scanf anomaly with %c and %s related behaviour - c

I need to use scanf to get a character and a string which would store the user's answer. (yes/no)
The code below skips scanf("%c", &elem).
while ( !strcmp ("yes", option))
{
printf("enter the elements \n\n");
elem=getchar();
printf("you have entered %c\n",elem);
enqueue(st, elem);
printf("please enter yes or no ");
scanf("%s[^\n]",option);
}
./out
enter the elements
a
you have entered a
enqueue elem= a
please enter yes or no yes
enter the elements
you have entered
enqueue elem=

You don't have any scanf("%c", &elem) in your code... btw the problem is with the enter for scanf. When you get an input by scanf, an enter character stays in the input buffer which will be read by your getchar() function in the second round. one simple way to solve it is to add a dummy getchar after your scanf line:
while ( !strcmp ("yes",option))
{
printf("enter the elements \n\n");
elem=getchar();
printf("you have entered %c\n",elem);
enqueue(st,elem);
printf("please enter yes or no ");
scanf("%s[^\n]",option);
getchar();
}
You can find more information about how to clear your input buffer here: How to clear input buffer in C?
I can recommend you consider two things:
For getting only a character, I personally found it much more easier to use getch and getche function in Windows, and equivalent of them for GCC-compatible environments. You can find samples of it online or on this line [What is Equivalent to getch() & getche() in Linux?
Always flush the input buffer after you read your input to prevent any similar problems to happen.
The input functions check the input buffer, which you can find at 0xb8000000, and check the first input there. If the buffer is empty, they wait for the user to enter the input, otherwise, they check the first element in the buffer and then examine that to what they expect to read. If they succeed, they read it and remove it from buffer. Otherwise, they fail to give you your input and depending on the function, the result is different.
For Example, consider the following line:
scanf("%d %d %f", &a, &b &c);
and give the input as:
a 2 4
The scanf will return 0, which means it reads zero inputs so 'a', 2, and 4 remains in your buffer. So your buffer looks like: [a, 2, 4]. As a result if you add the following line:
scanf("%c", &ch);
scanf will try to get a character from the buffer, and it reads character 'a' and put it in variable ch. So it doesn't get any input from user. And you end up with having 2 and 4 on your buffer again.

When you are pressing Enter/Return key to enter the element then a \n character is also passed to the buffer along with the element. This \n is read by your getchar on next call.
To consume this \n place this line after the getchar();
int ch;
while((ch = getchar()) != EOF && ch != '\n');

Take care mixing scanf() format specifiers "%c", "%s" and "%[]".
Correct usage of "%[^\n]": there is no s. If leading whitespace in not wanted to be saved, include a leading space as in " %[^\n]"
char option[100];
// scanf("%s[^\n]", option);
scanf(" %[^\n]", option);
// or better
scanf(" %99[^\n]", option);
// or pedantic
switch (scanf(" %99[^\n]", option)) {
case EOF: HandleEOForIOError(); break;
case 0: HandleNoData(); break; // might not be possible here.
case 1: HandleSuccess();
Correct usage of "%c". If leading whitespace in not wanted to be save, include a leading space as in " %c". This may be the case in OP's code so the preceding inputs Enter or '\n' is consumed.
char elem;
scanf(" %c", &elem);
Correct usage of "%s". Leading whitespace is not saved with or without a leading space.
char option[100];
scanf("%99s", option);
// Following works the same.
scanf(" %99s", option);

Related

Not able to take the user input when using if-else and switch case [ C language] [duplicate]

I have the following program:
int main(int argc, char *argv[])
{
int a, b;
char c1, c2;
printf("Enter something: ");
scanf("%d", &a); // line 1
printf("Enter other something: ");
scanf("%d", &b); // line 2
printf("Enter a char: ");
scanf("%c", &c1); // line 3
printf("Enter another char: ");
scanf("%c", &c2); // line 4
printf("Done"); // line 5
system("PAUSE");
return 0;
}
As I read in the C book, the author says that scanf() left a newline character in the buffer, therefore, the program does not stop at line 4 for user to enter the data, rather it stores the new line character in c2 and moves to line 5.
Is that right?
However, does this only happen with char data types? Because I did not see this problem with int data types as in line 1, 2, 3. Is it right?
The scanf() function skips leading whitespace automatically before trying to parse conversions other than characters. The character formats (primarily %c; also scan sets %[…] — and %n) are the exception; they don't skip whitespace.
Use " %c" with a leading blank to skip optional white space. Do not use a trailing blank in a scanf() format string.
Note that this still doesn't consume any trailing whitespace left in the input stream, not even to the end of a line, so beware of that if also using getchar() or fgets() on the same input stream. We're just getting scanf to skip over whitespace before conversions, like it does for %d and other non-character conversions.
Note that non-whitespace "directives" (to use POSIX scanf terminology) other than conversions, like the literal text in scanf("order = %d", &order); doesn't skip whitespace either. The literal order has to match the next character to be read.
So you probably want " order = %d" there if you want to skip a newline from the previous line but still require a literal match on a fixed string, like this question.
Use scanf(" %c", &c2);. This will solve your problem.
Another option (that I got from here) is to read and discard the newline by using the assignment-supression option. To do that, we just put a format to read a character with an asterisk between % and c:
scanf("%d%*c",&a); // line 1
scanf("%c%*c",&c1); // line 3
scanf will then read the next char (that is, the newline) but not assign it to any pointer.
In the end, however, I would second the FAQ's last option:
Or, depending on your requirements, you could also forget about scanf()/getchar(), use fgets() to get a line of text from the user and parse it yourself.
To echo what I have posted in another answer about C++: I suggest to toss scanf() away, to never use it, and to instead use fgets() and sscanf().
The reason for this is, that at least in Unix-like systems by default, the terminal your CLI program runs on does some processing of the user input before your program sees it. It buffers input until a newline is entered, and allows for some rudimentary line editing, like making backspace work.
So, you can never get a single character at a time, or a few single characters, just a full line. But that's not what e.g. scanf("%d") processes, instead it processes just the digits, and stops there, leaving the rest buffered in the C library, for a future stdio function to use. If your program has e.g.
printf("Enter a number: ");
scanf("%d", &a);
printf("Enter a word: ");
scanf("%s", word);
and you enter the line 123 abcd, it completes both scanf()s at once, but only after a newline is given. The first scanf() doesn't return when a user has hit space, even though that's where the number ends (because at that point the line is still in the terminal's line buffer); and the second scanf() doesn't wait for you to enter another line (because the input buffer already contains enough to fill the %s conversion).
This isn't what users usually expect!
Instead, they expect that hitting enter completes the input, and if you hit enter, you either get a default value, or an error, with possibly a suggestion to please really just give the answer.
You can't really do that with scanf("%d"). If the user just hits enter, nothing happens. Because scanf() is still waiting for the number. The terminal sends the line onward, but your program doesn't see it, because scanf() eats it. You don't get a chance to react to the user's mistake.
That's also not very useful.
Hence, I suggest using fgets() or getline() to read a full line of input at a time. This exactly matches what the terminal gives, and always gives your program control after the user has entered a line. What you do with the input line is up to you, if you want a number, you can use atoi(), strtol(), or even sscanf(buf, "%d", &a) to parse the number. sscanf() doesn't have the same mismatch as scanf(), because the buffer it reads from is limited in size, and when it ends, it ends -- the function can't wait for more.
(fscanf() on a regular file can also be fine if the file format is one that supports how it skims over newlines like any whitespace. For line-oriented data, I'd still use fgets() and sscanf().)
So, instead of what I had above, use something like this:
printf("Enter a number: ");
fgets(buf, bufsize, stdin);
sscanf(buf, "%d", &a);
or, actually, check the return value of sscanf() too, so you can detect empty lines and otherwise invalid data:
#include <stdio.h>
int main(void)
{
const int bufsize = 100;
char buf[bufsize];
int a;
int ret;
char word[bufsize];
printf("Enter a number: ");
fgets(buf, bufsize, stdin);
ret = sscanf(buf, "%d", &a);
if (ret != 1) {
fprintf(stderr, "Ok, you don't have to.\n");
return 1;
}
printf("Enter a word: ");
fgets(buf, bufsize, stdin);
ret = sscanf(buf, "%s", word);
if (ret != 1) {
fprintf(stderr, "You make me sad.\n");
return 1;
}
printf("You entered %d and %s\n", a, word);
}
Of course, if you want the program to insist, you can create a simple function to loop over the fgets() and sscanf() until the user deigns to do what they're told; or to just exit with an error immediately. Depends on what you think your program should do if the user doesn't want to play ball.
You could do something similar e.g. by looping over getchar() to read characters until a newline after scanf("%d") returned, thus clearing up any garbage left in the buffer, but that doesn't do anything about the case where the user just hits enter on an empty line. Anyway, fgets() would read until a newline, so you don't have to do it yourself.
Use getchar() before calling second scanf().
scanf("%c", &c1);
getchar(); // <== remove newline
scanf("%c", &c2);
Two workarounds. One is catch the extra whitespace trickly.
getchar(); // (1) add `getchar()`
scanf("%c", &c1);
scanf(" %c", &c1); // (2) add whitespace before %c
The other is using fgets() to instead scanf(). Note: gets() is unsafe, see why gets is dangerous
By contrast, I would prefer to recommend the second way. Because it's more readable and maintainable. For example, in first way, you must think a while before adding/moving some scanf().
/*Take char input using scanf after int input using scanf just use fflush(stdin) function after int input */
#include<stdio.h>
#include<conio.h>
void main()
{
int x;
char y;
clrscr();
printf(" enter an int ");
scanf("%d",&x);
fflush(stdin);
printf("\n Now enter a char");
scanf("%c",&y);
printf("\n X=%d and Y=%c",x,y);
getch();
}

While using structure I used gets to get string input for first case but it is not working for other case see? [duplicate]

I have the following program:
int main(int argc, char *argv[])
{
int a, b;
char c1, c2;
printf("Enter something: ");
scanf("%d", &a); // line 1
printf("Enter other something: ");
scanf("%d", &b); // line 2
printf("Enter a char: ");
scanf("%c", &c1); // line 3
printf("Enter another char: ");
scanf("%c", &c2); // line 4
printf("Done"); // line 5
system("PAUSE");
return 0;
}
As I read in the C book, the author says that scanf() left a newline character in the buffer, therefore, the program does not stop at line 4 for user to enter the data, rather it stores the new line character in c2 and moves to line 5.
Is that right?
However, does this only happen with char data types? Because I did not see this problem with int data types as in line 1, 2, 3. Is it right?
The scanf() function skips leading whitespace automatically before trying to parse conversions other than characters. The character formats (primarily %c; also scan sets %[…] — and %n) are the exception; they don't skip whitespace.
Use " %c" with a leading blank to skip optional white space. Do not use a trailing blank in a scanf() format string.
Note that this still doesn't consume any trailing whitespace left in the input stream, not even to the end of a line, so beware of that if also using getchar() or fgets() on the same input stream. We're just getting scanf to skip over whitespace before conversions, like it does for %d and other non-character conversions.
Note that non-whitespace "directives" (to use POSIX scanf terminology) other than conversions, like the literal text in scanf("order = %d", &order); doesn't skip whitespace either. The literal order has to match the next character to be read.
So you probably want " order = %d" there if you want to skip a newline from the previous line but still require a literal match on a fixed string, like this question.
Use scanf(" %c", &c2);. This will solve your problem.
Another option (that I got from here) is to read and discard the newline by using the assignment-supression option. To do that, we just put a format to read a character with an asterisk between % and c:
scanf("%d%*c",&a); // line 1
scanf("%c%*c",&c1); // line 3
scanf will then read the next char (that is, the newline) but not assign it to any pointer.
In the end, however, I would second the FAQ's last option:
Or, depending on your requirements, you could also forget about scanf()/getchar(), use fgets() to get a line of text from the user and parse it yourself.
To echo what I have posted in another answer about C++: I suggest to toss scanf() away, to never use it, and to instead use fgets() and sscanf().
The reason for this is, that at least in Unix-like systems by default, the terminal your CLI program runs on does some processing of the user input before your program sees it. It buffers input until a newline is entered, and allows for some rudimentary line editing, like making backspace work.
So, you can never get a single character at a time, or a few single characters, just a full line. But that's not what e.g. scanf("%d") processes, instead it processes just the digits, and stops there, leaving the rest buffered in the C library, for a future stdio function to use. If your program has e.g.
printf("Enter a number: ");
scanf("%d", &a);
printf("Enter a word: ");
scanf("%s", word);
and you enter the line 123 abcd, it completes both scanf()s at once, but only after a newline is given. The first scanf() doesn't return when a user has hit space, even though that's where the number ends (because at that point the line is still in the terminal's line buffer); and the second scanf() doesn't wait for you to enter another line (because the input buffer already contains enough to fill the %s conversion).
This isn't what users usually expect!
Instead, they expect that hitting enter completes the input, and if you hit enter, you either get a default value, or an error, with possibly a suggestion to please really just give the answer.
You can't really do that with scanf("%d"). If the user just hits enter, nothing happens. Because scanf() is still waiting for the number. The terminal sends the line onward, but your program doesn't see it, because scanf() eats it. You don't get a chance to react to the user's mistake.
That's also not very useful.
Hence, I suggest using fgets() or getline() to read a full line of input at a time. This exactly matches what the terminal gives, and always gives your program control after the user has entered a line. What you do with the input line is up to you, if you want a number, you can use atoi(), strtol(), or even sscanf(buf, "%d", &a) to parse the number. sscanf() doesn't have the same mismatch as scanf(), because the buffer it reads from is limited in size, and when it ends, it ends -- the function can't wait for more.
(fscanf() on a regular file can also be fine if the file format is one that supports how it skims over newlines like any whitespace. For line-oriented data, I'd still use fgets() and sscanf().)
So, instead of what I had above, use something like this:
printf("Enter a number: ");
fgets(buf, bufsize, stdin);
sscanf(buf, "%d", &a);
or, actually, check the return value of sscanf() too, so you can detect empty lines and otherwise invalid data:
#include <stdio.h>
int main(void)
{
const int bufsize = 100;
char buf[bufsize];
int a;
int ret;
char word[bufsize];
printf("Enter a number: ");
fgets(buf, bufsize, stdin);
ret = sscanf(buf, "%d", &a);
if (ret != 1) {
fprintf(stderr, "Ok, you don't have to.\n");
return 1;
}
printf("Enter a word: ");
fgets(buf, bufsize, stdin);
ret = sscanf(buf, "%s", word);
if (ret != 1) {
fprintf(stderr, "You make me sad.\n");
return 1;
}
printf("You entered %d and %s\n", a, word);
}
Of course, if you want the program to insist, you can create a simple function to loop over the fgets() and sscanf() until the user deigns to do what they're told; or to just exit with an error immediately. Depends on what you think your program should do if the user doesn't want to play ball.
You could do something similar e.g. by looping over getchar() to read characters until a newline after scanf("%d") returned, thus clearing up any garbage left in the buffer, but that doesn't do anything about the case where the user just hits enter on an empty line. Anyway, fgets() would read until a newline, so you don't have to do it yourself.
Use getchar() before calling second scanf().
scanf("%c", &c1);
getchar(); // <== remove newline
scanf("%c", &c2);
Two workarounds. One is catch the extra whitespace trickly.
getchar(); // (1) add `getchar()`
scanf("%c", &c1);
scanf(" %c", &c1); // (2) add whitespace before %c
The other is using fgets() to instead scanf(). Note: gets() is unsafe, see why gets is dangerous
By contrast, I would prefer to recommend the second way. Because it's more readable and maintainable. For example, in first way, you must think a while before adding/moving some scanf().
/*Take char input using scanf after int input using scanf just use fflush(stdin) function after int input */
#include<stdio.h>
#include<conio.h>
void main()
{
int x;
char y;
clrscr();
printf(" enter an int ");
scanf("%d",&x);
fflush(stdin);
printf("\n Now enter a char");
scanf("%c",&y);
printf("\n X=%d and Y=%c",x,y);
getch();
}

Switch statement outputs the case followed by the default [duplicate]

I have the following program:
int main(int argc, char *argv[])
{
int a, b;
char c1, c2;
printf("Enter something: ");
scanf("%d", &a); // line 1
printf("Enter other something: ");
scanf("%d", &b); // line 2
printf("Enter a char: ");
scanf("%c", &c1); // line 3
printf("Enter another char: ");
scanf("%c", &c2); // line 4
printf("Done"); // line 5
system("PAUSE");
return 0;
}
As I read in the C book, the author says that scanf() left a newline character in the buffer, therefore, the program does not stop at line 4 for user to enter the data, rather it stores the new line character in c2 and moves to line 5.
Is that right?
However, does this only happen with char data types? Because I did not see this problem with int data types as in line 1, 2, 3. Is it right?
The scanf() function skips leading whitespace automatically before trying to parse conversions other than characters. The character formats (primarily %c; also scan sets %[…] — and %n) are the exception; they don't skip whitespace.
Use " %c" with a leading blank to skip optional white space. Do not use a trailing blank in a scanf() format string.
Note that this still doesn't consume any trailing whitespace left in the input stream, not even to the end of a line, so beware of that if also using getchar() or fgets() on the same input stream. We're just getting scanf to skip over whitespace before conversions, like it does for %d and other non-character conversions.
Note that non-whitespace "directives" (to use POSIX scanf terminology) other than conversions, like the literal text in scanf("order = %d", &order); doesn't skip whitespace either. The literal order has to match the next character to be read.
So you probably want " order = %d" there if you want to skip a newline from the previous line but still require a literal match on a fixed string, like this question.
Use scanf(" %c", &c2);. This will solve your problem.
Another option (that I got from here) is to read and discard the newline by using the assignment-supression option. To do that, we just put a format to read a character with an asterisk between % and c:
scanf("%d%*c",&a); // line 1
scanf("%c%*c",&c1); // line 3
scanf will then read the next char (that is, the newline) but not assign it to any pointer.
In the end, however, I would second the FAQ's last option:
Or, depending on your requirements, you could also forget about scanf()/getchar(), use fgets() to get a line of text from the user and parse it yourself.
To echo what I have posted in another answer about C++: I suggest to toss scanf() away, to never use it, and to instead use fgets() and sscanf().
The reason for this is, that at least in Unix-like systems by default, the terminal your CLI program runs on does some processing of the user input before your program sees it. It buffers input until a newline is entered, and allows for some rudimentary line editing, like making backspace work.
So, you can never get a single character at a time, or a few single characters, just a full line. But that's not what e.g. scanf("%d") processes, instead it processes just the digits, and stops there, leaving the rest buffered in the C library, for a future stdio function to use. If your program has e.g.
printf("Enter a number: ");
scanf("%d", &a);
printf("Enter a word: ");
scanf("%s", word);
and you enter the line 123 abcd, it completes both scanf()s at once, but only after a newline is given. The first scanf() doesn't return when a user has hit space, even though that's where the number ends (because at that point the line is still in the terminal's line buffer); and the second scanf() doesn't wait for you to enter another line (because the input buffer already contains enough to fill the %s conversion).
This isn't what users usually expect!
Instead, they expect that hitting enter completes the input, and if you hit enter, you either get a default value, or an error, with possibly a suggestion to please really just give the answer.
You can't really do that with scanf("%d"). If the user just hits enter, nothing happens. Because scanf() is still waiting for the number. The terminal sends the line onward, but your program doesn't see it, because scanf() eats it. You don't get a chance to react to the user's mistake.
That's also not very useful.
Hence, I suggest using fgets() or getline() to read a full line of input at a time. This exactly matches what the terminal gives, and always gives your program control after the user has entered a line. What you do with the input line is up to you, if you want a number, you can use atoi(), strtol(), or even sscanf(buf, "%d", &a) to parse the number. sscanf() doesn't have the same mismatch as scanf(), because the buffer it reads from is limited in size, and when it ends, it ends -- the function can't wait for more.
(fscanf() on a regular file can also be fine if the file format is one that supports how it skims over newlines like any whitespace. For line-oriented data, I'd still use fgets() and sscanf().)
So, instead of what I had above, use something like this:
printf("Enter a number: ");
fgets(buf, bufsize, stdin);
sscanf(buf, "%d", &a);
or, actually, check the return value of sscanf() too, so you can detect empty lines and otherwise invalid data:
#include <stdio.h>
int main(void)
{
const int bufsize = 100;
char buf[bufsize];
int a;
int ret;
char word[bufsize];
printf("Enter a number: ");
fgets(buf, bufsize, stdin);
ret = sscanf(buf, "%d", &a);
if (ret != 1) {
fprintf(stderr, "Ok, you don't have to.\n");
return 1;
}
printf("Enter a word: ");
fgets(buf, bufsize, stdin);
ret = sscanf(buf, "%s", word);
if (ret != 1) {
fprintf(stderr, "You make me sad.\n");
return 1;
}
printf("You entered %d and %s\n", a, word);
}
Of course, if you want the program to insist, you can create a simple function to loop over the fgets() and sscanf() until the user deigns to do what they're told; or to just exit with an error immediately. Depends on what you think your program should do if the user doesn't want to play ball.
You could do something similar e.g. by looping over getchar() to read characters until a newline after scanf("%d") returned, thus clearing up any garbage left in the buffer, but that doesn't do anything about the case where the user just hits enter on an empty line. Anyway, fgets() would read until a newline, so you don't have to do it yourself.
Use getchar() before calling second scanf().
scanf("%c", &c1);
getchar(); // <== remove newline
scanf("%c", &c2);
Two workarounds. One is catch the extra whitespace trickly.
getchar(); // (1) add `getchar()`
scanf("%c", &c1);
scanf(" %c", &c1); // (2) add whitespace before %c
The other is using fgets() to instead scanf(). Note: gets() is unsafe, see why gets is dangerous
By contrast, I would prefer to recommend the second way. Because it's more readable and maintainable. For example, in first way, you must think a while before adding/moving some scanf().
/*Take char input using scanf after int input using scanf just use fflush(stdin) function after int input */
#include<stdio.h>
#include<conio.h>
void main()
{
int x;
char y;
clrscr();
printf(" enter an int ");
scanf("%d",&x);
fflush(stdin);
printf("\n Now enter a char");
scanf("%c",&y);
printf("\n X=%d and Y=%c",x,y);
getch();
}

C : Printing strings inputed from user [duplicate]

I have the following program:
int main(int argc, char *argv[])
{
int a, b;
char c1, c2;
printf("Enter something: ");
scanf("%d", &a); // line 1
printf("Enter other something: ");
scanf("%d", &b); // line 2
printf("Enter a char: ");
scanf("%c", &c1); // line 3
printf("Enter another char: ");
scanf("%c", &c2); // line 4
printf("Done"); // line 5
system("PAUSE");
return 0;
}
As I read in the C book, the author says that scanf() left a newline character in the buffer, therefore, the program does not stop at line 4 for user to enter the data, rather it stores the new line character in c2 and moves to line 5.
Is that right?
However, does this only happen with char data types? Because I did not see this problem with int data types as in line 1, 2, 3. Is it right?
The scanf() function skips leading whitespace automatically before trying to parse conversions other than characters. The character formats (primarily %c; also scan sets %[…] — and %n) are the exception; they don't skip whitespace.
Use " %c" with a leading blank to skip optional white space. Do not use a trailing blank in a scanf() format string.
Note that this still doesn't consume any trailing whitespace left in the input stream, not even to the end of a line, so beware of that if also using getchar() or fgets() on the same input stream. We're just getting scanf to skip over whitespace before conversions, like it does for %d and other non-character conversions.
Note that non-whitespace "directives" (to use POSIX scanf terminology) other than conversions, like the literal text in scanf("order = %d", &order); doesn't skip whitespace either. The literal order has to match the next character to be read.
So you probably want " order = %d" there if you want to skip a newline from the previous line but still require a literal match on a fixed string, like this question.
Use scanf(" %c", &c2);. This will solve your problem.
Another option (that I got from here) is to read and discard the newline by using the assignment-supression option. To do that, we just put a format to read a character with an asterisk between % and c:
scanf("%d%*c",&a); // line 1
scanf("%c%*c",&c1); // line 3
scanf will then read the next char (that is, the newline) but not assign it to any pointer.
In the end, however, I would second the FAQ's last option:
Or, depending on your requirements, you could also forget about scanf()/getchar(), use fgets() to get a line of text from the user and parse it yourself.
To echo what I have posted in another answer about C++: I suggest to toss scanf() away, to never use it, and to instead use fgets() and sscanf().
The reason for this is, that at least in Unix-like systems by default, the terminal your CLI program runs on does some processing of the user input before your program sees it. It buffers input until a newline is entered, and allows for some rudimentary line editing, like making backspace work.
So, you can never get a single character at a time, or a few single characters, just a full line. But that's not what e.g. scanf("%d") processes, instead it processes just the digits, and stops there, leaving the rest buffered in the C library, for a future stdio function to use. If your program has e.g.
printf("Enter a number: ");
scanf("%d", &a);
printf("Enter a word: ");
scanf("%s", word);
and you enter the line 123 abcd, it completes both scanf()s at once, but only after a newline is given. The first scanf() doesn't return when a user has hit space, even though that's where the number ends (because at that point the line is still in the terminal's line buffer); and the second scanf() doesn't wait for you to enter another line (because the input buffer already contains enough to fill the %s conversion).
This isn't what users usually expect!
Instead, they expect that hitting enter completes the input, and if you hit enter, you either get a default value, or an error, with possibly a suggestion to please really just give the answer.
You can't really do that with scanf("%d"). If the user just hits enter, nothing happens. Because scanf() is still waiting for the number. The terminal sends the line onward, but your program doesn't see it, because scanf() eats it. You don't get a chance to react to the user's mistake.
That's also not very useful.
Hence, I suggest using fgets() or getline() to read a full line of input at a time. This exactly matches what the terminal gives, and always gives your program control after the user has entered a line. What you do with the input line is up to you, if you want a number, you can use atoi(), strtol(), or even sscanf(buf, "%d", &a) to parse the number. sscanf() doesn't have the same mismatch as scanf(), because the buffer it reads from is limited in size, and when it ends, it ends -- the function can't wait for more.
(fscanf() on a regular file can also be fine if the file format is one that supports how it skims over newlines like any whitespace. For line-oriented data, I'd still use fgets() and sscanf().)
So, instead of what I had above, use something like this:
printf("Enter a number: ");
fgets(buf, bufsize, stdin);
sscanf(buf, "%d", &a);
or, actually, check the return value of sscanf() too, so you can detect empty lines and otherwise invalid data:
#include <stdio.h>
int main(void)
{
const int bufsize = 100;
char buf[bufsize];
int a;
int ret;
char word[bufsize];
printf("Enter a number: ");
fgets(buf, bufsize, stdin);
ret = sscanf(buf, "%d", &a);
if (ret != 1) {
fprintf(stderr, "Ok, you don't have to.\n");
return 1;
}
printf("Enter a word: ");
fgets(buf, bufsize, stdin);
ret = sscanf(buf, "%s", word);
if (ret != 1) {
fprintf(stderr, "You make me sad.\n");
return 1;
}
printf("You entered %d and %s\n", a, word);
}
Of course, if you want the program to insist, you can create a simple function to loop over the fgets() and sscanf() until the user deigns to do what they're told; or to just exit with an error immediately. Depends on what you think your program should do if the user doesn't want to play ball.
You could do something similar e.g. by looping over getchar() to read characters until a newline after scanf("%d") returned, thus clearing up any garbage left in the buffer, but that doesn't do anything about the case where the user just hits enter on an empty line. Anyway, fgets() would read until a newline, so you don't have to do it yourself.
Use getchar() before calling second scanf().
scanf("%c", &c1);
getchar(); // <== remove newline
scanf("%c", &c2);
Two workarounds. One is catch the extra whitespace trickly.
getchar(); // (1) add `getchar()`
scanf("%c", &c1);
scanf(" %c", &c1); // (2) add whitespace before %c
The other is using fgets() to instead scanf(). Note: gets() is unsafe, see why gets is dangerous
By contrast, I would prefer to recommend the second way. Because it's more readable and maintainable. For example, in first way, you must think a while before adding/moving some scanf().
/*Take char input using scanf after int input using scanf just use fflush(stdin) function after int input */
#include<stdio.h>
#include<conio.h>
void main()
{
int x;
char y;
clrscr();
printf(" enter an int ");
scanf("%d",&x);
fflush(stdin);
printf("\n Now enter a char");
scanf("%c",&y);
printf("\n X=%d and Y=%c",x,y);
getch();
}

When printing a string length using a character array with a for the first iteration does nothing, why? [duplicate]

I have the following program:
int main(int argc, char *argv[])
{
int a, b;
char c1, c2;
printf("Enter something: ");
scanf("%d", &a); // line 1
printf("Enter other something: ");
scanf("%d", &b); // line 2
printf("Enter a char: ");
scanf("%c", &c1); // line 3
printf("Enter another char: ");
scanf("%c", &c2); // line 4
printf("Done"); // line 5
system("PAUSE");
return 0;
}
As I read in the C book, the author says that scanf() left a newline character in the buffer, therefore, the program does not stop at line 4 for user to enter the data, rather it stores the new line character in c2 and moves to line 5.
Is that right?
However, does this only happen with char data types? Because I did not see this problem with int data types as in line 1, 2, 3. Is it right?
The scanf() function skips leading whitespace automatically before trying to parse conversions other than characters. The character formats (primarily %c; also scan sets %[…] — and %n) are the exception; they don't skip whitespace.
Use " %c" with a leading blank to skip optional white space. Do not use a trailing blank in a scanf() format string.
Note that this still doesn't consume any trailing whitespace left in the input stream, not even to the end of a line, so beware of that if also using getchar() or fgets() on the same input stream. We're just getting scanf to skip over whitespace before conversions, like it does for %d and other non-character conversions.
Note that non-whitespace "directives" (to use POSIX scanf terminology) other than conversions, like the literal text in scanf("order = %d", &order); doesn't skip whitespace either. The literal order has to match the next character to be read.
So you probably want " order = %d" there if you want to skip a newline from the previous line but still require a literal match on a fixed string, like this question.
Use scanf(" %c", &c2);. This will solve your problem.
Another option (that I got from here) is to read and discard the newline by using the assignment-supression option. To do that, we just put a format to read a character with an asterisk between % and c:
scanf("%d%*c",&a); // line 1
scanf("%c%*c",&c1); // line 3
scanf will then read the next char (that is, the newline) but not assign it to any pointer.
In the end, however, I would second the FAQ's last option:
Or, depending on your requirements, you could also forget about scanf()/getchar(), use fgets() to get a line of text from the user and parse it yourself.
To echo what I have posted in another answer about C++: I suggest to toss scanf() away, to never use it, and to instead use fgets() and sscanf().
The reason for this is, that at least in Unix-like systems by default, the terminal your CLI program runs on does some processing of the user input before your program sees it. It buffers input until a newline is entered, and allows for some rudimentary line editing, like making backspace work.
So, you can never get a single character at a time, or a few single characters, just a full line. But that's not what e.g. scanf("%d") processes, instead it processes just the digits, and stops there, leaving the rest buffered in the C library, for a future stdio function to use. If your program has e.g.
printf("Enter a number: ");
scanf("%d", &a);
printf("Enter a word: ");
scanf("%s", word);
and you enter the line 123 abcd, it completes both scanf()s at once, but only after a newline is given. The first scanf() doesn't return when a user has hit space, even though that's where the number ends (because at that point the line is still in the terminal's line buffer); and the second scanf() doesn't wait for you to enter another line (because the input buffer already contains enough to fill the %s conversion).
This isn't what users usually expect!
Instead, they expect that hitting enter completes the input, and if you hit enter, you either get a default value, or an error, with possibly a suggestion to please really just give the answer.
You can't really do that with scanf("%d"). If the user just hits enter, nothing happens. Because scanf() is still waiting for the number. The terminal sends the line onward, but your program doesn't see it, because scanf() eats it. You don't get a chance to react to the user's mistake.
That's also not very useful.
Hence, I suggest using fgets() or getline() to read a full line of input at a time. This exactly matches what the terminal gives, and always gives your program control after the user has entered a line. What you do with the input line is up to you, if you want a number, you can use atoi(), strtol(), or even sscanf(buf, "%d", &a) to parse the number. sscanf() doesn't have the same mismatch as scanf(), because the buffer it reads from is limited in size, and when it ends, it ends -- the function can't wait for more.
(fscanf() on a regular file can also be fine if the file format is one that supports how it skims over newlines like any whitespace. For line-oriented data, I'd still use fgets() and sscanf().)
So, instead of what I had above, use something like this:
printf("Enter a number: ");
fgets(buf, bufsize, stdin);
sscanf(buf, "%d", &a);
or, actually, check the return value of sscanf() too, so you can detect empty lines and otherwise invalid data:
#include <stdio.h>
int main(void)
{
const int bufsize = 100;
char buf[bufsize];
int a;
int ret;
char word[bufsize];
printf("Enter a number: ");
fgets(buf, bufsize, stdin);
ret = sscanf(buf, "%d", &a);
if (ret != 1) {
fprintf(stderr, "Ok, you don't have to.\n");
return 1;
}
printf("Enter a word: ");
fgets(buf, bufsize, stdin);
ret = sscanf(buf, "%s", word);
if (ret != 1) {
fprintf(stderr, "You make me sad.\n");
return 1;
}
printf("You entered %d and %s\n", a, word);
}
Of course, if you want the program to insist, you can create a simple function to loop over the fgets() and sscanf() until the user deigns to do what they're told; or to just exit with an error immediately. Depends on what you think your program should do if the user doesn't want to play ball.
You could do something similar e.g. by looping over getchar() to read characters until a newline after scanf("%d") returned, thus clearing up any garbage left in the buffer, but that doesn't do anything about the case where the user just hits enter on an empty line. Anyway, fgets() would read until a newline, so you don't have to do it yourself.
Use getchar() before calling second scanf().
scanf("%c", &c1);
getchar(); // <== remove newline
scanf("%c", &c2);
Two workarounds. One is catch the extra whitespace trickly.
getchar(); // (1) add `getchar()`
scanf("%c", &c1);
scanf(" %c", &c1); // (2) add whitespace before %c
The other is using fgets() to instead scanf(). Note: gets() is unsafe, see why gets is dangerous
By contrast, I would prefer to recommend the second way. Because it's more readable and maintainable. For example, in first way, you must think a while before adding/moving some scanf().
/*Take char input using scanf after int input using scanf just use fflush(stdin) function after int input */
#include<stdio.h>
#include<conio.h>
void main()
{
int x;
char y;
clrscr();
printf(" enter an int ");
scanf("%d",&x);
fflush(stdin);
printf("\n Now enter a char");
scanf("%c",&y);
printf("\n X=%d and Y=%c",x,y);
getch();
}

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