C编程传递字符功能

Question

 #include <stdio.h>
 #include <string.h>
 #define SIZE 200

 void check(char str[]);

 void check(char str[]){
        /* checking if the characters entered are digits */
        int i;
        int j=0;
        char tel[10];

        for(i = 0; str[i] != '\0'; i++){
            if(str[i] >= '0' && str[i] <= '9'){
                tel[j++] = str[i];
            }
        }
        tel[j] = '\0';

    /* checking if length of zip code  is less than 5 digits */
    if(strlen(tel) < 5)
            printf("Not enough digits on input!");


    /* if length of zip code is 5 digits */
    else if (strlen(tel) == 5){
        printf("\nProgram Output: ");
        printf("(");

        /* formating zip code with parenthesis around zip */
        for(i = 0; tel[i] != '\0' && i < 5; i++){
            printf("%c", tel[i]);
        }
        printf(")");
    };
}

int main(){

    /* variable declaration */
    char str[SIZE];

    printf("Enter a zip code: ");
    scanf("%s", str);

    check(str);

    return 0;
}

I am writing a simple C code that validates a 5 character zip code string from the user. If less than 5 characters, program outputs error message. If exactly 5 characters, places parenthesis around zip code.

When I test my code, only my error message works. My 'else if' parameter isn't being executed when I input 5 characters. Is it the way I'm passing the char input to my function?

Answer 1

looks like your input contains characters and characters are not taken care in your code.

1234a
output: Enter a zip code: Not enough digits on input!

12345
output:Program Output: (12345)

Take a look at your for loop inside check function

Answer 2

您的问题是您可能没有传递5个数字，请执行以下操作，添加其他对象以调试else printf ("%s %d", tel, strlen (tel));

Answer 3

So please try to use GCC compiler if not ! The other case of error will be wrong input! Program is working fine according to your expectation !

Answer 4

When I test my code, only my error message works. My 'else if' parameter isn't being executed when I input 5 characters. Is it the way I'm passing the char input to my function?

No. It's because you are using buffered output. printf stores its output in an internal buffer until either the buffer is full or it has to print a new line \\n . Your program is exiting before printf has a chance to flush the buffer to the output. You need to terminate your final printf with a \\n .

printf(")\n");

You do want a new line there, anyway, otherwise, if it did print, the next shell prompt will be on the same line.