How to scan in values and ignore the characters - c

Question

I am a newbie to C and I was looking over some questions where I pondered upon a question where we need to scan in values using the users input. Example
1 2 3 45 6 7. So Automatically we scan these values into a 2D array.
One thing that troubles me is what If the user inputs
1 2 3 2 3 Josh, how can we ignore Josh and only scan in the values into the array.
I looked at using getchar and use a flag variable but I am unable to figure out the conundrum of differentiating between the integer and character.
/* This is something that I tried */
#include <stdio.h>
int main(int argc, char *argv[]) {
int a;
int b;
int A[10];
while (((a = getchar()) != '\n') && (b = 0)) {
if (!(a >= "A" && a <= "Z")) {
scanf("%d", A[b]);
}
b++;
}
}
}

Answer 1

I think one good method for achieving what you want is using scanf with the format "%s", which will read everything as a string, effectively splitting the input according to white spaces. From the manual:
s
Matches a sequence of non-white-space characters; the next
pointer must be a pointer to character array that is long
enough to hold the input sequence and the terminating null
byte ('\0'), which is added automatically. The input string
stops at white space or at the maximum field width, whichever
occurs first.
To convert the string to integer, you can use atoi. From the manual:
The atoi() function converts the initial portion of the string
pointed to by nptr to int.
So, if it converts the initial portion of the string into an integer, we can use that to identify what is a number and what's not.
You can build a simple "word detector" for atoi.
Using the function isalpha from ctype.h you can do:
int isword(char *buffer)
{
return isalpha(*buffer);
}
And rewriting your reading program you have:
#include <stdio.h>
#include <ctype.h>
int isword(char *buffer)
{
return isalpha(*buffer);
}
int main(void)
{
char input[200];
int num;
while (1) {
scanf("%s", input);
if (!strcmp(input, "exit")) break;
if (isword(input)) continue;
num = atoi(input);
printf("Got number: %d\n", num);
}
return 0;
}
You should keep in mind that the name isword is fallacious. This function will not detect if buffer is, in fact, a word. It only tests the first character and if that is a character it returns true. The reason for this is the way our base function itoa works. It will return zero if the first character of the buffer is not a number - and that's not what you want. So, if you have other needs, you can use this function as a base.
That's also the reason I wrote a separate function and not:
if (!isalpha(input[0]))
num = itoa(input);
else
continue;
The output (with your input):
$ ./draft
1 2 3 2 3 Josh
Got number: 1
Got number: 2
Got number: 3
Got number: 2
Got number: 3
exit
$
About assigments and &&
while (((a = getchar()) != '\n') && (b = 0))
As I said in a comment, this loop will never work because you're making a logical conjunction(AND) with an assignment that will always return zero. That means the loop condition will always evaluate to false.
In C, assignments return the value assigned. So, if you do
int a = (b = 10);
a will have now hold the value 10. In the same way, when you do
something && (b = 0)
You're effectively doing
something && 0
Which will always evaluate to false (if you remember the AND truth table):
p q p && q
---------------
0 0 0
0 1 0
1 0 0
1 1 1

Answer 2

Your code is completely wrong. I suggest to delete it.
You could use scanf with %d to read in numbers. If it returns 0, there is some invalid input. So, scan and discard a %s and repeat this process:
int num = -1;
while(num != 0)
{
printf("Enter a number, enter 0 to exit:");
if(scanf("%d", &num) == 0) /* If scanf failed */
{
printf("Invalid input found!");
scanf("%*s"); /* Get rid of the invalid input (a word) */
}
}