Does the input buffer gets cleared after scanf reads?
#include <stdio.h>
int main(void)
{
int a, b;
scanf("%d", &a);//I inputted 3
scanf("%d", &b);//I inputted 4
}
So when I gave the input 4 was 3 present in the input buffer?
So when I gave the input 4 was 3 present in the input buffer?
No, the 3 was consumed.
You cannot re-read it (as int or otherwise).
If you input "3<enter>" the 3 is consumed and the buffer contains just the "<enter>". You then type "4<enter>" which is added to the buffer. The 2nd scanf (*) consumes the initial enter and the 4 leaving "<enter>" for the next input operation.
(*) the conversion specifier "%d" skips optional leading whitespace and (tries to) converts the rest of the input to integer (if no errors occur).
So when I gave the input 4 was 3 present in the input buffer?
No, it wasn't.
scanf() reads (and consumes) from the standard input until a match the specified format (in your case an integer) is found. That format is converted and consumed as well.
Related
I know the basics of scanf(); requiring a conversion specifier % and an address to the variable to store the input value &.
While doing an assignment, there was a task that involved inputting an arbitrary number of values, one example would be:
5 4 3 2 1
Into scanf, and then print out number of symbols corresponding to the integer typed (eg. '?'), so it would display the terminal as:
5 4 3 2 1
?????
????
???
??
?
In this case I had put the scanf in a while loop as so:
int i, num;
while(scanf("%d",&num)==1){
printf("%d",num);
for (i=0; i<num; i++){
printf("?");
}
printf("/n");
}
I am totally confused how scanf reads multiple integers in one input line (5 4 3 2 1) when many sources has specified that scanf only takes one integer until it reads a whitespace. From looking at this, my understanding is that the whitespace separating the integers indicate a new iteration following the previous integer?
When I tried to trace how printing works, it printed as:
5 4 3 2 1
5?????
4????
3???
2??
1?
...So my question is how does scanf 'save' all these integers in one line to 'num' and print EACH of the corresponding symbols to the values given AFTER input? Wouldn't integers be replacing the previous in the variable without an array?
Sorry if this question does not make sense - still quite new to coding. Thank you!
Your understanding of scanf is true, scanf only takes one integer until it reads a whitespace.
What makes you confused here is how the algorithm works. Your algorithm first reads the integer 5 and prints "immediately" five ?. It continues to read 4, print four ?, read 3, print three ?, and so on.
scanf stores the line in an internal buffer. If the buffer is not empty, it tries to read from the buffer and remove the portion that was successfully read; if the buffer is empty it waits for user input. On the first iteration of your outer loop (while), the buffer is empty. You input "5 4 3 2 1", so the buffer contains that. The first sscanf("%d",&num) reads the first integer from the buffer, which is 5; the buffer now contains " 4 3 2 1". On the second iteration you read 4, and the buffer will be " 3 2 1". This goes on until the buffer is exhausted. At this point your program continues; it just waits for more input.
scanf 'reads' characters from the standard input stream stdin (if there is nothing there yet it prompts the user to type something to be added to it), and converts them to a format depending on the conversion specifiers used in the format string.
You can think about stdin as a sequence of characters in the memory, and every time the user types something in the terminal and presses Enter what was typed is added to the stdin.
So in your case, when you type 5 4 3 2 1 and presses Enter in the terminal, stdin will receive the sequence of characters "5 4 3 2 1\n" (notice that the new line character was added when you pressed Enter).
Then the scanf("%d",&num), which is expecting only 1 decimal integer number, 'reads' and 'consumes' from stdin the relevant characters that form one continuous decimal integer number.
In this case 5 is read the first time scanf is executed, and " 4 3 2 1\n" is left in the stdin. Then as the condition is satisfied the loop statements are executed. And the second time scanf is executed the leading space is discarded by the %d specifier and 4 is read, leaving " 3 2 1\n" in the stdin. And so on until no decimal integers numbers can be read from the stdin...
Wouldn't integers be replacing the previous in the variable without an array?
There's no array required here at all.
The scanf() function takes the value (assuming white spaces included) and all of them are saved one by one into the variable num. If you use a debugger (suppose using GDB) and set a breakpoint in while(), you'll get to know by stepping into that the variable num is changed in each iteration and the for loop runs till num is reached.
An example of two iterations using a debugger is provided below (focus on num on the left corner):
Iteration 1:
Iteration 2:
I am very new to C programming. My sir gave this code to find the maximum of n numbers. When I do as Sir says things are perfect i.e Write a number when the line - Type the number of numbers and write numbers in a row like 7 8 9 10 when Type the numbers pop up.
#include <stdio.h>
main()
{
int n, max, number, i;
printf("Type the number of numbers");
scanf("%d", &n);
if(n>0)
{
printf("Type the numbers");
scanf("%d",&number);
max=number;
for(i=1; i<n; i++)
{
scanf("%d", &number);
if(number>max)
max=number;
}
printf("MAX=%d \n", max);
}
}
But if I write suppose - 5 8 9 10 7 6 - then the program understands it like --
It puts n = 5 then puts number = 8 then loop executes number changes to 9 then number changes to 10 till 6 and then gves max.
So how is scanf working here? It takes digit individually although they are written in a row with spaces?
From the horse's mouth:
7.21.6.2 The fscanf function
...7 A directive that is a conversion specification defines a set of matching input sequences, as
described below for each specifier. A conversion specification is executed in the
following steps:
8 Input white-space characters (as specified by the isspace function) are skipped, unless
the specification includes a [, c, or n specifier.284)
9 An input item is read from the stream, unless the specification includes an n specifier. An
input item is defined as the longest sequence of input characters which does not exceed
any specified field width and which is, or is a prefix of, a matching input sequence.285)
The first character, if any, after the input item remains unread. If the length of the input
item is zero, the execution of the directive fails; this condition is a matching failure unless
end-of-file, an encoding error, or a read error prevented input from the stream, in which
case it is an input failure.
10 Except in the case of a % specifier, the input item (or, in the case of a %n directive, the
count of input characters) is converted to a type appropriate to the conversion specifier. If
the input item is not a matching sequence, the execution of the directive fails: this
condition is a matching failure. Unless assignment suppression was indicated by a *, the
result of the conversion is placed in the object pointed to by the first argument following
the format argument that has not already received a conversion result. If this object
does not have an appropriate type, or if the result of the conversion cannot be represented
in the object, the behavior is undefined.
...
12 The conversion specifiers and their meanings are:
d Matches an optionally signed decimal integer, whose format is the same as
expected for the subject sequence of the strtol function with the value 10
for
the base argument. The corresponding argument shall be a pointer to
signed integer.
...
284) These white-space characters are not counted against a specified field width.
285) fscanf pushes back at most one input character onto the input stream. Therefore, some sequences
that are acceptable to strtod, strtol, etc., are unacceptable to fscanf.
The processing for scanf is exactly the same; the only difference is that scanf always reads from standard input.
Examples:
Suppose you type SpaceSpaceSpace123Enter in response to the first prompt; the input stream then contains the sequence {' ', ' ', ' ', '1', '2', '3', '\n'}. When you call scanf( "%d", &n );, scanf reads and discards the leading blank spaces, then reads and matches the sequence {'1', '2', '3'}, converts it to the integer value 123, and assigns the result to n. Since there was a successful conversion and assignment, scanf returns 1.
If the input stream contains the sequence {' ', ' ', ' ', '1', '2', '.', '3', '\n'}, scanf reads and discards the leading blanks, then reads and matches the sequence {'1', '2'}, converts it to the integer value 12, and assigns the result to n. The input stream will still contain {'.', '3', '\n'}. Since there was a successful conversion and assignment, scanf will return 1.
If the input stream contains the sequence {'.', '3', '\n'}, then there is no matching sequence of characters ('.' is not a valid character in a decimal integer). scanf will leave the . unread and leave the value of n unchanged. Since there was not a successful conversion and assignment, scanf returns 0 to indicate a matching failure.
If an end-of-file is signaled on the input stream before any matching characters have been read, or if there's some other input error, scanf does not assign any new value to n and returns EOF to indicate an input failure.
The "%d" in scanf("%d",&number); causes has 3 stages of scanning user text input into an int.
0 or more leading whites-space, like ' ', '\n', '\t' and some others are read and discarded.
Numeric text like "123", "-123", "+123" is read until until a non-numeric character is read. (or end-of-file, or a rare input error).
That non-numeric character is put back into stdin for subsequent input calls.
If step 2 is successful in reading at least 1 digit, the function returns 1. Good code checks the returned value.
if (scanf("%d",&number) != 1) Handle_UnexpectedInput();
The important thing is that '\n' is not so special with scanf("%d",&number);. It acts like a separator like another white-space or non-numeric text.
'\n' does cause the buffered stdin to accept the line of user input for processing by the various scanf() calls.
Here it is simplified explanation (not overly simplified, I hope) on how scanf works from a user point of view:
The arguments are divided in two parts:
The first part is a “format string”. The string is made of at least
one format specifier. In its simplest form a specifier begins with
% and it is followed by a letter that specifies the type of
variable you’re expecting (“%d” – I’m expecting an integer). The
number of specifiers must match the number of parameters and types
in the second part.
The second part is made of one or more addresses to locations memory
where the data you input will be stored. The pointed types must
match the specifiers.
When called, the function will repeat the following steps, starting with the first specifier and the first pointer, until the end of format string is detected:
Read and discard any white-space until a non-white-space character is found (white-space: space, tab, NL, at least);
Read characters up to first white-space or a character that do not match the expected input for current specifier;
Convert them to the type of current specifier and
Store the result in the location pointed by the current pointer.
There are three typical beginner mistakes which will result in undefined behavior (crash, most likely):
You forget the address-of operator &.
The specifier and the type do not match.
The number of specifiers do not match the number of pointers.
int d;
scanf( "%d", d ); // no &
scanf_s( "%s", &d ); // s do not match int
scanf_s( "%d%d", &d ); // too many specifiers
when you press the keyboard, you are filling a buffer on your computer. the scanf will read the amount of consecutive data untill it hit a space, so "1234 43", on the code scanf("%d") you are saying "read one number", and 1234 is one number, that's what it will read.
But if you have a loop that will execute that scanf again, the number "43" is currently in the reading buffer, and scanf will read it without stopping.
The manual for scanf doesn't explains that and it's a bit confusing for a newcomer to understand why the application is not stopping there to read a new number.
Let me explain as simply as possible...
scanf reads bytes from the standard input stream stdin.
Let's say the input you give is "23 67 21 99\n" (The \n is from when you pressed Enter).
Then each next call to scanf will start reading from this input buffer and it will interpret what is sees as whatever you tell it ("%d", etc) while separating inputs by an empty character. This could be a new line, a space, a tab, etc.
While there are still bytes to be read, scanf will not wait for you to input. That is what is happening here.
let's keep it simple. I assume you don't know anything about buffer or stdin.
scanf is used to take input from user. Whenever you type a number and press 'space' or ' enter' on keyboard the number is entered into program for further purposes. When you type scanf("%d",&n); it mean take integer input from the user and store it on the address of variable n.
I got this piece of code
#include<stdio.h>
#include <stdlib.h>
int main()
{
int i;
int number, S;
float MO;
S = 0;
for(i=0;i<10;i++)
{
printf("AAA %d\n", i );
scanf("%d\n", &number);
S = S + number;
}
MO = S/10;
printf("%d , %f \n",S , MO );
return 0;
}
when the execution starts, AAA 0 is printed.I then give my first number.After that, i am expecting to see AAA 1 , but this will be printed only after i give my second number.
Checked this here
C/C++ printf() before scanf() issue
but seems i can get none of these solutions work for me
The answers claiming that this has something to do with flushing input or output are wrong. The problem has nothing to do with this. The appearance of the \n character at the end of the scanf() template string instructs scanf() to match and discard whitespace characters. It will do so until a non-whitespace character is encountered, or end-of-file is reached. The relevant part of the C11 Standard is §7.21.6.2 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. The directive never
fails.
In OP's case, a second number must be placed in the input stream so that the first can be assigned. The second number then remains in the input stream until the next call to scanf(). In the case of the example given by Stephan Lechner, taking input from a file, there is a number in the input stream after each number to be assigned, until the last number (if there are exactly ten numbers), and then the EOF causes scanf() to return. Note that OP could also have signalled EOF from the keyboard after each input. Or, OP could enter all numbers on one line, with an extra number to signal end of input:
1 2 3 4 5 6 7 8 9 10 11
The solution is simply to remove the \n from the end of the scanf() template string. Whitespace characters at the end of such a template string are tricky, and almost never what is actually desired.
Just remove the \n from the scanf format string:
scanf("%d", &number);
"\n" waits for non-white-space. Instead use scanf("%d", &number) #Michael Walz and check its return value.
Let us break down scanf("%d\n", &number);
"%d" Scan for numeric text. This specifier will allow leading white-space. Once some non-numeric character is found, put that character back into stdin. If valid numeric text for an integer was found, convert to an int and save to number.
"\n" Scan and discard 0 or more white-spaces such as '\n', ' ', and others. Continue scanning until non-white-space is encountered, put that character back into stdin.
Now return the number of fields scanned (1).
Notice step 2. User had to type in data after the Return or '\n' to get scanf() to return.
The issue is with the buffering of stdout which doesn't force a flush unless a \n is in the output or fflush is called specifically
IF "for loop" is like this,
for(i=0;i<5;i++)
{
scanf("%d",&a[i]);
}
Then,
Input can be given either as,
1
2
3
4
5
Hitting Enter Every time for the next Input.
OR
1 2 3 4 5
With space character between different input values.
Now, How does this thing work?
I mean if loop is running,
then how does it recognizes that Input is given for the next loop cycle wihtout hitting the Enter key.
How does this thing work?
This works because scanf() reads only till a space or newline or tab is encountered.
So, when you give input 1 2 3 4 5, at the first iteration scanf() reads 1 and a space is encountered and scanf() terminates keeping the value 1 to a[0]. At the next iteration2` is read the same way.
The scanf (and family) function with the "%d" format reads white-space delimited input. If the white-space is a newline or a single space doesn't matter.
The Enter key simply tells the input stream system that input is ready (stdin which is used by scanf is line buffered by default), exactly what that input is doesn't matter.
I have some trouble working with scanf in a while loop.
I wanted to make a program that would ask the user to write three integers and save them in an array of three positions. If the user writes something which is not an integer, the program should continue asking for an integer until (s)he enters it. But it didn't work properly.
So I tried to simplify the problem with this code:
#include <stdio.h>
int main()
{
int num1=1;
int num2=2;
int num3=3;
printf ("write a number\n");
scanf("%i", &(num1));
printf("%i\n",num1);
printf ("write a number2\n");
scanf("%i", &(num2));
printf("%i\n",num2);
printf ("write a number3\n");
scanf("%i", &(num3));
printf("%i\n",num3);
}
If the inputs are 3 integers, there's no problem. But if you write a character, for example a, for the first integer, the other 2 values are not scanned and it simply writes:
a
2
3
The last two values are the initialization values.
Can anyone tell me what I have to do?
The scanf function does not have to read after it encounters the first invalid character in the input.
The %i specifier allows a as a hexadecimal, but it MUST be preceded by 0x.
If a was the first character in the input, and it was supposed to match to %i, then scanf wouldn't have to read anything afterwards - it can stop at the first invalid character..
References:
http://www.gidnetwork.com/b-64.html
For each conversion specifier scanf tries to locate the appropriate data item. scanf reads the item, stopping when it encounters a character that can't possibly belongs to the item. If any item is not read successfully then scanf returns immediately without looking at the rest of the format string.
When you enter a 5 10, scanf finds a for the specifier %i. It immediately returns and stops reading other inputs 5 and 10.