I write my code in C language but when I write like this
int main()
{
char one,two,three;
int num=0;
scanf("%d",&num);
scanf("%c %c %c",&one,&two,&three);
printf("%c %c %c",one,two,three);
}
When I input chars a b c that shows me
a b
and does not show c
It happens when I input num. If I set a variable of num to be 3 it shows me
a b c
like my input
How to solve this?
When your first "scanf" processes, in addition to whatever number you're inputting you're also putting a newline character "/n" in as well. This is being read by your second "scanf" such that &one is your new line.
To fix this just change this line:
scanf("%c %c %c",&one,&two,&three);
To this:
scanf(" %c %c %c",&one,&two,&three);
The extra space in the second line will prevent the newline from being used by the second "scanf" statement.
Related
#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.
I tried a simple program..function returning integer and character pointer..after i run that code i found some weird acting by scanf..I tried to print message(enter a,b:) read two integer inputs and message(enter c,d:) read two char inputs.but at run time..i found that input for the char c is read rightafter i enter the inputs for a,b..
for eg:
enter a,b: 10
20
enter c,d: g
it gets only one input(for d) and input for c is newline after 20..
for eg 2:
enter a,b: 10
20a
enter c,d: g
it gets only one input(for d) and input for c is a after 20..
why is this happening..please clarify it
int* add(int *a,int *b)
{
return (*a>*b?a:b);
}
char* charret(char *c,char *d)
{
return (*c>*d?c:d);
}
int main()
{
int a,b;
char c,d;
printf("\n\t\tFUNCTION RETURNING INTEGER POINTER\n\t");
printf("Enter the Number A and B:");
scanf("%d %d",&a,&b);
printf("\tEnter the character c :");
scanf("%c %c",&c,&d);
printf("The Biggestt Value is : %d\n\t",*add(&a,&b));
printf("\n\tThe character c= %c hi d= %c",c,d);
// scanf("%c",&d);
printf("\n\tThe Biggestt Value is : %c", *charret(&c,&d));
getch();
return 0;
}
%c will read any character, including the newline character from your previous entry. If you want to read the first non-whitespace character, add a space before %c in your format string:
scanf(" %c %c",&c,&d);
/* ^ added space */
This will cause scanf() to eat any number of whitespaces before reading the character.
For most scanf() specifiers, any leading whitespace is skipped. %c is an exception to this, because it reads a single character value, including whitespace characters. Keep in mind when you press Enter, you've sent a '\n' to the input buffer.
scanf("%d %d",&a,&b);
Reads in two numbers. The \n at the end, from pressing Enter, is left in the buffer.
scanf("%c %c",&c,&d);
Reads in two characters, the first of which will be the \n left in the buffer. One way to get around this is:
while (getch() != '\n');
This will eat everything up to an including a newline. You can put that after the scanf() lines you know will leave a newline behind.
#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.
#include <stdio.h>
int main()
{
char C, B;
int x;
printf("What comes after G\n");
scanf("%c", &C);
printf("What comes after O\n");
scanf("%c", &B);
printf("What is your age?\n");
scanf("%d", &x);
printf("You said %c comes after G, %c after T and you're %d years old? Right?", C, B, x);
return 0;
}
The problem is whenever you run the code it skips the second question "What comes after O" and then asks "What is your age?"
The only way I could avoid the program skip the 2nd question was by adding a space to the code
printf("What comes after O\n");
scanf(" %c", &B);
You can see the space between " and %c
Can you please explain this to me?
You need to eat up the white space (i.e. new line) - as per the manual page http://linux.die.net/man/3/scanf
You can use scanf to eat the single character without assigning it to anything like this::
scanf( "%[^\n]%*c", &C ) ;
%[^\n] tells the scanf to read every character that is not '\n'. That leaves the '\n' character in the input buffer, then the * (assignment suppression) will consume the a single character ('\n') but would not assign it to anything.
The reason for this problem is newline character \n leftover by the previous scanf after pressing Enter. This \n is left for the next call of scanf.
To avoid this problem you need to place a space before %c specifier in your scanf.
scanf(" %c", &C);
...
scanf(" %c", &B);
...
scanf(" %c", &X);
A space before %c is able to eat up any number of newline characters.
The problem is you are using scanf to get the character ..and a new line will be added at the end of each input from user . So the second time only the new line will be stored in the 'B' Because of the first input given by you ..
Instead of scanf , change it to getchar - your problem should get solved
#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.