Still in learning mode and may be the following question is a really dumb but I dont have any idea why it is happening..
#include<stdio.h>
int main()
{
/* code to accept string and then a character from stdin */
char str[20], inp;
/*take string from stdin */
printf("string:\n");
scanf("%s",str);
fflush(stdin);
/*input a character */
printf("char:\n");
scanf("%c",&inp);/* code does not reach this point and exits */
}
As mentioned in the comment, after I input the string , for eg. 'strng' the code just exits after printing char: but it does not wait for me input the character. As per my understanding, I have given the size of the array large enough to store the string and if the string entered is smaller than the size of the str array, the compiler will automatically add null character at the end of the string and proceed further. Am I missing something or is there a mistake in my code. Please suggest.
Thanks.
Try removing fflush(stdin);
and put a space before %c in scanf(" %c",&inp);
First of all fflush(stdin) is wrong. Many people recommend it but it is plain and simple undefined.
The problem is caused by scanf leaving \n in the input buffer because "%s" doesn't read whitespace characters. When scanf("%c"..) is reached, it is immediately "satisfied" and fills inp with \n and calls it a day.
The problem is common enough, see these C FAQs:
Scanf interlace
Scanf problems
One (possibly dangerous) solution is to discard \n input:
while((c = getchar()) != '\n' && c != EOF)
;
Another solution might be to use fgets and parse that, or possibly read one character at a time with getc, or maybe tweak the second scamf to discard whitespace characters.
Put a space before the %c in the second scanf like this:
scanf(" %c",&inp)
And as stated by others fflush is defined only for output streams.
Related
This question already has answers here:
scanf() leaves the newline character in the buffer
(7 answers)
Closed 6 years ago.
I have this block of code (functions omitted as the logic is part of a homework assignment):
#include <stdio.h>
int main()
{
char c = 'q';
int size;
printf("\nShape (l/s/t):");
scanf("%c",&c);
printf("Length:");
scanf("%d",&size);
while(c!='q')
{
switch(c)
{
case 'l': line(size); break;
case 's': square(size); break;
case 't': triangle(size); break;
}
printf("\nShape (l/s/t):");
scanf("%c",&c);
printf("\nLength:");
scanf("%d",&size);
}
return 0;
}
The first two Scanf's work great, no problem once we get into the while loop, I have a problem where, when you are supposed to be prompted to enter a new shape char, it instead jumps down to the printf of Length and waits to take input from there for a char, then later a decimal on the next iteration of the loop.
Preloop iteration:
Scanf: Shape. Works Great
Scanf: Length. No Problem
Loop 1.
Scanf: Shape. Skips over this
Scanf: length. Problem, this scanf maps to the shape char.
Loop 2
Scanf: Shape. Skips over this
Scanf: length. Problem, this scanf maps to the size int now.
Why is it doing this?
scanf("%c") reads the newline character from the ENTER key.
When you type let's say 15, you type a 1, a 5 and then press the ENTER key. So there are now three characters in the input buffer. scanf("%d") reads the 1 and the 5, interpreting them as the number 15, but the newline character is still in the input buffer. The scanf("%c") will immediately read this newline character, and the program will then go on to the next scanf("%d"), and wait for you to enter a number.
The usual advice is to read entire lines of input with fgets, and interpret the content of each line in a separate step. A simpler solution to your immediate problem is to add a getchar() after each scanf("%d").
The basic problem is that scanf() leaves the newline after the number in the buffer, and then reads it with %c on the next pass. Actually, this is a good demonstration of why I don't use scanf(); I use a line reader (fgets(), for example) and sscanf(). It is easier to control.
You can probably rescue it by using " %c" instead of "%c" for the format string. The blank causes scanf() to skip white space (including newlines) before reading the character.
But it will be easier in the long run to give up scanf() and fscanf() and use fgets() or equivalent plus sscanf(). All else apart, error reporting is much easier when you have the whole string to work with, not the driblets left behind by scanf() after it fails.
You should also, always, check that you get a value converted from scanf(). Input fails — routinely and horribly. Don't let it wreck your program because you didn't check.
Try adding a space in the scanf.
scanf(" %d", &var);
// ^
// there
This will cause scanf() to discard all whitespace before matching an integer.
Use the function
void seek_to_next_line( void )
{
int c;
while( (c = fgetc( stdin )) != EOF && c != '\n' );
}
to clear out your input buffer.
The '\n' character is still left on the input stream after the first call to scanf is completed, so the second call to scanf() reads it in. Use getchar().
When you type the shape and ENTER, the shape is consumed by the first scanf, but the ENTER is not! The second scanf expects a number so, the ENTER is skipped because is considered a white space, and the scanf waits for a valid input ( a number) that, again, is terminated by the ENTER. Well, the number is consumed, but the ENTER is not, so the first scanf inside the while uses it and your shape prompt is skipped... this process repeats. You have to add another %c in the scanfs to deal with the ENTER key. I hope this helps!
You can also use
scanf("%c%*c", &c);
to read two characters and ignore the last one (in this case '\n')
I'm just asking what does the getchar do in this code and how does it work? I don't understand why the getchar affects the code, to me it seems as if its just getting the value but nothing is being done with the value.
int c=0;
while (c>=0)
{
scanf("%d", &c);
getchar();
}
Some possibilities of why getchar() might have been used there:
1) If it's done to ignore whitespaces (typically used when scanning chars with %c), it's not needed here because %d ignores whitespaces anyway.
2) Other possibility is that after this loop, some further scanning is done where the last \n left by the last call to scanf() might be a problem. So, getchar() might be used to ignore it.
3) In case you enter characters do not match %d, scanf() will fail. In that the characters you entered are left in the input stream and you'll never be able to read an int again (For example, if you input abcdddgdfg without that getchar() call). So, getchar() here will consume all those
chars (one per iteration) and eventually you'll be able to read int (using %d) again.
But this is all really not needed; it's just an attempt to fix flaws of scanf(). Reading inputs using scanf() and getting it correct is really difficult. That's why it's always recommended to use fgets() and parse using sscanf() or using strto*() functions if you are just scanning integers.
See: Why does everyone say not to use scanf? What should I use instead?
In this code, getchar is being called for its side effects: it reads a character from standard input and throws it away.
Probably this is reading input from the user. scanf will consume a number, but leave the newline character after the number untouched. The getchar consumes the newline and throws it away. This isn't strictly necessary in this loop, because the next scanf will skip over whitespace to find the next number, but it might be useful if the code after the loop isn't expecting to have a newline as the first thing on stdin.
This code is buggy, because it doesn't check for EOF, because it doesn't do anything sensible when the input is not a number or when there's more text on the line after the number, and because it uses scanf, which is broken-as-specified (for instance, it's allowed to crash the program if the input overflows the range of an int). Better code would be something like
char *linep = 0;
size_t asize = 0;
char *endp;
long c;
while (getline(&linep, &asize, stdin) > 0) {
errno = 0;
c = strtol(linep, &endp, 10);
if (linep == endp || *endp != '\0' || errno) {
puts("?Redo from start");
continue;
}
if (c == 0) break;
do_something_with(c);
}
free(linep);
Most likely the code is for reading in a list of integers, separated by a new line.
scanf will read in an integer, and put it into variable c.
The getchar is reading in the next character (assuming a new line)
Since it doesn't check, there is some potential that it wasn't a new line, or that scanf failed as the what it tried to read wasn't a number.
getchar(); is simply reading and consuming the character after the number, be it a space, comma, new line or the beginning of another integer or anything else.
IMO, this is not robust code. Good code would 1) at least test the result of scanf() and 2) test or limit the consumption of the following character to prevent "eating" a potential sign of the following number. Remember code cannot control what a user types, but has to cope with whatever is entered.
v space
"123 456"
v comma
"123,456"
v negative sign
"123-456"
This question already has answers here:
scanf() leaves the newline character in the buffer
(7 answers)
Closed 6 years ago.
I have this block of code (functions omitted as the logic is part of a homework assignment):
#include <stdio.h>
int main()
{
char c = 'q';
int size;
printf("\nShape (l/s/t):");
scanf("%c",&c);
printf("Length:");
scanf("%d",&size);
while(c!='q')
{
switch(c)
{
case 'l': line(size); break;
case 's': square(size); break;
case 't': triangle(size); break;
}
printf("\nShape (l/s/t):");
scanf("%c",&c);
printf("\nLength:");
scanf("%d",&size);
}
return 0;
}
The first two Scanf's work great, no problem once we get into the while loop, I have a problem where, when you are supposed to be prompted to enter a new shape char, it instead jumps down to the printf of Length and waits to take input from there for a char, then later a decimal on the next iteration of the loop.
Preloop iteration:
Scanf: Shape. Works Great
Scanf: Length. No Problem
Loop 1.
Scanf: Shape. Skips over this
Scanf: length. Problem, this scanf maps to the shape char.
Loop 2
Scanf: Shape. Skips over this
Scanf: length. Problem, this scanf maps to the size int now.
Why is it doing this?
scanf("%c") reads the newline character from the ENTER key.
When you type let's say 15, you type a 1, a 5 and then press the ENTER key. So there are now three characters in the input buffer. scanf("%d") reads the 1 and the 5, interpreting them as the number 15, but the newline character is still in the input buffer. The scanf("%c") will immediately read this newline character, and the program will then go on to the next scanf("%d"), and wait for you to enter a number.
The usual advice is to read entire lines of input with fgets, and interpret the content of each line in a separate step. A simpler solution to your immediate problem is to add a getchar() after each scanf("%d").
The basic problem is that scanf() leaves the newline after the number in the buffer, and then reads it with %c on the next pass. Actually, this is a good demonstration of why I don't use scanf(); I use a line reader (fgets(), for example) and sscanf(). It is easier to control.
You can probably rescue it by using " %c" instead of "%c" for the format string. The blank causes scanf() to skip white space (including newlines) before reading the character.
But it will be easier in the long run to give up scanf() and fscanf() and use fgets() or equivalent plus sscanf(). All else apart, error reporting is much easier when you have the whole string to work with, not the driblets left behind by scanf() after it fails.
You should also, always, check that you get a value converted from scanf(). Input fails — routinely and horribly. Don't let it wreck your program because you didn't check.
Try adding a space in the scanf.
scanf(" %d", &var);
// ^
// there
This will cause scanf() to discard all whitespace before matching an integer.
Use the function
void seek_to_next_line( void )
{
int c;
while( (c = fgetc( stdin )) != EOF && c != '\n' );
}
to clear out your input buffer.
The '\n' character is still left on the input stream after the first call to scanf is completed, so the second call to scanf() reads it in. Use getchar().
When you type the shape and ENTER, the shape is consumed by the first scanf, but the ENTER is not! The second scanf expects a number so, the ENTER is skipped because is considered a white space, and the scanf waits for a valid input ( a number) that, again, is terminated by the ENTER. Well, the number is consumed, but the ENTER is not, so the first scanf inside the while uses it and your shape prompt is skipped... this process repeats. You have to add another %c in the scanfs to deal with the ENTER key. I hope this helps!
You can also use
scanf("%c%*c", &c);
to read two characters and ignore the last one (in this case '\n')
I want to execute a statement based on the input of the user:
#include <stdio.h>
#include <string.h>
void main() {
char string_input[40];
int i;
printf("Enter data ==> ");
scanf("%s", string_input);
if (string_input[0] == '\n') {
printf("ERROR - no data\n");
}
else if (strlen(string_input) > 40) {
printf("Hex equivalent is ");
}
else {
printf("Hex equivalent is ");
}
}
When I run it, and just press enter, it goes to a new line instead of saying "ERROR - no data".
What do I do?
CANNOT USE FGETS as we have not gone over this in class.
Use
char enter[1];
int chk = scanf("%39[^\n]%c", string_input, enter);
but string_input will not have a '\n' inside. Your test
if (string_input[0] == '\n') {
printf("ERROR - no data\n");
}
will have to be changed to, for example
if (chk != 2) {
printf("ERROR - bad data\n");
}
use fgets instead of scanf. scanf doesn't check if user enters a string longer than 40 chars in your example above so for your particular case fgets should be simpler(safer).
Can you use a while loop and getch, then test for the <Enter> key on each keystroke?
scanf won't return until it sees something other than whitespace. It also doesn't distinguish between newlines and other whitespace. In practice, using scanf is almost always a mistake; I suggest that you call fgets instead and then (if you need to) use sscanf on the resulting data.
If you do that, you really ought to deal with the possibility that the user enters a line longer than the buffer you pass to fgets; you can tell when this has happened because your entire buffer gets filled and the last character isn't a newline. In that situation, you should reallocate a larger buffer and fgets again onto the end of it, and repeat until either you see a newline or the buffer gets unreasonably large.
You should really be similarly careful when calling scanf or sscanf -- what if the user enters a string 100 characters long? (You can tell scanf or sscanf to accept only a limited length of string.)
On the other hand, if this is just a toy program you can just make your buffer reasonably long and hope the user doesn't do anything nasty.
fgets does what you need. Avoid using scanf or gets. If you can't use fgets try using getchar
The problem is that "%s" attempts to skip white-space, and then read a string -- and according to scanf, a new-line is "whitespace".
The obvious alternative would be to use "%c" instead of "%s". The difference between the two is that "%c" does not attempt to skip leading whitespace.
A somewhat less obvious (or less known, anyway) alternative would be to use "%[^\n]%*[\n]". This reads data until it encounters a new-line, then reads the new-line and doesn't assign it to anything.
Regardless of which conversion you use, you want (need, really) to limit the amount of input entered so it doesn't overflow the buffer you've provided, so you'd want to use "%39c" or "%39[^\n]". Note that when you're specifying the length for scanf, you need to subtract one to leave space for the NUL terminator (in contrast to fgets, for which you specify the full buffer size).
What platform are you running on?
Is the character sent when your press the ENTER key actually '\n', or might it be '\r'? Or even both one after the other (ie. "\r\n").
I'm having a weird problem
i'm trying to read a string from a console with scanf()
like this
scanf("%[^\n]",string1);
but it doesnt read anything. it just skips the entire scanf.
I'm trying it in gcc compiler
Trying to use scanf to read strings with spaces can bring unnecessary problems of buffer overflow and stray newlines staying in the input buffer to be read later. gets() is often suggested as a solution to this, however,
From the manpage:
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.
So instead of using gets, use fgets with the STDIN stream to read strings from the keyboard
That should work fine, so something else is going wrong. As hobbs suggests, you might have a newline on the input, in which case this won't match anything. It also won't consume a newline, so if you do this in a loop, the first call will get up to the newline and then the next call will get nothing. If you want to read the newline, you need another call, or use a space in the format string to skip whitespace. Its also a good idea to check the return value of scanf to see if it actually matched any format specifiers.
Also, you probably want to specify a maximum length in order to avoid overflowing the buffer. So you want something like:
char buffer[100];
if (scanf(" %99[^\n]", buffer) == 1) {
/* read something into buffer */
This will skip (ignore) any blank lines and whitespace on the beginning of a line and read up to 99 characters of input up to and not including a newline. Trailing or embedded whitespace will not be skipped, only leading whitespace.
I'll bet your scanf call is inside a loop. I'll bet it works the first time you call it. I'll bet it only fails on the second and later times.
The first time, it will read until it reaches a newline character. The newline character will remain unread. (Odds are that the library internally does read it and calls ungetc to unread it, but that doesn't matter, because from your program's point of view the newline is unread.)
The second time, it will read until it reaches a newline character. That newline character is still waiting at the front of the line and scanf will read all 0 of the characters that are waiting ahead of it.
The third time ... the same.
You probably want this:
if (scanf("%99[^\n]%*c", buffer) == 1) {
Edit: I accidentally copied and pasted from another answer instead of from the question, before inserting the %*c as intended. This resulting line of code will behave strangely if you have a line of input longer than 100 bytes, because the %*c will eat an ordinary byte instead of the newline.
However, notice how dangerous it would be to do this:
scanf("%[^n]%*c", string1);
because there, if you have a line of input longer than your buffer, the input will walk all over your other variables and stack and everything. This is called buffer overflow (even if the overflow goes onto the stack).
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char *text(int n);
int main()
{
char str[10];
printf("enter username : ");
scanf(text(9),str);
printf("username = %s",str);
return 0;
}
char *text(int n)
{
fflush(stdin);fflush(stdout);
char str[50]="%",buf[50],st2[10]="[^\n]s";
char *s;itoa(n,buf,10);
// n == -1 no buffer protection
if(n != -1) strcat(str,buf);
strcat(str,st2);s=strdup(str);
fflush(stdin);fflush(stdout);
return s;
}