_integer( user_input );
That’s it!