Isdigit功能为什么不能正常工作？

Question

I wrote this short code to test my understanding of the isdigit function:

int inChar;
printf("enter input:");
scanf(" %d", &inChar);

if (isdigit(inChar))
   printf("Your input was a number"); 
else
   printf("Your input was not a number.\n");

When I test this program and I enter a number, C returns the else statement (Your input was not a number.). So regardless of if I enter a number or a letter, the program returns the else statement.

Why is this so?

Answer 1

isdigit() checks if a single character that was passed to it by converting the char value an unsigned char . So, you can't directly pass any int value and expect it to work.

Man isdigit() says:

   isdigit()
          checks for a digit (0 through 9).

To check single digit, you can modify:

char inChar;
printf("enter input:");
scanf(" %c", &inChar);

if (isdigit((unsigned char)inChar)) {
   printf("Your input was a number"); 
}
else {
printf("Your input was not a number.\n");
}

If you have an array (a string containing a number) then you can use a loop.

Answer 2

The function's purpose is to classify characters (like '3' ). Running it on something that's read using %d doesn't make sense.

You should read a single char using %c . Remember to check that reading succeeded.

Answer 3

The C library function void isdigit(int c) checks if the passed character is a decimal digit character.

If you badly wanna try it with an int you can init in this way

int inChar = '2';

The following code gave expected results.

int main()
{
    char inChar;
    printf("enter input:");
    scanf(" %c", &inChar);
    if (isdigit(inChar))
        printf("Your input was a number. \n");
    else
        printf("Your input was not a number.\n");
    return 0;   
}

Output:

vinay-1> ./a.out
enter input:1
Your input was a number