CODEWITHSUNDEEP
c
Babain
Babain

Reputation: 1

Why is the while loop not working?

im new to c language and i am facing some problems with my code.. I have to make a program that computes the prices for renting cars but the while loop is not functioning properly.The scanf above the while condition is %s and not %c because it wont let me give a characther when it is %c.Thank you in advance.

int main(){
    float skms, ekms, tkms, price;
    char catcode;
    int days;

    do{
        printf("Selected rental program:\n");
        scanf("%c",&catcode);
        printf("Enter number of days:\n");
        scanf("%d",&days);
        printf("Starting kilometers display:\n");
        scanf("%f",&skms);
        printf("Kilometers display at end of rent:\n");
        scanf("%f",&ekms);
        tkms=(ekms/10)-(skms/10);
        switch(catcode){
                case'a':
                case'A':
                    price=ammountDueA(days, tkms);
                    printf("%.2f\n",price);
                    break;
                case'b':
                case'B':
                    price=ammountDueB(days, tkms);
                    printf("%.2f\n",price);
                    break;
                case'c':
                case'C':
                    price=ammountDueC(days, tkms);
                    printf("%.2f\n",price);
        }

        printf("Select rental program.('Q' or 'q' to exit)\n");



        scanf("%s",&catcode);


    }while(catcode==!'Q'&&catcode==!'q');
    printf("%c",catcode);


    return 0;
}

Upvotes: 0

Views: 81

Answers (2)

3442
3442

Reputation: 8576

There are two problems in your code:

First of all, the specifiers in scanf...

scanf("%s", &catcode);

This won't work. The "%s" specifier expects its corresponding argument to be a char* to a buffer of unspecified length (i.e: this has the same problem as gets()). You probably meant "%c", whose corresponding argument shall point to a single character. Now, your format specifier is not recommended to be "%c", but " %c", because that extra space indicatesscanf()to ignore all whitespace it can before"%c"`. So, that line becomes...

scanf(" %c", &catcode);

Now, you're right, there's a problem in your while loop:

while(catcode==!'Q'&&catcode==!'q');

Let's "expand" this with whitespace, to make what you (apparently) intended more obvious (that's reason #1 to use spaces between operators!)...

while(catcode ==! 'Q' && catcode ==! 'q');

For you it appears okay. Not for me. The C language does not define any operator ==!, rather, it defines operator == (that compares for equality), and operator ! (that negates its single operand). The compiler would have understood that line as...

while(catcode == !'Q' && catcode == !'q');

Now, what does that mean? If the operand is non-zero, the ! operator returns zero (in this case, the null character, '\0'). Otherwise, it returns a non-negative value (usually 1 (does the standard mandates this?)). Because 'Q' is not '\0', and same for 'q', that means that your loop is effectively...

while(catcode == '\0' && catcode == '\0'); // Think of '\0' as 0

Absolute non-sense expressed in code! The solution is to simply use operator != (compares for inequality)...

while(catcode != 'Q' && catcode != 'q');

I hope this has led some light on you!

Upvotes: 1

William Pursell
William Pursell

Reputation: 212564

By "not working", I suppose you mean that it never enters a second iteration, but that is exactly what it is supposed to do. catcode==!'Q'&&catcode==!'q' is true if catcode is equal to !'Q' and equal to !'q'. 'Q' is some non-zero integer (depends on your system, but is probably 81), and !'Q' is zero. Similarly, !'q' is zero. catcode isn't zero, so the loop terminates.

Upvotes: 1

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