This question already has an answer here:
Does scanf() take '\n' as input leftover from previous scanf()?
(1 answer)
Closed 9 years ago.
Im trying to make a basic C console application calculator. Yet when i execute it the second scanf command is skipped and the third is run instead. This is a problem as here i need to get operation of the user +, -, *, or /. How do i stop this from happening?
float num1;
char sign;
float num2;
float total;
printf("~~~ Calculator ~~~\n");
printf("Please enter the first number: ");
scanf("%f", &num1);//Get value of num1 from user
printf("\nNow please enter the operation , either +, -, *, or / : ");
scanf("%c", &sign);//Get value of sign from user
printf("\n\nFinaly enter the second number: ");
scanf("%f", &num2);
Edit: Actually after trying various suggestions it seams a space before the %c was the correct way and cleanest way of fixing things. Thanks for the help.
Put a blank character before %c so it will consume the new line character like this:
scanf(" %c", &sign);//Get value of sign from user
After you hit enter, the first scanf consumes the actual number and leaves the \n. The latter is consumed in the next scanf.
scanf("%f\n", &order)
Do that in order to consume the \n you are entering.
As #JonathanLeffler mentioned, it's better approach to use " %c" (Note the space) in the next line in order to consume the newline character.
After the program asking for the first number you enter a number the you hit return. So the the number is stored in the variable num1. But the new line character '\n' still in the buffer you need to clear the buffer before the second scanf is executed Otherwise the signe variable will accept that '\n' as an input because it's a character and it will not prompt for an input.
You should always clear the buffer before asking to input a char
int c;
do {
c = getchar(); // clearing the buffer
} while (c!='\n' && c!=EOF);
printf("\nNow please enter the operation , either +, -, *, or / : ");
scanf("%c", &sign);//Get value of sign from user
Now it shouold work
Actually, this happens because there is still a value in the stdin stream of the program.
An easy solution would be using:
fix: DON'T USE -> fflush(stdin);
Consume the newline char by using scanf("%f\n")...
before the second call to scanf...
Best!
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 question already has an answer here:
How to read / parse input in C? The FAQ
(1 answer)
Closed 4 years ago.
I am trying to figure out the best way to get an integer and a character from a user
Here is what I have so far:
#include <stdio.h>
int main()
{
int a;
char b;
printf("enter the first number: \n");
scanf("%d", &a);
printf("enter the second char: \n");
scanf("%c", &b);
printf("Number %d",a);
printf("Char %c",b);
return 0;
}
The output is not shown correctly. Is there any problem with this?
Your input and output statements are fine. Just replace printf("Number %d",a); with printf("Number %d\n",a); to better format the output. Also you should change your second scanf statement to scanf(" %c", &b);. This will deal with the newline character entered after the number is inputted.
After you enter the number, you pressed the Enter key. Since the scanf function works on the input stream, when you try to process the next char after reading the number, you are not reading the character you typed, but the '\n' character preceding that. (i.e. because the Enter key you pressed added a '\n' character to your input stream, before you typed your char)
You should change your second call to scanf with the following.
scanf(" %c", &b);
Notice the added space character in the formatting string. That initial space in the formatting string helps skip any whitespace in between.
Additionally, you may want to add \n at the end of the formatting strings of both printf calls you make, to have a better output formatting.
Here you need to take care of hidden character '\n' , by providing the space before the %c in scanf() function , so the "STDIN" buffer will get cleared and scanf will wait for new character in "STDIN" buffer .
modify this statement in your program : scanf("%c",&b); to scanf(" %c",&b);
This question already has answers here:
scanf() leaves the newline character in the buffer
(7 answers)
Closed 4 years ago.
If I try something such as:
int anint;
char achar;
printf("\nEnter any integer:");
scanf("%d", &anint);
printf("\nEnter any character:");
scanf("%c", &achar);
printf("\nHello\n");
printf("\nThe integer entered is %d\n", anint);
printf("\nThe char entered is %c\n", achar);
It allows entering an integer, then skips the second scanf completely, this is really strange, as when I swap the two (the char scanf first), it works fine. What on earth could be wrong?
When reading input using scanf, the input is read after the return key is pressed but the newline generated by the return key is not consumed by scanf, which means the next time you read a char from standard input there will be a newline ready to be read.
One way to avoid is to use fgets to read the input as a string and then extract what you want using sscanf as:
char line[MAX];
printf("\nEnter any integer:");
if( fgets(line,MAX,stdin) && sscanf(line,"%d", &anint)!=1 )
anint=0;
printf("\nEnter any character:");
if( fgets(line,MAX,stdin) && sscanf(line,"%c", &achar)!=1 )
achar=0;
Another way to consume the newline would be to scanf("%c%*c",&anint);. The %*c will read the newline from the buffer and discard it.
You might want to read this:
C FAQ : Why does everyone say not to use scanf?
The other answers are correct - %c does not skip whitespace. The easiest way to make it do so is to place whitespace before the %c:
scanf(" %c", &achar);
(Any whitespace in the format string will make scanf consume all consecutive whitespace).
It doesn't skip the second scanf(); the second scanf() reads the newline left behind by the first scanf(). Most format codes skip white space; the %c format does not skip white space.
calling getchar() before scanf will also purge the stored line break. More lightweight but more situational
char input_1;
char input_2;
getchar();
scanf("%c", &input_1);
getchar();
scanf("%c", &input_2);
will flush the line breaks, more useful in consecutive lines of code where you know it's only one queued value and not a string
Try also _flushall() after each printf call. . Basically, by default MS’s C++ buffers stream output, and the the flushing causes the output stream to empty.
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.
This question already has answers here:
C: Multiple scanf's, when I enter in a value for one scanf it skips the second scanf [duplicate]
(7 answers)
Closed 8 years ago.
This is part of a university lab and the TA tells me there is an error but I haven't a clue. When I run it it asks me for the first char but then runs through the program and doesn't ask me at the second scanf.
#include <stdio.h>
int main(void) {
char sen, ben;
printf("Type in a character: ");
scanf("%c", &sen);
printf("The key just accepted is %d", sen);
printf("\nType in another character: ");
scanf("%c", &ben);
printf("The key just accepted is %d", ben);
}
Actually this is C not C++. Save it as file.c.
Try this:
#include <stdio.h>
int main(void) {
char sen, ben;
printf("Type in a character: ");
sen = getchar();
printf("The key just accepted is %d", sen);
printf("\nType in another character: ");
getchar();
ben = getchar();
printf("The key just accepted is %d", ben);
}
Explanation: when you enter the first character and press enter it takes enter's ASCII code as the second.
I suggest not to use scanf. But it works both ways if you put a getchar to "take" the enter.
Adding a space before %c in the second scanf will solve the issue.
This is done because scanf does not consume the \n character after you enter the first character and leaves it in the stdin.As the Enter key(\n) is also a character,it gets consumed by the next scanf call.The space before the %c will discard all blanks like spaces.
When you are scanning a character(%c) using scanf,add a space before %c as it would help reduce confusion and help you. Therefore, in both the scanfs , you can add the space.
When you pressed your key and then hit enter, you typed in two keys. The first was the desired key ,a for example, and the second was the key <enter> typically written as \n. So, your second scanf captures the result \n.
Since printing out the \n character doesn't result in something that is easy to see on the screen, it will appear like your program is just skipping the second scanf and printing out only the fixed parts of the printf without a easily viewable value.
One way to get around this problem is to consume all the key strokes just before the key you want to capture. This is done by accepting more input after the character up until you see a newline character \n. Once you see that character, then you do your next read.
// flush extra input up the to carriage return
char flush = 0;
while (flush != '\n') {
scanf("%c", &flush);
}
// now read my desired input
scanf("%c", &ben);
that's because nobody accepts '\n'. call scanf like this scanf("%c%*c", &sen). %*c means you want to omit one character, which is '\n'.
btw, void main() is allowed. main function is not the real entry point of executable, so it's ok to do that. but it seems not everybody likes it.