如何获取在scanf中用空格输入的3个字符到一个char数组中？

Question

I am previously a java programmer, but I'm now doing a C course at university (computer science major).

I need the user to be able to enter 3 chars,the first 2 being numbers, and the last 1 being either 'v' or 'h'. For example "1 2 v".

I need the user to be able to enter it with the spaces in between each character.

This is my current code:

void manageInput(char box[][width]){
    char move[4];
    char input[16]; 
    while(1){

            scanf("%s", input);
            int i = 0;
            while(input[i] != 0){

                    if(input[i] != ' ' && input[i] != "\n"){
                            move[i] = input[i];
                    }
                    i++;
            }
            printf("%s\n", move);

            makeMove(box, move);
            printBox(box, height, width);
            // TODO
            if(move[0] == 'x'){
                    exit(0);
            }
    }

}

However if I run it, it works fine when I enter the chars with out spaces like "12v", but If I enter "1 2 v", it will print out "1", call printBox, then print out "2", then print out box again, and so on.

If someone could explain what I'm doing wrong here, I would appreciate it.

Answer 1

If someone could explain what I'm doing wrong here, I would appreciate it.

The short story is: Your code doesn't fulfill your requirements. It simply doesn't do what you want it to do.

Your requirements are:

• All fields must be one character. This requirement isn't fulfilled by your code. Your code will mistakenly accept multiple characters per field.
• There must be one space (exactly one space?) between the fields. This requirement isn't fulfilled by your code. There might be multiple spaces between the fields, and your code will mistakenly accept that.

In fact, your code invokes undefined behaviour by accessing the move array out of bounds. Consider that as a consequence of one of the above scenarios i might become some value higher than 3. What might happen in this code: move[i] = input[i]; ?

Your code is also way too complex. All of your functionality can be performed by scanf alone. It's a very powerful function, when you know how to use it correctly... I suggest reading and understanding the manual multiple times, when you have an opportunity. You'll learn a lot !

I notice something you neglected to mention from within the logic you have presented: It's expected that the first field might also be 'x' , which corresponds to an exit usecase. This is a bad design; the caller has no opportunity to clean up... but I'll run with it. You really should use return (and return an int value or something, corresponding to error/success) instead.

Let us caste that last paragraph aside, because we can simply consider 'x' to be invalid input (and exit as a result), and I don't want to change the contracts of your functions; I'll leave that to you. The expression described so far appears to be int x = scanf("%1[0123456789]%*1[ ]%1[0123456789]%*1[ ]%1[vh]", a, b, c); .

Note that it is expected that a , b and c will have enough space to store a string of one byte in length. That is, their declaration should look like: char a[2], b[2], c[2]; .

Make sure you check the return value ( x , in the example)! If x is 3, it's safe to assume that the three variables a , b and c are safe to use. If x is 2, it's safe to assume that a and b are safe to use, and so on... If x is EOF or 0 , none of them are safe to use.

By checking the return value, you can reject input that doesn't match that precise pattern, that is:

• Fields that aren't exactly one byte in width will be rejected.
• Too many or too few spaces will be rejected.

Something else popped up that you have neglected to mention, and it's also present within your code: Chux mentioned that you'll likely be expecting the input to be terminated with a '\\n' (newline) character. This can also be implemented in a number of ways using scanf :

• scanf("%1*[\\n]"); will attempt to read and discard precisely one '\\n' character, but there's no way to ensure that was successful. getchar would be more appropriate for that purpose; something along the lines of if (getchar() != '\\n') { exit(EXIT_FAILURE); } if (getchar() != '\\n') { exit(EXIT_FAILURE); } might make sense, if you wish to ensure that the lines of input are perfectly formed and bomb out when they aren't... #define BOMB_OUT ?
• scanf("%*[^\\n]"); scanf("%*c"); makes more sense; If you're interested in reading one item per line , then it makes sense to discard everything remaining on the line, and then the newline character itself. Note that your program should always tell the user when it's discarding or truncating input. You could also use getchar for this.

---

void manageInput(char box[][width]){
    for (;;) {
        char a[2], b[2], c[2];
        int x = scanf("%1[0123456789]%*1[ ]%1[0123456789]%*1[ ]%1[vh]", a, b, c);
        if (x != 3) {
            /* INVALID INPUT should cause an error value to be returned!
             * However, this function has no return value (which makes it
             * poorly designed)... Calling `exit` gives no opportunity for
             * calling code to clean up :(
             */
            exit(EXIT_FAILURE);
        }

        if (getchar() != '\n') {
#           ifdef BOMB_OUT
            exit(EXIT_FAILURE);
#           else
            scanf("%*[^\n]");
            getchar();
            puts("NOTE: Excess input has been discarded.");
#           endif
        }

        char move[4] = { a[0], b[0], c[0] };
        printf("%s\n", move);
        makeMove(box, move);
        printBox(box, height, width);
        // TODO
        if(move[0] == 'x'){
            exit(0);
        }
    }
}

Answer 2

%s reads a whitespace-delimited string with scanf, so if that's not what you want, it's not the thing to use. %c reads a single character, but does not skip whitespce, so you probably also want a (space) in your format to skip whitespace:

char input[3];

scanf(" %c %c %c", intput, input+1, input+2);

will read 3 non-whitespace characters and skip any whitespace before or between them. You should also check the return value of scanf to make sure that it is 3 -- if not, there was less than 3 characters in your input before an end-of-file was reached.

Answer 3

It's usuall a bad idea to read string via scanf because of potential buffer overflow. Consider using fscanf or better fgets as in

fgets(input, 15, stdin);

Note the extra byte for '\\0'.

---

Also, you're comparing char to string here: input[i] != "\\n" . It should be input[i] != '\\n' instead.

---

And btw you can just use something like

int x, y;
char d;
scanf("%d%d%c", &x, &y, &d);

Answer 4

This looks like two simple bugs.

You need to use separate indexes for move[] and input[]

    int i = 0;
    while(input[i] != 0){

            if(input[i] != ' ' && input[i] != "\n"){
                    move[i] = input[i];
            }
            i++;
    }

Imagine input of 1 2 v

input[0] != 0, so we enter the loop

it's not ' ' or '\\n' either, so we copy input[0] to move[0]

so far so good

You increment i, and discover that input[1] == ' '

But then you increment i again

You discover that you are interested in input[2] (2) - so you copy it to move[2], rather than move[1]. Oops!

Then to make things worse, you never put an end-of-string character after the last valid character of move[].