How could one handle the following scenario. The User is asked to input his command:

1. the user inputs only a digit for example 1
2. the user inuts a string for example hello
3. the user inuts a string for example hello 1 2

The Terminal could look like

Enter: 1 -- programm do somehting(not important)
Enter: hello -- programm do somehting(not important)
Enter: hello 1 2 -- programm do somehting(not important) 

My Code

   printf("Enter:");
   scanf("%[^\n]s", command_player);
   getchar()

  for(size_t i = 0; i < strlen(command); i  ) 
        {
          if(isdigit(command[i]))
          {
            has_digit = 1;
          }
          if(isalpha(command[i]))
          {
            has_letter = 1;
          }
        }

and then 

if(has_digit == 1 && has_letter == 0)
//do something
if(has_digit == 0 && has_letter == 1)
//do something
if(has_digit == 1 && has_letter == 1)
//do something

However I have the problem that if I enter in one of the ifs another datatype as it is stated my programm crashes

CodePudding user response：

You can achieve your goal by reading the input line with fgets() and trying to parse it with sscanf() in different ways:

#include <stdio.h>

int main() {
    char buf[256];
    char word[256];
    int n1, n2;
    char endc;

    // loop for ever: equivalent to while (1) {}
    for (;;) {
        // output the prompt        
        printf("Enter: ");

        // flush the output to make sure if is visible:
        // output is normally line buffered, but the prompt does
        // not end with a newline so it is still in the stdout buffer
        // most systems will flush the output when a read operation
        // is requested, but some don't so `fflush(stdout)` ensures
        // the prompt is visible on all systems.
        flush(stdout);

        // read a line of input, and break from the loop at end of file
        if (!fgets(buf, sizeof buf, stdin))
            break;

        // try matching different input patterns:
        // `sscanf()` returns the number of successful conversions,
        if (sscanf(buf, "%d %c", &n1, &endc) == 1) {
            // `sscanf()` returned `1` if there is exactly a number
            // preceded and/or followed by optional whitespace
            // but no further character.
            printf("user entered a single number %d\n", n1);
        } else
        if (sscanf(buf, " %5[a-zA-Z] %c", word, &endc) == 1) {
            // this second `sscanf()` returns 1 if there is exactly
            // a single word (a sequence of uppercase or lowercase
            // letters) with optional initial and training whitespace.
            printf("user entered a single word %s\n", word);
        } else
        if (sscanf(buf, " %5[a-zA-Z] %d %d %c", word, &n1, &n2, &endc) == 3) {
            // this third `sscanf()` returns 3 if there is exactly
            // a word followed by 2 numbers with optional initial 
            // and training whitespace.
            printf("user entered a word %s and 2 numbers %d and %d\n",
                   word, n1, n2);
        } else {
            printf("user input does not match a known pattern: %s", buf);
        }
    }
    return 0;
}

The format strings can be extended to include other characters for the words and modified to match other input patterns.