this is a follow on question from one I asked recently:
C Programming help - providing user with option to exit a program
I now have a new problem. I can get the program to exit if a user enters any letter which is great, but now if a number is entered nothing happens. The while loop doesn't seem to run..
Can you please have a look at my code and see if you can spot what's wrong, thanks. Also, ideally i'd like xterm's window to close if the user wishes to exit. I'd be greatful if anyone could show me how to do this. Anyway here's the code:
#include <stdio.h>
#include <stdlib.h>
int main()
{
float number;
float sum = 0;
printf ("Please enter number or enter any letter to exit:\n");
scanf ("%f", &number);
// if user ENTERS a letter, program will terminate
if(1!=scanf ("%f", &number))
{
getchar();
printf ("Exiting the program...\n");
exit(0);
}
while (1)
{
sum += number;
printf ("Sum: %.2f\n", sum);
printf ("Please enter number or enter any letter to exit:\n");
scanf ("%f", &number);
// if user ENTERS a letter, program will terminate
if(1!=scanf ("%f", &number))
{
getchar();
printf ("Exiting the program...\n");
exit(0);
}
}
return 0;
}
As #BrianCain commented: you are calling scanf a second time in your if statement. If you entered a letter for the first scanf, it forces the second to immediately fail; if you entered a number for the first, then the second is waiting for you to enter another number.
I removed two of your scanf functions and it seems to fix the problem
#include <stdio.h>
#include <stdlib.h>
int main()
{
float number;
float sum = 0;
printf ("Please enter number or enter any letter to exit:\n");
// if user ENTERS a letter, program will terminate
if(1!=scanf ("%f", &number))
{
getchar();
printf ("Exiting the program...\n");
exit(0);
}
while (1)
{
sum += number;
printf ("Sum: %.2f\n", sum);
printf ("Please enter number or enter any letter to exit:\n");
// if user ENTERS a letter, program will terminate
if(1!=scanf ("%f", &number))
{
getchar();
printf ("Exiting the program...\n");
exit(0);
}
}
return 0;
}
The scanf in your if erased the value of the previous one you did.
While the previous two answers are essentially correct, the simple (and clear) version would be:
if(number == 1)
break;
Place this right after your original scanf() and get rid of the other if(..) ..; condition entirely. This will test to see if the number '1' was entered, and if so will break the infinite loop, allowing the program to continue down to the return 0; statement.
It will also deal with what could be confusing style, which is having an exit condition (the final return 0;) that's essentially impossible to get to, or should be.
Related
I am rewriting the Guessing Game code from 'C Programming for Absoulute Beginners' to verify that the user has entered in a digit, using the isdigit() function.
The rest of the code works, in terms of error checking; but the moment that the user enters in a non-digit, the code goes into an infinite loop.
#include <stdio.h>
#include <stdlib.h>
#define NO 2
#define YES 1
main()
{
int guessGame;
guessGame = 0;
int iRandomNum = 0;
int iResponse = 0;
printf("\n\nWould you like to play \"The Guessing Game\"?\n\n");
printf("\nType '1' for Yes or '2' for No!\n\n");
scanf("%d", &guessGame);
do{
if(guessGame == YES){
iRandomNum = (rand() % 10) + 1;
printf("\nGuess a number between 1 and 10:\n\n ");
scanf("%d", &iResponse);
if(!isdigit(iResponse)){
printf("\nThank you\n");
printf("\nYou entered %d\n", iResponse);
if(iResponse == iRandomNum){
printf("\nYou guessed right\n");
printf("\nThe correct guess is %d!\n", iRandomNum);
printf("\nDo you wish to continue? \n");
printf("\nType '1' for Yes or '2' for No!\n\n");
scanf("%d", &guessGame);
} else {
printf("\n\nSorry, you guessed wrong\n");
printf("\nThe correct guess was %d!\n", iRandomNum);
printf("\n\nDo you wish to continue? \n");
printf("\nType '1' for Yes or '2' for No!\n\n");
scanf("%d", &guessGame);
}
}
else {
printf("\nYou did not enter a digit\n");
printf("\n\nPlease enter a number between 1 and 10:\n\n");
scanf("%d", &iResponse);
}
}
else {
printf("\nThe window will now close. Try again later!\n");
exit(0);
}
}while(guessGame != NO);
}
The code goes into infinite loop as scanf() is unable to read an integer. The character you entered remains in the keyboard buffer.No more reading of integers is possible as long as the character is present in the buffer. scanf() simply returns the number of items read as 0 each time. Hence,the program does not wait for the user to enter data and infinite loop results.
scanf() returns number of items successfully read. So,you can simply check for the return value of scanf(), if its 1 then scanf() has correctly read an integer.
check = scanf("%d", &iResponse);
if(check == 1){
printf("\nThank you\n");
printf("\nYou entered %d\n", iResponse);
and flush the buffer if wrong input is entered
else {
while (getchar() != '\n'); //flush the buffer
printf("\nYou did not enter a digit\n");
printf("\n\nPlease enter a number between 1 and 10:\n\n");
//scanf("%d", &iResponse);
}
no need to ask for input here, while loop will continue and prompt for input in the beginning
Trying taking the input in the form of string .. also u will have to compare the input in the form 'number' :)
I've been trying to write a program that asks at end of the calculation if we wish to exit it based on input provided. But if I input n, it waits for another input for it to exit the program and after the second input it exits. Is there any way to avoid this? And also the reason why this happens. Here is the complete code.
#include<stdio.h>
int main()
{
float num1, num2;
char op, cont;
while(1)
{
scanf(" %f %c %f",&num1,&op,&num2);
if (op=='+')
printf("%.3f \n",num1+num2);
else if(op=='-')
printf("%.3f \n",num1-num2);
else if(op=='*')
printf("%.3f \n",num1*num2);
else if(op=='/')
{
if(num2==0)
printf("Look where you put your zeroes\n");
else
printf("%.3f\n",num1/num2);
}
else if(op=='%')
printf("%d\n",(int)num1%(int)num2);
else
{
printf("What was the crap you just entered?\n");
}
printf("Continue using calc?[y/n]");
scanf(" %c\n",&cont);
if (cont=='n')
break;
}
return 0;
}
After the following changes I have your program working:
char op, cont='y';
while(cont == 'y')
scanf(" %c", &cont);
Detailed explanation :Simple C scanf does not work?
The problem occured because of the last scanf. For excuting first scanf for num1 op num2 it requires a return (Enter) and after that while executing scanf for char count it takes that last retun key (Enter) as input .. So again it asks for num1 op num2 . this can be stopped by many ways , to stop it via scanf use "scanf(℅c℅*c,&count)" .%*c will ignore that return key input.
I've written a program to compute the running sum of any numbers input by a user. I need to provide the user with an option to exit the program at any stage, which I'm not sure how to do. I was looking into it and think getchar() is what I need to use but I'm not sure, there seem to be a few ways to do it.
I basically want the user to be able to hit "e" on the keyboard if they want to exit the program, and it will terminate. The comments in the code are just ideas I had so I've left them there. Help appreciated, thanks. Code:
#include <stdio.h>
#include <stdlib.h>
int main()
{
float number;
float sum = 0;
int i = 1;
//char exit [2] = {'e'};
//void exit (int status);
printf ("Please enter number or enter \"e\" to exit at any stage:\n");
scanf ("%f", &number);
// if user inputs string e, program will terminate
/* if (number == 'e')
{
printf ("Exiting the program...\n");
exit(0);
} */
while (i == 1)
{
sum += number;
printf ("Sum: %.2f\n", sum);
printf ("Please enter number:\n");
scanf ("%f", &number);
// if user inputs string e, program will terminate
/* if (number == 'e')
{
printf ("Exiting the program...\n");
exit(0);
} */
}
return 0;
}
replace
scanf ("%f", &number);
to
if(1!=scanf ("%f", &number)){
if (getchar() == 'e'){
printf ("Exiting the program...\n");
exit(0);
}
}
#define f(x) (x*(x+1)*(2*x+1))/6
void terminate();
main()
{
int n,op;
char c;
printf("Enter n value\n");
scanf("%d",&n);
op=f(n);
printf("%d",op);
printf("want to enter another value: (y / n)?\n");
scanf("%c",&c); // execution stops here itself without taking input.
getch();
if(c=='y')
main();
else
terminate();
getch();
}
void terminate()
{
exit(1);
}
In the program above , I want to take input from the user until he enters an n value.
For this I'm trying to call main() function repeatedly . If it is legal in C , I want to know why the program terminates at the scanf("%c",&c) as shown in commented line.
Someone , please help.
You should never call main from within your program. If you need to run it more then once use a while loop inside it.
Your execution stops because by default stdin in a terminal is line buffered. Also you are not using the return value from getch.
int main()
{
int n,op;
char c;
do {
printf("Enter n value\n");
scanf("%d",&n);
op=f(n);
printf("%d",op);
printf("want to enter another value: (y / n)?\n");
scanf("%c",&c);
} while (c == 'y')
return 0;
}
You first have
scanf("%d",&n);
which you have to press the Enter key for it to accept the number.
Later you have
scanf("%c",&c);
There is a problem here, which is that the first call to scanf leaves that Enter key in the input buffer. So the later scanf call will read that.
This is easily solved, by changing the format string for the second scanf call just a little bit:
scanf(" %c",&c);
/* ^ */
/* | */
/* Note space here */
This tells the scanf function to skip leading whitespace, which includes newlines like the Enter key leaves.
It's legal, but you'll have a STACKOVERFLOW after a while (pun intended).
What you need is a loop:
while (1) {
printf("Enter n value\n");
scanf("%d",&n);
op=f(n);
printf("%d",op);
printf("want to enter another value: (y / n)?\n");
scanf("%c",&c); // execution stops here itself without taking input.
getch();
if(c != 'y')
break;;
}
hello guys I coded something like kfc menu,and I got it to work(finally),but when I input something other than numbers for "menu",eg:the letter "A", I just can't get it to loop again to normal,instead it finishes the program
#include <stdio.h>
#include <stdlib.h>
int main()
{
char counter='y';
float totalprice=0;
while (counter=='Y' || counter=='y')
{
int menu;
float price=0;
printf("\nplease select from menu:");
scanf (" %i", &menu);
switch(menu)
{
case 1: {
printf("\none hotbox1 =RM10.50");
totalprice=totalprice+10.50;
break;
}
case 2: {
printf ("\none hotbox2=RM10.60");
totalprice=totalprice+10.60;
break;
}
case 3:{
printf ("\none hotbox3=RM10.70");
totalprice=totalprice+10.70;
break;
}
default : {
printf ("\nplease enter proper number please:");
scanf("%2f", &menu);
break;
}
}
printf("\n\nadd order?(Y/N):");
scanf (" %c", &counter);
}
printf("\n\nThe total price is: %f", totalprice);
return 0;
}
You should use fgets() (reference here) first and then sscanf() (reference here), checking it's return value to see if it's a number.
char inputBuffer[MAX_BUFFER];
do
{
fgets(inputBuffer, MAX_BUFFER, stdin);
}
while(sscanf(inputBuffer, "%d", &menu) != 1)
You scanf with %f in the default case, I am fairly certain that is for floats. Use %d.
Remove scanf("%2f", &menu);
Switch in C does not support char in switch-case. Before you start switch-case validate the user input. If it is a number go into switch case otherwise display a user message to enter only numeric value
I recommend that you debug this by printing out the value of "counter" at various points in the loop (i.e. after you read it in, at the bottom of the loop, etc.). This will give you visibility into what your code is doing.
You can try something like this
#include <stdio.h>
int main()
{
char counter;
float totalprice=0;
int menu=0;
do
{
printf("\n1. one hotbox1=RM10.50");
printf("\n2. one hotbox2=RM10.60");
printf("\n3. one hotbox3=RM10.70");
printf("\nplease select from menu:");
scanf ("%d", &menu);
switch(menu)
{
case 1:
printf("\none hotbox1 =RM10.50");
totalprice=totalprice+10.50;
break;
case 2:
printf ("\none hotbox2=RM10.60");
totalprice=totalprice+10.60;
break;
case 3:
printf ("\none hotbox3=RM10.70");
totalprice=totalprice+10.70;
break;
default :
printf ("\nplease enter proper number please:");
scanf("%d", &menu);
}
printf("\n\nadd more order?(Y/N):");
fflush(stdin); //to empty the input stream.
scanf("%c",&counter);
}while(tolower(counter) != 'n'); //tolower returns the lowercase character.
printf("\n\nThe total price is: %.2f", totalprice);
return 0;
}
When scanf("%i", &menu) tries to read an integer from the input, it finds A, which it cannot interpret as a number, so it does not read it(*). Then the next scanf continues reading the input from when the other left off and happily reads the letter 'A'. Since you loop as long as the read letter is either 'y' or 'Y' (which A is neither), it exits the loop.
(*) read up on the documentation of scanf to see how to tell if it encountered an error.
Note: scanf("%i", &menu) should be scanf("%d", &menu) as%d` is the formatting symbol for integers.
One solution would be to change the loop condition to:
while (counter!='N' && counter!='n')
{
...
}
This way you end the loop only if an explicit 'N' or 'n' is inputted.
Note: it won't help if you accidentally type 'n' for the menu item, so see the comment about the error handling above.