Dev-C++ Input skipped - c

#include<stdio.h>
#include<conio.h>
main()
{
int i;
char c, text[30];
float f;
printf("\nEnter Integer : ");
scanf("%d",&i);
printf("\nEnter Character : ");
c = getch();
printf("\nEnter String:");
gets(text);
printf("\nEnter Float:");
scanf("%f",&f);
printf("\nInteger : %d",i);
printf("\nCharacter : %c8",c);
printf("\nString : %s",text);
printf("\nFloat : %f",f);
getch();
}
Why is this simple program not able to read a string using the gets() function? What else should I use to correct it? Well it it worked in Turbo C in my old 32-bit PC but not here...

Scanf or other input parsing functions take only required quantity of characters as specified in the call from stdin and reject others.As a result these rejected values,during next read of stdin enter into the variables along with the newline characters and thus skipping inputs for a few calls.So its better to call a clear routine that cleans stdin and stops garbage entering into other variables.
Although your code is quite vulnerable still it has solution:-
#include<stdio.h>
int clear()
{
while ((getchar())^'\n');
}
int main()
{
int i;
char c, text[30]={0};
float f;
printf("\nEnter Integer : ");
scanf(" %d",&i);
printf("\nEnter Character : ");
scanf(" %c",&c);
printf("\nEnter String:");
clear();
gets(text);
printf("\nEnter Float:");
scanf(" %f",&f);
printf("\nInteger : %d",i);
printf("\nCharacter : %c",c);
printf("\nString : %s",text);
printf("\nFloat : %f",f);
getchar();
}

With some little research, I guess that the problem comes with scanf(). scanf() reads a line without the end of line character '\n' which seems to stay in the buffer and actually red by the next statement.
alternatively you can use fgets() and sscanf() as follows:
To read a character I used:
fgets(text,sizeof(text),stdin);
sscanf(text,"%c",&c); /* or: c = text[0]; */
to read an integer I have used
fgets(text,sizeof(text),stdin);
sscanf(text,"%d",&i);
I had a major problem with gets() in a C course I had (to which DevC++) was advised as a compiler. However, I totally recall I didn't follow the advice and it turned out that the behavior of fgets() is also compiler dependent.
The man page for gets() has this:
BUGS
Never use gets(). Because it is impossible to tell without knowing the data in advance how many characters gets() will read, and because gets() will continue to store characters past the end of the buffer, it is extremely dangerous to use. It has been used to break computer security. Use fgets() instead.

When you type 42 (or whatever) as the first integer, you actually type three characters: 4, 2 and then the newline character that comes from pressing ENTER. Your first scanf reads an integer, which means that it only reads the 4 and the 2, leaving the newline character in the input buffer.
When your program gets to gets, it reads a the very short line that consists just of that newline character.
You can fix it by reading and throwing away the newline character just after scanf, something like this:
printf("\nEnter Integer : ");
scanf("%d",&i);
while (getchar() != '\n')
;

Related

Why is fgets waiting for input before it's even called?

I'm trying to write a simple program to read an integer and then a string, then print both to standard output. Ideally, the execution should look something like this:
Input the number.
> 10
Input the string.
> a string
number: 10
string: a string
However, when I run the program, it freezes after the call to scanf() until more input is provided.
Input the number.
> 10
a string
Input the string.
>
number: 10
string: a string
Why is it waiting for input before fgets() is ever called?
#include <stdio.h>
int main()
{
int number;
char string[32];
printf("Input the number.\n> ");
scanf("%d\n", &number);
printf("\nInput the string.\n> ");
fgets(string, 32, stdin);
printf("\nnumber: %d\nstring: %s\n", number, string);
}
From a previous post...
https://stackoverflow.com/a/5918223/2203541
#include <stdio.h>
int main()
{
int number;
int c;
char string[32];
printf("Input the number.\n> ");
scanf("%d", &number);
do
{
c = getchar();
} while (c != '\n');
printf("\nInput the string.\n> ");
fgets(string, 32, stdin);
printf("\nnumber: %d\nstring: %s\n", number, string);
}
"Why is fgets waiting for input before it's even called"
fgets() does not act until it is called, but if when called, and if pointing to stdin and there is content remaining in the stdin stream, it will consume it immediately. If that contents contained EOF, n-1, OR a newline (read link) execution flow will continue.
The problem here is that scanf() (called prior to fgets()) is notorious for doing exactly what it is asked to do. For example, upon user entering 12<return> two recognizable items are entered into stdin, digits and a newline but by using "%d" as the format specifier, only the first of those items is consumed, leaving the \n hanging, until the very next call to fgets(), which accepts it as input, allowing execution flow to resume immediately (as described above), causing the apparent skip you are seeing.
[This is one (of several) examples that will provide a work around for the scanf() issue:
Change this:
printf("Input the number.\n> ");
scanf("%d", &number);//leaves the newline
To this:
char c;
...
printf("Input the number.\n> ");
scanf("%d%c", &number, &c);//consumes the newline
From comments:
"is there a way to use fgets to read an integer"
Yes, I prefer fgets() coupled with your favorite string to number converter. (There are several) The simplest is this:
char cNum[10];
int num;
printf("Input the number.\n> ");
if(fgets(cNum, sizeof cNum, stdin))
{
num = atoi(cNum);
}
else //handle error
See also strtol() for a more robust solution.
Alright, so I played around with it some more and did a little more studying on scanf() format syntax, and figured out a solution. Apparently, putting the whitespace character at the end of my scanf call there tells it to keep reading until it finds something AFTER the whitespace, so of course it would hang up there until you give more input.
Remove the whitespace character in the scanf formatter, and add a leading space to the following scanf call.
The reason I had used fgets originally was so I could specify a buffer length to avoid overflow. Apparently, the same effect can be achieved using %32s in the scanf call. The fixed code looks like this:
#include <stdio.h>
int main()
{
int number;
char string[32];
printf("Input the number.\n> ");
scanf("%d", &number);
printf("\nInput the string.\n> ");
scanf(" %32s", &string);
printf("\nnumber: %d\nstring: %s\n", number, string);
}

inside while loop used fflush does not work

#include<stdio.h>
int main(){
char arr[20];
int a;
int count=0;
while(1){
printf("enter string with space:");
fflush(stdin);
scanf("%[^\n]%*c", arr);
printf("enter integer:");
scanf("%d",&a);
count++;
if(count==4){
break;
}
}
}
I want to read string with space and I have to use this code. I wrote this in while loop and I want to read string and after integer number and it does not work.
Output:
enter string with space:hey hey
enter integer:159
enter string with space:enter integer:
But output should be:
enter string with space:hey hey
enter integer:159
enter string with space:.... heyy
enter integer: 1235
fflush() is not defined input streams such as stdin. You will find that it works on Microsoft's C library, but not GNU.
Your scanf() call is incorrect. The first format specifier is incorrect, and you have no format specifier for string ;
scanf("%*[^\n]%*c") ;
will discard any remaining buffered characters to the end of the line inclusive. It accepts no input - it is not clear what string is or what you want to do with it. I'd suggest using a separate input call for that, for clarity.
Another somewhat less arcane solution is:
int c ;
while ((c = getchar()) != '\n' && c != EOF) { }
enter string with space:enter integer:
This happens because after you enter integer input, \n is still left in the stream. %[ does not skip leading whitespace unlike most other format specifiers. Change scanf("%[^\n]%*c", arr); to scanf(" %[^\n]%*c", arr); so it manually skips the leading whitespace before taking input.
As for fflush(stdin), that is undefined behavior (see Using fflush(stdin)). #Clifford already provides two nice solutions to that issue.

C Program :Error when reading character from stdin using scanf

Currently im trying to learn simple C Programs. But, i came into this situation :
#include<conio.h>
#include<stdio.h>
void main()
{
char c;
int tryagain=1;
while(tryagain>0){
printf("Enter the Character : ");
scanf("%c",&c);
printf("You entered the character \"%c\" and the ascii value is %d",c,c);
getch();
clrscr();
tryagain=0;
printf("You want to Trry again Press 1 : ");
scanf("%d",&tryagain);
clrscr();
}
}
The program is fine when user first enter a character. And, when it ask to continue. And, user enter 1 then it is behaving weired. It automatically input blank character and prints the ascii and goto the same place.
How can i resolve this? And, specially, Why is the reason for this?
And, Im sorry about my poor english!
Thank you in Advance.
When you use
scanf("%d",&tryagain);
the number is read into tryagain but the newline character, '\n', is still left on the input stream. The next time you use:
scanf("%c",&c);
the newline character is read into the c.
By using
scanf("%d%*c",&tryagain);
the newline is read from the input stream but it is not stored anywhere. It is simply discarded.
The issue is that you are reading a single number in the second scanf, but user inputs more than a single number there, the user also input a new line character by pressing .
User enters "1\n". Your scanf reads "1", leaving out "\n" in the input stream. Then the next scanf that reads a character reads "\n" from the stream.
Here is the corrected code. I use getc to discard the extra new line character that is there.
#include <stdio.h>
void main()
{
char c;
int tryagain = 1;
while (tryagain > 0) {
printf("Enter a character: ");
scanf("%c", &c);
printf("You entered the character \"%c\" and the ascii value is %d\n", c, c);
tryagain = 0;
printf("If you want to try again, enter 1: ");
scanf("%d", &tryagain);
// get rid of the extra new line character
getc(stdin);
}
}
Also, as a side note, you use conio.h which is not part of standard C, it's MS-DOS header file, thus it's not portable C you are writing. I have removed it from my code, but you might wish to keep it.

scanf() function doesn't work? [duplicate]

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 may be a simple question, but i searched a lot and still didn't figure it out.
I compiles below snip code by gcc and run program from terminal. In correct, It allow to enter an int and a char but it doesn't. It doesn't wait to enter the char??
Anyone here can help me will be kind. thanks in advance!
#include <stdio.h>
int main()
{
char c;
int i;
// a
printf("i: ");
fflush(stdin); scanf("%d", &i);
// b
printf("c: ");
fflush(stdin); scanf("%c", &c);
return 0;
}
%d will read consecutive digits until it encounters a non-digit. %c reads one character. Probably what's happening is that you're giving it a number (several digits) followed by a new line. %c then reads in that new line. You were probably intending for the fflush(stdin); to discard anything that hadn't yet been read, but unfortunately, that's undefined behavior.
The solution is to discard all whitespace before reading the character:
scanf(" %c", &c);
Note the space at the start. That means to discard all whitespace.
You can use the getchar() to achieve what you want.
or consume the extra newline by using:-
scanf(" %c", &c);
^^^ <------------Note the space
Reason:- Your next scanf for reading the character just reads/consumes the newline and hence never waits for user input
Instead of fflush(stdin); scanf("%c", &c);
1.use scanf with extra space
scanf(" %c",&c);
or
2.use getchar() two times , first time reads '\n' which is entered after giving integer input and second time call ask you for give input as c:
getchar();
c=getchar();
would help you.
First of all, scanf works when used as directed. I think the following code does what you want. Stdout is flushed so that user is prompted to enter an integer or a character. Using %1s allows white space like \n.
int main()
{
char c[2];
int i;
printf("i: ");
fflush(stdout);
scanf("%d", &i);
printf("c: ");
fflush(stdout);
scanf("%1s", &c);
printf("\ni = %d, c = %c", i, c[0]);
return 0;
}
This code was tested/run on an Eclipse/Microsoft C compiler.
That fflush() is not guaranteed to do anything, and gcc/g++ doesn't. Not on Linux, anyway.
I thought I invented the following way to flush the rest of a line...until I saw it as an example in the ISO C spec (90 or 99...forgot which, but it's been there a long time either way...and I'll bet most readers here have seen it before.)
scanf("%*[^\n]%*c"); /* discard everything up to and including the next newline */
You can put that in your own "flush" function to save typing or pasting that all over the place.
You should still follow the suggestions to to put a space in scanf(" %c", &c);.
That will patiently wait for a non-whitespace character in case of a leading space or a double hit of the enter key.

Odd loop does not work using %c [duplicate]

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.
I am leaning C programming. I have written an odd loop but doesn't work while I use %c in scanf().Here is the code:
#include<stdio.h>
void main()
{
char another='y';
int num;
while ( another =='y')
{
printf("Enter a number:\t");
scanf("%d", &num);
printf("Sqare of %d is : %d", num, num * num);
printf("\nWant to enter another number? y/n");
scanf("%c", &another);
}
}
But if I use %s in this code, for example scanf("%s", &another);, then it works fine.Why does this happen? Any idea?
The %c conversion reads the next single character from input, regardless of what it is. In this case, you've previously read a number using %d. You had to hit the enter key for that number to be read, but you haven't done anything to read the new-line from the input stream. Therefore, when you do the %c conversion, it reads that new-line from the input stream (without waiting for you to actually enter anything, since there's already input waiting to be read).
When you use %s, it skips across any leading white-space to get some character other than white-space. It treats a new-line as white-space, so it implicitly skips across that waiting new-line. Since there's (presumably) nothing else waiting to be read, it proceeds to wait for you to enter something, as you apparently desire.
If you want to use %c for the conversion, you could precede it with a space in the format string, which will also skip across any white-space in the stream.
The ENTER key is lying in the stdin stream, after you enter a number for first scanf %d. This key gets captured by the scanf %c line.
use scanf("%1s",char_array); another=char_array[0];.
use getch() instead of scanf() in this case. Because scanf() expects '\n' but you are accepting only one char at that scanf(). so '\n' given to next scanf() causing confusion.
#include<stdio.h>
void main()
{
char another='y';
int num;
while ( another =='y')
{
printf("Enter a number:\t");
scanf("%d", &num);
printf("Sqare of %d is : %d", num, num * num);
printf("\nWant to enter another number? y/n");
getchar();
scanf("%c", &another);
}
}

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