If I want to receive a one character input in C, how would I check to see if extra characters were sent, and if so, how would I clear that?
Is there a function which acts like getc(stdin), but which doesn't prompt the user to enter a character, so I can just put while(getc(stdin)!=EOF);? Or a function to peek at the next character in the buffer, and if it doesn't return NULL (or whatever would be there), I could call a(nother) function which flushes stdin?
Edit
So right now, scanf seems to be doing the trick but is there a way to get it to read the whole string, up until the newline? Rather than to the nearest whitespace? I know I can just put "%s %s %s" or whatever into the format string but can I handle an arbitrary number of spaces?
You cannot flush the input stream. You will be invoking undefined behavior if you do. Your best bet is to do:
int main() {
int c = getchar();
while (getchar() != EOF);
return 0;
}
To use scanf magic:
#include <stdio.h>
#include <stdlib.h>
#define str(s) #s
#define xstr(s) str(s)
#define BUFSZ 256
int main() {
char buf[ BUFSZ + 1 ];
int rc = scanf("%" xstr(BUFSZ) "[^\n]%*[^\n]", buf);
if (!feof(stdin)) {
getchar();
}
while (rc == 1) {
printf("Your string is: %s\n", array);
fflush(stdout);
rc = scanf("%" xstr(LENGTH) "[^\n]%*[^\n]", array);
if (!feof(stdin)) {
getchar();
}
}
return 0;
}
You can use getline to read a whole line of input.
Alternatively (in response to your original question), you can call select or poll on stdin to see if there are additional characters to be read.
I had a similar problem today, and I found a way that seems to work. I don't know the details of your situation, so I don't know if it will work for you or not.
I'm writing a routine that needs to get a single character from the keyboard, and it needs to be one of three specific keystrokes (a '1', a '2', or a '3'). If it's not one of those, the program needs to send and error message and loop back for another try.
The problem is that in addition to the character I enter being returned by getchar(), the 'Enter' keystroke (which sends the keystroke to the program) is saved in an input buffer. That (non-printing) newline-character is then returned by the getchar() facility in the error-correction loop, resulting further in a second error message (since the newline-character is not either a '1', a '2', nor a '3'.)
The issue is further complicated because I sometimes get ahead of myself and instead of entering a single character, I'll enter the filename that one of these options will request. Then I have a whole string of unwanted characters in the buffer, resulting in a long list of error messages scrolling down the screen.
Not cool.
What seems to have fixed it, though, is the following:
c = getchar(); // get first char in line
while(getchar() != '\n') ; // discard rest of buffer
The first line is the one that actually uses the character I enter. The second line disposes of whatever residue remains in the input buffer. It simply creates a loop that pulls a character at a time from the input buffer. There's no action specified to take place while the statement is looping. It simply reads a character and, if it's not a newline, goes back for the next. When it finds a newline, the loop ends and it goes on to the next order of business in the program.
We can make a function to clear the keyboard buffer, like this:
#include <stdio.h>
void clear_buffer(){
char b;
//this loop take character by character in the keyboard buffer using
//getchar() function, it stop when the variable "b" was
//enter key or EOF.
while (((b = getchar()) != '\n') && (b != EOF));
}
int main()
{
char input;
//get the input. supposed to be one char!
scanf("%c", &input);
//call the clearing function that clear the buffer of the keyboard
clear_buffer();
printf("%c\n",input); //print out the first character input
// to make sure that our function work fine, we have to get the
// input into the "input" char variable one more time
scanf("%c", &input);
clear_buffer();
printf("%c\n",input);
return 0;
}
Use a read that will take a lot of characters (more than 1, maybe 256), and see how many are actually returned. If you get more than one, you know; if you only get one, that's all there were available.
You don't mention platform, and this gets quite tricky quite rapidly. For example, on Unix (Linux), the normal mechanism will return a line of data - probably the one character you were after and a newline. Or maybe you persuade your user to type ^D (default) to send the preceding character. Or maybe you use control functions to put the terminal into raw mode (like programs such as vi and emacs do).
On Windows, I'm not so sure -- I think there is a getch() function that does what you need.
Why don't you use scanf instead of getc, by using scanf u can get the whole string.
Related
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"
I was always bad at inputting characters in C and this is another example. Though I understood (maybe) what's happening but I can't figure out the solution.
I have the following code
scanf("%ld %ld",&n,&m);
for(i=0;i<n;i++)
scanf("%ld",&array[i]);
for(i=0;i<m;i++)
{
fflush(stdin);
//inputting a character 'R' but it is picking '\n' from past buffer
scanf("%c",&query);
//As a result of above problem, it is also acting wierd for same reason
scanf("%ld",&d);
printf("%c %ld",query,d);
printf("\nI=%ld\n",i);
}
Please help me figure out the reason why its happening and what is the solution.
Using scanf with %d (or %ld) only extracts the number from the input stream; it leaves the newline in the stream.
So when you write scanf("%c", it reads that newline.
To fix this (if your intent is that scanf("%c" reads the first character of the next line), you need to flush the input of the previous line. One way to do that is:
int ch; while ( (ch == getchar()) != EOF && ch != '\n' ) { }
Your line fflush(stdin); causes undefined behaviour - don't do that. The fflush function is only for output streams.
Also , it is a really good idea to check the return value of scanf. If it was not what you expected then you may wish to take some action, instead of pretending that a number was entered.
Since you are tired of input issues, I can give you a method that can help to simplify your live.
I can observe that:
You have problems in handling end-of-lines.
Sometimes you need to input numbers and sometimes you need characters or another kind of input. So, you (think that you) are forced to use formatted input.
My advice is that you separate the issue of reading input from the issue of interpreting data entered from input.
The standard C brings only a few functions to handle input/output operations, in the standard header <stdio.h>.
If you are not interested in very sofisticated I/O results, the standard library is enough.
However, the functions of <stdio.h> usually have the effect that input is read one line at the time, which includes the end-of-line character: '\n'.
What you can do, then, it's what follows:
Read a line with fgets(..., stdin) and put the result in a buffer (not so long), used only for this purpose.
Once you have read an entire line, no more issues with end-of-line will bother you.
Then, re-read this line, that it's held in a buffer, and apply to it all the formatted input that you need.
A short example:
#include <stdio.h>
int main(void) {
char buffer[200] = ""; // Initialize array to 0's
long int n, m;
char c;
fgets(buffer, sizeof(buffer), stdin);
sscanf(buffer,"%ld %ld",&n,&m);
// Now you have processed the "integer number" input,
// read input characters again, withou any "flushes" and extrange things:
fgets(buffer, sizeof(buffer), stdin);
sscanf(buffer,"%c", &c);
fgets(buffer, sizeof(buffer), stdin);
// and so on...
}
Thus, every time you need to separate a section of input from a previous one, just do a new line reading with fgets(..., stdin), which stores the input in buffer, and then process the buffer with sscanf(), which applies the format string to the buffer instead of the input itself (in its flesh).
Note: This method can have a little problem: If the string input has more than sizeof(buffer) characters (in the example: 200), the line is not completely read. This situation can be handled by checking if the character before last in buffer is not equal to '\n' nor '\0'. In such a case, you would make automatically some kind of "flushing input" operation (reading and discarding characters till the next end-of-line is found).
I have to write a program in C that handles the newline as part of a string. I need a way of handling the newline char such that if it is encountered, it doesn't necessarily terminate the input. So far I've been using fgets() but that stops as soon as it reaches a '\n' char. Is there a good function for processing the input from the console that doesn't necessarily end at the newline character?
To clarify:
I need a method that doesn't terminate at the newline char because in this particular exercise when the newline char is encountered it's replaced with a space char.
fgets gets a line from a stream. A line is defined as ending with a newline, end-of-file or error, so you don't want that.
You probably want to use fgetc. Here's a code example of a c program file fgetc.c
#include <stdio.h>
int main (void) {
int c;
while ((c = fgetc(stdin)) != EOF) fputc(c, stdout);
}
compile like this:
cc fgetc.c -o fgetc
use like this (notice the newline character '\n'):
echo 'Hello, thar!\nOh, hai!' | ./fgetc
or like this:
cat fgetc.c | ./fgetc
Read the fgetc function manual to find out more: man fgetc
If I understand your question correctly you want to read from the standard input until user has finished typing ( which ain't be a newline of course ). This can be done by setting a flag like EOF while getting input. One way which I came out with is this:
#include <stdio.h>
int main(void)
{
char ch;
char str[100];
int i = 0;
setbuf (stdout,NULL);
while ( (ch = getchar()) != EOF)// user can input until the EOF which he or she enters to mark the end of his/her typing or more appropriately input.
{
str[i] = ch;// you can store all the input character by character in a char array
i++;
}
printf ("%s",str);// then you can print it at last as a whole
return 0;
}
BEGINNER's NOTE- EOF can vary from system to system so check it and enter the proper EOF for your system.
If you are simply reading blocks of information without the need for scanf(), then fread() may be what you are after. But on consoles you can read to the \n, notice the \n, then continue reading more if you decide that \n is not for you.
scanf works when used as directed. Specifically, it treats \n as white space.
Depending on how your application is coded (i.e. how buffers are defined), a \n prompts the system to flush the buffer and feed data into scanf. This should occur as a default without you having to do anything.
So the real question is, what kind of data or characters do you need from the console? In some cases scanf will remove white space and NOT pass blanks, tabs, new lines, into your program. However, scanf can be coded to NOT do this!
Define how data should be entered and I can guide you on how to code scanf.
I can't get around this problem. I need the user to type a string then hit enter, then another string. When he/she is done hit enter once more (this last string would only have \n character so I know when to stop).
char * buff = malloc (100);
printf("Type in strings, to finish hit enter\n");
do{
scanf (" %[^\n]",buff);
//do some other stuff with the string
} while(*buff);
printf("You have finished typing strings\n");
This approach I came up with is no use for me, since the [^\n] command is telling the function to read everything but the \n meaning that the \n is kept in the console buffer. If I simply do
while(*buff)
{
scanf ("%s",buff);
}
if I hit enter it does nothing.
Any other approach?
Yeah, scanf is actually looking for characters. What you want is gets (to get a line).
[edit] as pointed out by Daniel Fischer:
gets has (at last) been removed from the language. Even before, the
man page has for a long time said Never use gets()
Looks like my advice was not the best. I guess this means use fgets, as it protects against buffer overrun. Unlike gets, the newline character(s) will also be stored in the string and it is the programmer's responsibility to check for them.
const size_t bufsize = 100;
char buf[bufsize];
while( fgets(buf, bufsize, stdin) != NULL )
{
if( buf[0] == '\n' ) break;
/* Do something with your string... */
}
Besides what paddy pointed out (use gets to get a whole line), the condition
while(*buff);
won't ever be false, unless the user input would be a null character (as in ASCII value 0). The line break character (\n) has an ASCII value of 10, which evaluates to true in conditional statements.
Try this:
while(strlen(*buff) > 0);
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").