scanf validation sits and waits for another input. Why? - c

Question

I was working on this sample exercise, and everything works as I would like it to, but there is one behavior I don't understand.
When providing input: if I make consecutive invalid entries everything seems to work great. But if I enter a number different from 1,2,3 in the case of the first question, or 1,2 in the case of the second question, the program just sits there until a new input is given. If another invalid entry is made, it goes back to the error "invalid entry" message, and if an appropriate number is entered, everything moves along fine.
I do not understand why it stops to wait for a second input...anyone?
Thanks guys.
#include <stdio.h>
static int getInt(const char *prompt)
{
int value;
printf("%s",prompt);
while (scanf("%d", &value) !=1)
{
printf("Your entry is invalid.\nGive it another try: %s", prompt);
getchar();
scanf("%d", &value);
}
return value;
}
int main() {
int wood_type, table_size, table_price;
printf("Please enter " );
wood_type = getInt("1 for Pine, 2 for Oak, and 3 for Mahogany: ");
printf("Please enter ");
table_size = getInt("1 for large, 2 for small: ");
printf("\n");
switch (wood_type) {
case 1:
table_price = (table_size == 1)? 135:100;
printf("The cost of for your new table is: $%i", table_price);
break;
case 2:
table_price = (table_size == 1)? 260:225;
printf("The cost of for your new table is: $%i", table_price);
break;
case 3:
table_price = (table_size == 1)? 345:310;
printf("The cost of for your new table is: $%i", table_price);
break;
default:
table_price = 0;
printf("The cost of for your new table is: $%i", table_price);
break;
}
}

Answer 1

You most likely need to flush your input buffer (especially with multiple scanf calls in a function). After scanf, a newline '\n' remains in the input buffer. fflush does NOT do this, so you need to do it manually. A simple do...while loop works. Give it a try:
edit:
static int getInt(const char *prompt)
{
int value;
int c;
while (printf (prompt) && scanf("%d", &value) != 1)
{
do { c = getchar(); } while ( c != '\n' && c != EOF ); // flush input
printf ("Invalid Entry, Try Again...");
}
return value;
}
The blank line you get if you enter nothing is the normal behavior of scanf. It is waiting for input (some input). If you want your routine to immediately prompt again in the case the [Enter] key is pressed, then you need to use another routine to read stdin like (getline or fgets). getline is preferred as it returns the number of characters read (which you can test). You can then use atoi (in <stdlib.h>) to convert the string value to an integer. This will give you the flexibility you need.
example:
int newgetInt (char *prompt)
{
char *line = NULL; /* pointer to use with getline () */
ssize_t read = 0; /* number of characters read */
size_t n = 0; /* numer of chars to read, 0 no limit */
static int num = 0; /* number result */
while (printf ("\n %s ", prompt) && (read = getline (&line, &n, stdin)) != -1)
{
if ((num = atoi (line)))
break;
else
printf ("Invalid Input, Try Again...\n");
}
return num;
}

Answer 2

If some invalid input is entered, it stays in the input buffer.
The invalid input must be extracted before the scanf function is completed.
A better method is to get the whole line of input then work on that line.
First, put that input line into a temporary array using fgets(),
then use sscanf() (safer than scanf because it guards against overflow).
#include <stdio.h>
int main(int argc, const char * argv[]) {
char tempbuff[50];
int result, d , value;
do
{
printf("Give me a number: ");
fgets( tempbuff, sizeof(tempbuff), stdin ); //gets string, puts it into tempbuff via stdin
result = sscanf(tempbuff, "%d", &value); //result of taking buffer scanning it into value
if (result < 1){ //scanf can return 0, # of matched conversions,
//(1 in this case), or EOF.
printf("You didn't type a number!\n");
}
}while (result < 1);
//some code
return 0;
}
Knowledge from: http://www.giannistsakiris.com/2008/02/07/scanf-and-why-you-should-avoid-using-it/