Analyze a variable in C and then decide what to do - c

Question

Hi I'm still a C beginner and a beginner coder,
I've been trying to develop some sort of a rudimental interface via command line and would like to implement that when the user inserts something that isn't a number the loop breaks and moves on
(my dream would be that it checks when the input is equal to the word new) Anyhow here's my code snippet:
do {
printf("\nInsert the score you obtained: ");
scanf("%d", &avg);
} while ( isdigit(avg) == 0 );
Basically until the input is a number it will keep going and when something that isn't an integer is inserted it should exit the loop.

Answer 1

To receive something that isn't an integer, you have to scanf something that isn't an integer.
char* input;
do {
printf("\nInsert the score you obtained, or exit to quit: ");
scanf("%[^/n]", input);
...
Now that you have a variable that can hold both "quit" and some number (represented as characters), you need to make a decision based on what kind of input you received.
char input[80];
do {
printf("\nInsert the score you obtained, or exit to quit: ");
scanf("%[^/n]", &input);
while ( strncmp(input, "exit", 4) == 0 );
this means also that you'll have to possibly convert the "string" of the number into a number value.
char input[80];
do {
printf("\nInsert the score you obtained, or exit to quit: ");
scanf("%[^/n]", &input);
int number = convert_to_number(input);
while ( strncmp(input, "exit", 4) == 0 );
but the conversion should only happen if it looks like a number
do {
printf("\nInsert the score you obtained, or exit to quit: ");
scanf("%[^/n]", &input);
if ( looks_like_number(input) ) {
int number = convert_to_number(input);
}
} while ( strncmp(input, "exit", 4) == 0 );
and finally you might need to collect multiple inputs, as the loop will accept them now
do {
printf("\nInsert the score you obtained, or exit to quit: ");
scanf("%[^/n]", &input);
if ( looks_like_number(input) ) {
sum = add(sum, convert_to_number(input));
}
} while ( strncmp(input, "exit", 4) == 0 );
This is only one solution of a very large number of possible ways to do this.
This solution also could be improved (like trimming white space around the "exit" flag, so a user could type " exit" or "exit " to leave. One could also make the loop not a loop, as the question seems to imply entering one number or "exit" once, while a loop will permit many numbers until exit was typed. And one could remove the silly add function which was only used to demonstrate with a word a very clear idea, when + might do.
Even if the above loop / approach isn't a perfect fit for your need, please look at it carefully because the ideas within it can be used to make your program a success.
Also, don't use isdigit for checking for an entire number. Look into sscanf which is like printf but for reading stuff out of strings. sscanf your number off the string, and check the return value to see if you found a number within the string.

Answer 2

isdigit (prototyped as int isdigit (int Test_Character);) is not the right function to use with testing whether an int is a digit or a number. isdigit() is used rather to test if a single character is a decimal digit (0-9). The input to this function must be a character with integer value between 0 and 255. ( refer to ASCII table to view what these value are.)
The right thing to do here is to determine if scanf converted an input value correctly.
The function scanf() is prototyped with a return value, indicating success or failure, but your code does not currently check it.
So, in summary, your while loop can be written using the return of scanf():
int main(void)
{
int count = 0;
int avg= 0;
do
{
printf("\nInsert the score you obtained: ");
count = scanf("%d", &avg);
}while(count > 0); //stays in loop for any numeric value
// } while ( isdigit(avg) == 0 );//Note; input containing decimal points will be truncated
//to contain only integer portion
return 0;
}
As long as the input is an integer numeric value, scanf will return count as a positive value (the number of values scanned correctly.)
For example:
123, -123, 8, 0 will all allow the loop to continue.
a, dfg will all result in a failed attempt to convert the input to a numeric, and return EOF (-1)
Process as string representation of numeric: (similar behavior with alternative coding approach.)
I promised to show how to use strtol():
By changing the input format specifier from "%d" to "%s", scanf() will read the input value of 123 as a string: "123". When input is brought into your program as a string representation of a numeric, it must be converted to a number a numeric type before it can be used. strtol() can be used to do this conversion.
Here is a modification of your original program illustrating how it can be done:
int main(void)
{
int count = 0;
char avg[20] = {0};//changed to string variable
long val = 0;
bool success = FALSE;
do
{
printf("\nInsert the score you obtained: ");
count = scanf("%s", avg);//changed format specifier to read strings
if(count)
{
success = parseLong(avg, &val);//see correct usage of strtol in function
}
}while(success); //Stays in loop for any numeric input.
//Note: input containing decimal values
//will be truncated to contain integer portion only.
return 0;
}
bool parseLong(const char *str, long *val)
{
char *temp;
BOOL rc = TRUE;
errno = 0;
*val = strtol(str, &temp, 0);
if (temp == str || ((*val == LONG_MIN || *val == LONG_MAX) && errno == ERANGE))
rc = FALSE;
return rc;
}