#include <stdio.h>
main()
{
int num;
char another="y";
for(;another=="y";)
{
printf("no. is ");
scanf("%d", &num);
printf("sq. of %d is %d", num,num*num);
printf("\nWant to enter another no. : y/n");
scanf("%c", &another);
}
}
I have C code like this. According to me, this should work like: Enter the no and give square. But its nor running in infinite loop. But it is running only once. Why?
I am using GCC4.8.1 compiler on windows 64 bit.
Because on second iteration scanf assign \n to anotherinstead of assigning y.
EXPLANATION: When you press Enter key after typing the input, then one more character goes to the buffer along with the typed input. This character is produced by Enter and is \n. Suppose you typed y and then pressed the Enter key then the buffer would contain y\n, i.e, two characters, y and \n.
When scanf("%d", &num); is executed then it reads the number typed in and leaves behind the \n character in the buffer for next call of scanf. This \n is read by the next scanf call scanf("%c", &another); irrespective of what you have typed in your console.
To eat up this new line char, use a space before %c specifier in scanf.
scanf(" %c", &another);
^Notice the space before %c.
And change
for(;another=="y";) {...} // Remove the double quote.
to
for(;another=='y';) {...} // Single quote is used for `char`s.
The test in the loop is wrong:
another=="y"
this compares the value of another, a single character, with the value of a string literal, which will be reprented as a pointer to the character y. It should be:
another == 'y'
You should have gotten compiler warnings for this, since it's very strange to compare a small integer with a pointer.
Related
This question already has answers here:
scanf() leaves the newline character in the buffer
(7 answers)
Closed 3 years ago.
I'm the learning the very basics of C programming right now, and I'm practicing the scanf() function. This very, very simple program takes in a number and a letter and uses the printf() function to display the number and letter.
If I ask the user to enter the number first, the program works, i.e., asks for a number, asks for a letter, and prints the input. If I ask for the letter first, the program asks for a letter but then doesn't ask for a number.
I've tried multiple ways and reordered it, but it doesn't seem to work.
This works:
#include<stdio.h>
void main(){
int number;
char letter;
printf("Enter letter...");
scanf("%c", &letter);
printf("Enter number....");
scanf("%d", &number);
printf("Number entered: %d and letter entered: %c.\n", number, letter);
}
But, this combination doesn't work:
#include<stdio.h>
void main(){
int number;
char letter;
printf("Enter number....");
scanf("%d", &number);
printf("Enter letter...");
scanf("%c", &letter);
printf("Number entered: %d and letter entered: %c.\n", number, letter);
}
The output I get for the first program is:
Enter letter...a
Enter number....9
Number entered: 9 and letter entered: a.
Which is correct
But the second case doesn't work, and I don't get why it wouldn't work -- skips the "enter letter" part
the output is
Enter number....9
Enter letter...Number entered: 9 and letter entered:
.
Context: I entered "a" for letter and "9" for number in the above example.
It turns out there's a surprising difference between %d and %c. Besides the fact that %d scans potentially multiple digits while %c scans exactly one character, the surprising difference is that %d skips any leading whitespace, while %c does not.
And then there's another easily-overlooked issue when you're using scanf to read user inputs, which is, what happens to all those newlines -- the \n characters -- that get inserted when the user hits the ENTER key to input something?
So here's what happened. Your first program had
printf("Enter letter...");
scanf("%c", &letter);
printf("Enter number....");
scanf("%d", &number);
The user typed a letter, and ENTER, and a number, and ENTER. The first scanf call read the letter and nothing else. The \n stayed in the input stream. And then the second scanf call, with %d, skipped the \n (because \n is whitespace) and read the number, just like you wanted.
But in your second program you had the inputs in the other order, like this:
printf("Enter number....");
scanf("%d", &number);
printf("Enter letter...");
scanf("%c", &letter);
Now, the user types a number and hits ENTER, and the first scanf call reads the number and leaves the \n on the input stream. But then in the second scanf call, %c does not skip whitespace, so the "letter" it reads is the \n character.
The solution in this case is to explicitly force the whitespace-skipping that %c doesn't do by default. Another little-known fact about scanf is that a space in a format string doesn't mean "match one space character exactly", it means "match an arbitrary number of whitespace characters". So if you change your second program to:
printf("Enter number....");
scanf("%d", &number);
printf("Enter letter...");
scanf(" %c", &letter);
Now, the space character in " %c" in the second scanf call will skip over the \n that was left over after the user typed the number, and the second scanf call should read the letter it's supposed to.
Finally, a bit of editorializing. If you think this is a bizarre situation, if you think the exception to the way %c works is kind of strange, if you think it shouldn't have been this hard to read a number followed by a letter, if you think my explanation of what's going on has been far longer and more complicated than it ought to have been -- you're right. scanf is one of the uglier functions in the C Standard Library. I don't know any C programmers who use it for anything -- I don't believe I've ever used it. Realistically, its only use is for beginning C programmers to get data into their first programs, until they learn other, better ways of performing that task, ways that don't involve scanf.
So my advice to you is not to spend too much time trying to get scanf to work, or learning about all of its other foibles. (It has lots.) As soon as you're comfortable, start learning about the other, better ways of doing input, and leave scanf comfortably behind forever.
Try this
#include <stdio.h>
int main(void) {
int number;
char letter;
printf("Enter letter...");
scanf("%s", &letter);
printf("Enter number....");
scanf("%d", &number);
printf("Number entered: %d and letter entered: %c.\n", number, letter);
return 0;
}
If you change the %c to %s then you get the correct output.
Add a space before %c. So, change this:
scanf("%c", &letter);
to this:
scanf(" %c", &letter);
As I have written in caution when using scanf, this will make scanf eat the whitespaces and special characters (otherwise it will consider them as inputs).
Here, it will consume the newline character, on other words, the Enter you press, after typing your input!
To be exact, in your example, think of what the user (in this case you) do:
You type 9
You press Enter
You type 'a'
You press Enter
Now, when you input something, from your keyboard in this case, this will go into the Standard Input buffer, where it will patiently await to be read.
Here, scanf("%d", &number); will come and read a number. It finds 9 in the first cell of the STDIN buffer, it reads it, thus deleting it from the buffer.
Now, scanf("%c", &letter); comes, and it reads a character. It finds the newline character, that's the first Enter you pressed, and the function is now happy - it was told to read a character, and that's exactly what it did. Now that newline character gets deleted from the buffer (now what's left in there is 'a' and a newline character - these two are not going to be read, since there is no other function call. left for that).
So what changes if I write scanf(" %c", &letter); instead?
The first scanf will still read the number 9, and the buffer will now have a newline character, the 'a' character, and another newline character.
Now scanf(" %c", &letter);` is called, and it goes to search for a character to read in the STDIN buffer, only that now it will first consume any special characters found.
So there it goes to the buffer, it firstly encounters the newline character, it consumes it.
Then, it will encounter 'a', which is not a special character, and therefore it will read normally, and stored to the passed variable in scanf.
The last newline character will remain in the STDIN buffer, untouched and unseen, until the program terminates and the buffer gets deallocated.
Tip: You probably meant to write int main(void), instead of void main(). Read more in What should main() return in C and C++?
Specifying scanf the following way
scanf("%c", &letter);
does not skip white spaces and can read for example a new line character stored in the input buffer when the user pressed Enter entering previous data.
Use instead
scanf(" %c", &letter);
^^^
to skip white spaces.
From the C Standard (7.21.6.2 The fscanf function)
8 Input white-space characters (as specified by the isspace function)
are skipped, unless the specification includes a [, c, or n specifier.
and
5 A directive composed of white-space character(s) is executed by
reading input up to the first non-white-space character (which remains
unread), or until no more characters can be read.
Pay attention to that according to the C Standard the function main without parameters shall be declared like
int main(void)
From the C Standard (5.1.2.2.1 Program startup)
1 The function called at program startup is named main. The
implementation declares no prototype for this function. It shall be
defined with a return type of int and with no parameters:
int main(void) { /* ... */ }
This is a menu driven program asking for user's choice.
Why are if conditions not executed?
Output is attached.
Creating a program asking for user's input:
void main()
{
float a,b,ans=0;char ch,choice;
choice='y';
while(choice=='Y'||choice=='y')
{
printf("Enter two numbers \n");
scanf("%f %f",&a,&b);
printf("1.+for Addition\n");
printf("2.-for subtraction \n");
printf("3.*for multiplication \n ");
printf("4./for Division \n");
printf("Enter your choice of operation \n");
scanf("%c",&ch);
if(ch=='+')
ans=a+b;
else if (ch=='-')
ans=a-b;
else if(ch=='*')
ans=a*b;
else if(ch=='/')
ans=a/b;
else
{
printf("wrong choice entered\n");
}
printf("Answer is %f \n",ans);
printf("Do you want to coninue (Y/N)\n");
scanf("%c",&choice);
}
printf("program Terminated\n");
}
Output:
/* Enter two numbers
1010
22
1.+for Addition
2.-for subtraction
3.*for multiplication
4./for Division
Enter your choice of operation
wrong choice entered
Answer is 0.000000
Do you want to coninue (Y/N)
n
program Terminated
*/
The above is the output screen.
It doesn't perform operations.
When you input first 2 numbers, they are placed into variables a and b. BUT after entering those 2 numbers, you pressed enter. Computer sees that as new input and place it in first next appropriate variable that requires input. In this case it's your variable ch, and instead of +,-./ or *, ch has value of "new line". If you try to write value of ch on standard output as an integer, it will write number 10. It's ASCII character of new line. Simply adding getchar() after inputting first 2 numbers will collect that new line sign, and your next scanf will work properly.
By the way, you have same problem with your last input scanf("%c",&choice); because pressing enter after previous operation decision, will also cause your program not to work properly. Do the same thing for this part, or simply leave blank character before %c.
Try the following
scanf(" %c",&ch);
^^
and
scanf(" %c",&ch);
^^
Otherwise a next character is read that can be a white space character.
Take into account that according to the C Standard function main without parameters shall be declared like
int main( void )
scanf() does not consume trailing newlines. The skipped scanf() receives the newline from the previous line typed by the user and terminates without receiving more input as you would expect...
scanf() is a bit cumbersome with newlines. A possible solution would be to use fgets() to get a line from the console and then employ sscanf() to parse the received string.
Another, more targeted, solution would be to use " %c" in the format string of the last scanf() call. The %c format specifier does not consume leading whitespace on its own, which is why it gets the remaining newline, rather than a character typed by the user.
#include<stdio.h>
int main(void)
{
char a;
char b;
printf("A is ");
scanf("%c",&a);
printf("B is ");
scanf("%c",&b);
}
The reason is that when you enter first character then after pressing Enter, a newline character (\n) is also passed to the input buffer along with that character. Since scanf("%c",&a); reads a single character at a time, it left behind \n in the buffer for next call of scanf. This \n is read by your second scanf call.
Put a space before %c in scanf to consume that newline character.
scanf(" %c",&a);
Make the scanf like this
scanf(" %c", &b);
when you are after giving the value for first variable, you will give enter. Here %c will take that as input. So avoid this, make the whitespace before the control string. It will skip the whitespace character(newline, tab, space). And ask the input from the user.
You can verify that new line is taken as a input for second variable like this,
printf("%d", b); // You will get the ascii value of new line.
when you give the enter newline placed in the input buffer, then that value will taken by the scanf. So need of scanf ( getting the input) is done. so it doesn't ask the input from you.
#include<stdio.h>
int main()
{
char a, b;
scanf("%c", &a);
scanf("%c", &b);
printf("%c %c",a,b);
return 0;
}
When I run this program, I only get output as a & I don't get the prompt to enter 2nd character. Why?
In this line,
scanf("%c", &a);
you are actually taking a %d from the stdin (standard input) but at the time you entered a character from stdin, you also typed ENTER from your keyboard which means that now you have two characters in stdin; the character itself & \n. So, the program took first character as the one you entered & second character as \n.
You need to use
scanf("%c\n", &a);
so that scanf eats the newline (that came by pressing ENTER) too.
As rodrigo suggested, you can use these too.
scanf(" %c", &a); or scanf("%c ", &a);
The way you are thinking that second character is printed is wrong. It's actually being printed but it's \n so your prompt might be coming to the next line.
Your code will work if you enter both characters without using ENTER.
shadyabhi#archlinux /tmp $ ./a.out
qw
q wshadyabhi#archlinux /tmp $
Note, when you used this, the only thing in STDIN was q & w. So, the first scanf ate q & the second one w.
Because when you press the enter key, the resulting newline is read as a separate character into b. Try this instead:
#include<stdio.h>
int main()
{
char a, b;
scanf("%c %c", &a, &b);
printf("%c %c",a,b);
return 0;
}
The %c is a format string which accepts only a single character. I think you pressed Enter key as soon as you pressed an alphabet key. The Enter key is also recognized as a character. So the next variable is taking the enter key which has a value of "\0".
The computer is still printing the character from the second variable but its invisible since nothing is getting printed. If you keenly observe, there will be a new line.
Enter two characters one after the other and you will be getting the right output.
/* Write macro for the following :
1. Arithmetic Mean of two no.
2. Absolute value of a no.
3. To convert a Uppercase letter to lower case.
4. To obtain bigger of two numbers.
*/
#include<stdio.h>
#define am(a,b) ((a+b)/2)
#define abs(a) (a>=0?a:-a)
#define ul(ch) (ch>=65 && ch<=96 ? ch+32 : ch)
#define bigger(a,b) (a>=b?a:b)
int main () {
int x,y;
char c;
printf("\nEnter two numbers:");
scanf("%d%d",&x,&y);
printf("\nThe arithmetic mean of two numbers is %f",(float)am(x,y));
printf("\nEnter the number:");
scanf("%d",&x);
printf("\nThe absolute value of the number is %d",abs(x));
printf("\nEnter the character:");
scanf("%c",&c);
printf("\nThe letter in lower case is %c",ul(c));
printf("\nEnter two numbers:");
scanf("%d%d",&x,&y);
printf("\nThe bigger of two numbers is %d",bigger(x,y));
return 0;
}
Everything is working fine except that program does not stop for taking character input.
Here is the snapshot of the output ....
Enter two numbers:4
5
The arithmetic mean of two numbers is 4.000000
Enter the number:-7 **/*After hitting enter here it reaches line no. 7 */**
The absolute value of the number is 7
Enter the character:
The letter in lower case is
Enter two numbers:4 **/*line no. 7*/**
6
The bigger of two numbers is 6
It is because the %d skips white space, but %c does not -- or in other words.
The %d will skip any proceeding white space in your input stream, and the input pointer will be then just after the last digit -- which is most likely you newline. So when you come to ask for the %c you will actually already have input data -- that is your newline -- and that is what you will read.
change your scanf to ask it to skip white space by just inserting a space before the %c, so
scanf(" %c",&c);
I believe the problem here is that your scanf("%c",&c) is grabbing the carriage return entered when you hit enter to put in the -7.
Put a getchar (or another scanf("%c",&c)) right before the scanf and you shouldn't have that problem.
%c reads any character including whitescape, so it will "eat" the newline character.
Use: scanf(" %c",&c);
A common issue with scanf is that it doesn't consume the newline caused by pressing enter. I usually get around it by using the following macro after a call to scanf
#define consumeBuffer() while (getchar() != '\n');
Of course this is not always what you want, but for most cases it will do the trick.
That's because after your first scanf, the enter key is still in the input buffer, and the next scanf will store the enter value in x. Then your next printf will print it - effectively moving to a new line.
To fix this you can just add a getchar() call after each scanf.