dev-cpp-users

[Dev-C++] Basic C question: scanf

[Mary S. <ms...@in...>]

I've written a short program using Dev-C++ 4, but it does not execute as I
would expect. The code is as follows:

#include <stdio.h>

int main(int argc, char *argv[])
{
  double salary, fed_withhold, state_withhold;
  double salary_sum = 0, fed_sum = 0, state_sum = 0;

  printf("Enter salaries: ");
  while (scanf("%lf", &salary) == 1){
        fed_withhold = salary * .17;
        state_withhold = salary * .03;
        printf("salary: %.2f fed tax: %.2f state tax: %.2f\n",
                        salary, fed_withhold, state_withhold);
        salary_sum += salary;
        fed_sum += fed_withhold;
        state_sum += state_withhold;
  }
  printf("\nSalary total: %.2f Fed tax total: %.2f State tax total: %.2f\n",
                   salary_sum, fed_sum, state_sum);

  system("PAUSE");
  return 0;
}

When the program is run, the prompt "Enter salaries:" is printed. After
that, I enter a series of numbers, then press the Enter key. The printf in
the loop is executed for each number input, but then the program hangs. I
have to either interrupt the program (<cntrl>-z worked for this) or type in
alpha characters and hit Enter. At that point the printf after the while
loop is executed.

What have I got wrong or what is it that I don't understand?

Thanks,

Mary

Re: [Dev-C++] Basic C question: scanf

[Nurhidayat <dh...@ma...>]

At 20:47 11/07/2001 -0700, you wrote:
>When the program is run, the prompt "Enter salaries:" is printed. After
>that, I enter a series of numbers, then press the Enter key. The printf in
>the loop is executed for each number input, but then the program hangs. 

No, it doesn't hang. 
After the printf in the loop is executed, scanf still waiting legal input. 

  while (scanf("%lf", &salary) == 1) {
     ...
  }

So to finish scanf, you have to give illegal input or non numeric
character, like 'x' (or another alphabet character you like) then press
[Enter].

Try to give input like this:

Enter salaries: 100 200 150 600 x [Enter]



___Nurhidayat___

Re: [Dev-C++] Basic C question: scanf

[Vassili <v.v...@bt...>]

The reason it hangs, or seems to hang, is because your in a loop. The
scanf() is waiting for an input. When you type in an alpha character, the
return value isn't 1, so you exit the loop. You might be better off using a
"yes" or "no" style system to exit the loop. In essence, something like:

char ch = 'Y';   // Initialise to enter loop

while( (ch == 'Y') || (ch == 'y') ) {
  scanf();

  // Rest of code

  printf("Continue ?");
  ch = getchar();
}

Best of luck

Vas

----- Original Message -----
From: "Mary Sipple" <ms...@in...>
To: <dev...@li...>
Sent: Thursday, July 12, 2001 4:47 AM
Subject: [Dev-C++] Basic C question: scanf


> I've written a short program using Dev-C++ 4, but it does not execute as I
> would expect. The code is as follows:
>
> #include <stdio.h>
>
> int main(int argc, char *argv[])
> {
>   double salary, fed_withhold, state_withhold;
>   double salary_sum = 0, fed_sum = 0, state_sum = 0;
>
>   printf("Enter salaries: ");
>   while (scanf("%lf", &salary) == 1){
>         fed_withhold = salary * .17;
>         state_withhold = salary * .03;
>         printf("salary: %.2f fed tax: %.2f state tax: %.2f\n",
>                         salary, fed_withhold, state_withhold);
>         salary_sum += salary;
>         fed_sum += fed_withhold;
>         state_sum += state_withhold;
>   }
>   printf("\nSalary total: %.2f Fed tax total: %.2f State tax total:
%.2f\n",
>                    salary_sum, fed_sum, state_sum);
>
>   system("PAUSE");
>   return 0;
> }
>
> When the program is run, the prompt "Enter salaries:" is printed. After
> that, I enter a series of numbers, then press the Enter key. The printf in
> the loop is executed for each number input, but then the program hangs. I
> have to either interrupt the program (<cntrl>-z worked for this) or type
in
> alpha characters and hit Enter. At that point the printf after the while
> loop is executed.
>
> What have I got wrong or what is it that I don't understand?
>
> Thanks,
>
> Mary
>
>
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