scanf() method does not working in while loop?

Question

I've been out of this loop for 5 hours. My scanf method does not work.
Here is my loop.
I could not execute library's strcmp so I wrote myself.

int string_compare(char str1[], char str2[])//Compare method
{    
    int ctr=strlen(str1);
    int ctr2=strlen(str2);
    int counter=0;
    if(strlen(str1)!=strlen(str2) ){//if their lengths not equal -1     
        return -1;
    } else
    {
        for (int i = 0; i < strlen(str1); ++i)
        {
            if(str1[i]==str2[i]){ //looking for their chars
                counter++;
            }
        }

        if(counter==strlen(str1)){  
            return 0;
        } else
        {
            return -1;
        }
    }
}

char str1[100]; //for users command
char newString[10][10]; //after spliting command i have

while(string_compare(newString[0],"QUIT") != 0){
    printf("Enter Commands For Execution\n");
    scanf("%10[0-9a-zA-Z ]s\n",str1);
    int i,j,ctr;
    j=0; ctr=0;
    for(i=0;i<=(strlen(str1));i++)
    {
        // if space or NULL found, assign NULL into newString[ctr]
        if(str1[i]==' '||str1[i]=='\0')
        {
            newString[ctr][j]='\0';
            ctr++;  //for next word
            j=0;    //for next word, init index to 0
        } else
        {
            newString[ctr][j]=str1[i];
            j++;
        }
    }

    if(string_compare(newString[0],"QUIT") == 0){
        printf("Quitting\n");
        break;
    }

    if(string_compare(newString[0],"MRCT") == 0){
        printf("hey\n");
    }
    if(string_compare(newString[0],"DISP") == 0){
        printf("hey2\n");
    }
}

When I execute my c file,
the loop asks me to enter command such as "MRCT".

It forever prints

Enter Command
hey
Enter Command
hey

My way of using scanf() does not work here.

Here is my string-compare method int string_compare(char str1[], char str2[]) { int ctr=strlen(str1); int ctr2=strlen(str2); int counter=0; if(strlen(str1)!=strlen(str2) ){ return -1; } else{ for (int i = 0; i < strlen(str1); ++i) { if(str1[i]==str2[i]){ counter++; } } if(counter==strlen(str1)){ return 0; } else{ return -1; } } } — dud3, Feb 28 '18 at 22:55
No worries, don't post the code in a comment, it is unreadable, edit your question and post it there — Stephen Docy, Feb 28 '18 at 22:57
Hmmm, should you be passing in `newString[0]` to see if it is `QUIT` or `newString[ctr - 1]`? In your input, is `QUIT` the last word entered or the first? I'm thinking last, as in `MRCT DISP QUIT`? — Stephen Docy, Feb 28 '18 at 23:00
In fact, the code in the while loop will only ever process the first word in the input, over and over — Stephen Docy, Feb 28 '18 at 23:06
Edited. By the way someone told me that i should use gets method. It worked. newString contains user input for example, if user enters MRCT 5, i have MRCT at newString[0] and 5 at newString[1]. — dud3, Feb 28 '18 at 23:07
@dud3 "By the way someone told me that i should use gets method." Using `gets()` is a [terrible choice](https://stackoverflow.com/questions/1694036/why-is-the-gets-function-so-dangerous-that-it-should-not-be-used) - and a hint to the quality of that advise. Use `fgets()`. — chux - Reinstate Monica, Feb 28 '18 at 23:36

Martin York · Answer 1 · 2018-02-28T23:40:00.150

The scanf() stops scanning after the first failure.

So here:

scanf("%10[0-9a-zA-Z ]s\n",str1);

The scan tries to interpret three things:

A string %[] into a variable str1
the character s
The character '\n' which will result in a sequence of white space characters being read from the input.

Note if the string is zero length it will fail at the string and not interpret the s or the \n character.

One: I suspect that s is an error and you don't want it.

Two: Don't use "%[^\n]\n" to read a line. As this fails if the line is empty (just has the \n character).

if (scanf("%10[0-9a-zA-Z ]", str1) == 1)
{
    // Always check that the value was read.
    // Then deal with it.
    ....
}
scanf("%*[^\n]"); // Ignore any remaining character above 10.
                  // Note this may still fail so don't add \n on the end
                  // Deal with end of line separately.

char c;
if (scanf("%c", &c) == 1 && c == '\n')  // Now read the end of line character.
{
    // End of line correctly read.
}

score -1 · Accepted Answer · answered Feb 28 '18 at 23:05

-1

Use: scanf("%[^\n]\n" , str1); to get a line using scanf function.

answered Feb 28 '18 at 23:05

Ebram Shehata

446
5
20

1

`"%[^\n]\n"` fails to read anything if the line begins with a `"\n"`. Otherwise `"%[^\n]\n"` will consume a _line_ and all following white-space and blocks until following non-white-space detected. Would not pass a code review. Use `fgets()` to read a line. – chux - Reinstate Monica Feb 28 '18 at 23:21

scanf() method does not working in while loop?

2 Answers2