I am new to C programming. I wrote a simple switch case but it is not executing as expected . Can some one tell me what is wrong here??
#include <stdio.h>
int main() {
int i;
char yes;
bool flag = true;
while(flag) {
printf("Enter the value");
scanf("%d",&i);
switch(i) {
case 1:
printf("Hi");
break;
case 2:
printf("Hello");
break;
}
printf("Enter Y or N to continue");
scanf("%c",&yes);
if (yes == 'N') {
flag = false;
}
}
return 0;
}
The result I am expecting is:
Enter the Value
1
Hi
Enter Y or N to continue
Y
Enter the Value
2
Hello
Enter Y or N to continue
N
But the result I am getting is :
Enter the value 1
HiEnter Y or N to continueEnter the value N
HiEnter Y or N to continue
When you hit Enter after typing in the first number, scanf read all numeric characters from the input stream except the newline character produced by that Enter hit. The newline character is not a part of the number. It is left in the input stream, unread, waiting for someone else to read it.
The next scanf("%c",&yes); discovered that pending newline charcter and it read it without waiting. The %c format specifier does not skip whitespace in the input, it just reads the first character it sees.
Replace your scanf with
scanf(" %c",&yes);
to make it skip whitespace. That way it will ignore that pending newline and actually wait for you to enter something.
In all your printf you need to add \n at the end.
For example on usage, see here: printf
This should work for you:
(You forgot all '\n' in your printf statements and add a space in your char scanf statements)
#include <stdio.h>
int main() {
int i;
char yes;
int flag = 1;
while(flag) {
printf("Enter the value\n");
scanf("%d",&i);
switch(i){
case 1:
printf("Hi\n");
break;
case 2:
printf("Hello\n");
break;
}
printf("Enter Y or N to continue\n");
scanf(" %c", &yes);
if (yes == 'N')
flag = 0;
}
return 0;
}
Output:
Enter the Value
1
Hi
Enter Y or N to continue
Y
Enter the Value
2
Hello
Enter Y or N to continue
N
It's not a problem with the switch statement. It's a problem with your output - there aren't line breaks ('\n'). For example, instead of printf("Hi"); you might want to have printf("Hi\n");, which adds a line space at the end.
Related
This question already has answers here:
getchar does not stop when using scanf
(5 answers)
Closed 3 years ago.
I started C just a while ago (same as coding), so I`m a noob.
My Goal:
to state that the user hasn't entered a or b and then wait for the user to press enter to return to the calculator menu.
My problem:
getchar() doesn't wait for me to press enter. (Case 3)
#include <stdlib.h>
int main()
{
for (int i = 0;i == 0;){
int options=0,enteredA=0, enteredB=0;
float *a, A, *b, B, *c, C;
a=&A,b=&B,c=&C;
printf ("Calculator\nAvailable options:\n[1]Enter value for A\n[2]Enter value for B\n[3]Addition\n[9]Exit\n");
scanf ("%d",&options);
system ("clear");
switch (options) {
case 1:
printf("Enter a value for A:");
scanf ("%f",&*a);
enteredA++;
break;
case 2:
printf("Enter a value for B:");
scanf ("%f",&*b);
enteredB++;
break;
case 3:
if ((enteredA==0) | (enteredB== 0)){
printf("A and B are not initialized yet. Please enter a value in the menu.\nPress [Enter] to continue to the menu:\n");
fflush(stdin);
getchar();
break;
} else{
printf("%f+%f=%f\n",*a,*b,*c=*a+*b);
fflush(stdin);
getchar();
break;
}
break;
case 9:i++;break;
}
system("clear");
}
printf("Calculator Shut Down");
return 0;
}
In the following line:
scanf ("%d",&options);
you actually enter a number, and a newline character. The scanf function reads only the number. It leaves the newline (\n) in the input stream.
When you call getchar(), it will find a newline in the input stream. Hence, it will read it without waiting for user input. It only wait for user input if it didn't find anything in the input stream.
A possible workaround for this is to call getchar two times instead of one.
The first call will read the already existing newline in the stream. The second call won't find anything in the input stream. So, it will wait for user input as you expect.
I have some small comments that aren't related to your question:
You use scanf ("%f",&*a);. Why not just scanf("%f", a); or scanf("%f", &A); ?
Why you even create a pointer a for the variable A ?
I don't think you really need the variable c as well.
You don't need the variable i in the loop as well.
At the beginning of the loop, you keep re-initializing enteredA and enteredB variables to zero. That way, the condition in case 3: will be always true. You need to move these variables outside of the loop.
Your code also missing a #include <stdio.h>.
I'd simplify things like the following:
#include <stdio.h>
#include <stdlib.h>
int main()
{
int enteredA = 0, enteredB = 0;
while (1)
{
int options;
float A, B;
printf ("Calculator\nAvailable options:\n[1]Enter value for A\n[2]Enter value for B\n[3]Addition\n[9]Exit\n");
scanf("%d", &options);
getchar(); // The extra getchar to read the newline left in the stdin.
system ("clear");
switch (options)
{
case 1:
printf("Enter a value for A:");
scanf("%f", &A);
enteredA++;
break;
case 2:
printf("Enter a value for B:");
scanf ("%f", &B);
enteredB++;
break;
case 3:
if (enteredA ==0 || enteredB == 0)
{
printf("A and B are not initialized yet. Please enter a value in the menu.\nPress [Enter] to continue to the menu:\n");
}
else
{
printf("%f + %f = %f\n", A, B, A + B);
}
getchar();
break;
case 9:
printf("Calculator Shut Down");
return 0;
}
system("clear");
}
}
This question already has answers here:
scanf() leaves the newline character in the buffer
(7 answers)
Closed 5 years ago.
C beginner here and I'm trying to learn to write a menu program. Below is my code
//A test file to test functions
#include <stdio.h>
#include <stdlib.h>
int enterChangeChar();
void printMenu();
char C = ' ';
int main()
{
char inputVal;
while (1)
{
printMenu();
scanf("%c", &inputVal);
switch (inputVal)
{
case 'c': enterChangeChar(); break;
case 'q': exit(0);
}
}
return 0;
}
int enterChangeChar()
{
int input;
printf("Would you like to Enter or Change the Character value?\n\n");
printf("Enter 1 to Assign OR 2 to Change the value of C: \n");
scanf("%d", &input); //input a integer value
if(input == 1)
{
printf("Enter a character: \n");
scanf(" %c", &C);
printf("You entered: %c", C);
}
else if (input == 2)
{
printf("Change Value option selected\n");
}
return 0;
}
void printMenu()
{
printf("Select a menu choice from the options below: \n\n");
printf(" OPTIONS INPUT\n");
printf("Enter/Change Character 'c'\n");
printf("Quit Program 'q'\n\n");
printf("Enter your CHOICE:\n");
}
when I run the program. The menu is printed twice after I input a character. Why is the program doing so?
Select a menu choice from the options below:
OPTIONS INPUT
Enter/Change Character 'c'
Quit Program 'q'
Enter your CHOICE:
c
Would you like to Enter or Change the Character value?
Enter 1 to Assign OR 2 to Change the value of C:
1
Enter a character:
v
You entered: v
Select a menu choice from the options below:
OPTIONS INPUT
Enter/Change Character 'c'
Quit Program 'q'
Enter your CHOICE:
Select a menu choice from the options below:
OPTIONS INPUT
Enter/Change Character 'c'
Quit Program 'q'
Enter your CHOICE:
q
EDIT: just added the printMenu() function code which I missed out earlier.
Edit 2: Maybe my question is not clear enough. The problem is not with getchar and scanf. The problem is that the menu prints twice after user inputs a character even though it's only called once in the whole program.
Edit 3: Final Edit: Guys I got it now. \n was in the input buffer and whenever choice was given, it was counted as input so the menu was printed multiple times. Thanks for the help.
There are unused newline characters '\n' in your input stream that need to be drained.
scanf() often causes confusion because it will leave the trailing '\n' in the input stream. In your case, the newline is getting picked up by the subsequent getchar().
int main()
{
while (1)
{
char inputVal;
printMenu();
do {
inputVal = getchar();
} while (inputVal == '\n');
switch (inputVal)
{
case 'c': enterChangeChar(); break;
case 'q': exit(0);
}
}
return 0;
}
The problem is that when i type any character except for y or n it display this message two times instead to one)
This program is 'Calculator'
Do you want to continue?
Type 'y' for yes or 'n' for no
invalid input
#include<stdio.h>
#include<conio.h>
#include<stdlib.h>
void main ()
{
//program
//first to get two numbers
//second to get choice
int x=0,y=0,n=0;
char choice;
//clrscr(); does no work in devc++
system("cls"); //you may also use system("clear");
while(x==0)
{
puts("\t\tThis program is 'Calculator'\n\n");
puts("Do you want to continue?");
puts("Type 'y' for yes or 'n' for no ");
scanf("%c",&choice);
x++;
if(choice=='y')
{
y++;
puts("if this worked then we would continue to calculate the 2 no");
}
else if(choice=='n')
exit(0);
else
{
puts("invalid input");
x=0;
}
}
getch();
}
`
it looping twice because enter(\n) character is stored in buffer use scanf like this(add space before %c)
scanf(" %c",&choice);
That is because of the trailing new line after you enter y or n and hit enter key.
Try this out:
scanf("%c",&choice);
while(getchar()!='\n'); // Eats up the trailing newlines
If you input any character other than 'y' or 'n', control enters the :
else
{
puts("invalid input");
x=0;
}
block, which resets x to 0, Now the loop condition :
while(x == 0)
is true and hence it enters the loop again.
Also you may want to skip the trailing newline character while reading like :
scanf(" %c", &choice );
#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;;
}
#include<stdio.h>
#include<stdlib.h>
main(){
int b,c,r,d;
char a;
while(1){
printf("Enter the operator\n");
scanf("%c",&a);
if(a=='+') d=1;
if(a=='-') d=2;
if(a=='&') d=3;
if(a=='|') d=4;
if(a=='.') d=5;
printf("Enter the operands\n");
scanf("%d",&b);
scanf("%d",&c);
switch(d){
case 1:r=c+b;
break;
case 2:r=c-b;
break;
case 3:r=c&b;
break;
case 4:r=c|b;
break;
case 5:exit(0);
deafult:printf("Enter a valid operator");
}
printf("Result = %d\n",r);
}
}
Output:
Enter the operator
+
Enter the operands
8
7
Result = 15
Enter the operator
Enter the operands
scanf("%d",... will read a number (skipping whitespace beforehand) but leave the newline on the input stream. scanf("%c",... will read the first character, and does not skip whitespace.
One simple modification is to use
scanf(" %c", &a);
This will tell scanf to skip any whitespace before the character.
That because of the function scanf width param "%c", after the 1st time loop, at line scanf("%d",&c);, like +, there's a end-line character in the input stream, then the second loop, scanf get the end-line character as the input and parse it to a;
To fix this, you can add a scanf("%c"); line right after scanf("%d",&c);
have a look at
scanf error in c while reading a character