Thread: Loop control

I do not understand how to pause my display loop. I thought the second scanf would do the job, but it doesn't. Here is the code

Code:

  int main ()
{
	int op1;					// user input operand1
	int op2;					// user input operand2
	char operator;				// user input operator
	int result;					// answer for calculation					
	int status;					// open or close calculator program

for (status = 0 ;status !='q'; )
	{
	printf("\n\t\tCALCULATOR\n\n");
	printf("   Make your selection by entering 2 numbers\n");
	printf("   and an operator as shown in the form below\n\n");
	
	printf("\ti1 + i2		i1 plus i2\n");
	printf("\ti1 - i2		i1 minus i2\n");
	printf("\ti1 * i2		i1 multiply times i2\n");
	printf("\ti1 / i2		i1 divided by i2\n");
	printf("\ti1 ^ i2		i1 to the power i2\n");
	printf("\ti1 G i2		the greatest common divisor (gcd) of i1 and i2\n");
	printf("\ti1 L i2		the least common multiple (lcm) of i1 and i2\n");
	printf("\ti1 R i2		the square root of one quarter of the sum of i1 and i2\n");
	printf("\ti1 Q i2		quit the calculator program\n\n");
	
	printf("  The above menu lists available operations on the left and a brief explanation\n");
	printf("  on the right.  Use the format of the operations on the left to enter a formula\n");
	printf("  \n");
	printf("Type NUMBER   OPERATOR   NUMBER and press RETURN:\n");
	scanf("%d %c %d", &op1, &operator, &op2);

	 switch (operator)
		{
		case '+': result = plus(op1,op2);		break;
		case '-': result = minus(op1,op2);		break;
		case '*': result = multi(op1,op2);		break;
		case '/': result = divd(op1,op2);		break;
		case '^': result = powr(op1,op2);		break;	
		case 'g': result = gcd(op1,op2);		break;
		case 'l': result = lcm(op1,op2);		break;
		case 'r': result = qtrsumroot(op1,op2);	break;
		default: printf("Please re-enter your request\n"); break;
		}
printf("your formula is %d %c %d.  The answer is %d\n\n", op1, operator, op2, result); 	

printf("Press any key to reset calculator.  Press Q to quit.\n");
scanf("%c". &status);
	}

}

thanks for any insight.
Keith

thanks pinko,
I updated the type of status to char and revised the notation on the scanf line. There must be a logic error. I am thinking that the for(status='0'; status !='q'; ) line at the start of the function will be updated by the scanf line at the end of the function. The loop passes by the scanf. hmmmm....

Salem,
Thank you, I did not realize that the input stream needs to be cleared. I very much appreciate you writing -- taking time to reply. This gives me the direction needed to understand what is going on. Of course I had heard that there are better ways than scanf and that scanf has limits, but I didnt' know exactly what that means nor how it would affect me in this instance.
Thanks again for your direction,
Keith

These changes made a vast improvment in the functionality and operability of the display screen that caused me a problem using scanf. My thanks to SALEM for assisting with specific code and explanations of the nature of the problem using scanf.

Code:

int main ()
{
					
char status;	// open or close calculator program
char buff[100];           // user input workspace 

for (status = 0 ;status !='q'; )
	{
	
	fgets( buff, sizeof(buff), stdin );
	sscanf( buff, "%d %c %d", &op1, &operator, &op2);


	 switch (operator)
	{
	}

printf("Press ENTER to reset calculator.  Press Q to quit.\n");
fgets( buff, sizeof(buff), stdin );
sscanf( buff, "%c", &status);
}

}