How to read multiple lines of string from stdin in C? - c

I am a novice in C programming. Suppose I want to read multiple lines of string from stdin. How can I keep reading until a line only containing EOL?
example of input
1+2\n
1+2+3\n
1+2+3+4\n
\n (stop at this line)
It seems that when I hit enter(EOL) directly, scanf won't execute until something other than just EOL has been entered. How can I solve that problem?
I'll be really grateful if someone can help me with this. Thank you.

If you want to learn C, you should avoid scanf. The only use cases where scanf actually makes sense are in problems for which C is the wrong language. Time spent learning the foibles of scanf is not well spent, and it doesn't really teach you much about C. For something like this, just read one character at a time and stop when you see two consecutive newlines. Something like:
#include <stdio.h>
int
main(void)
{
char buf[1024];
int c;
char *s = buf;
while( (c = fgetc(stdin)) != EOF && s < buf + sizeof buf - 1 ){
if( c == '\n' && s > buf && s[-1] == '\n' ){
ungetc(c, stdin);
break;
}
*s++ = c;
}
*s = '\0';
printf("string entered: %s", buf);
return 0;
}

to read multiple lines of string from stdin. How can I keep reading until a line only containing EOL?
Keep track of when reading the beginning of the line. If a '\n' is read at the beginning, stop
getchar() approach:
bool beginning = true;
int ch;
while ((ch = getchar()) != EOF) {
if (beginning) {
if (ch == '\n') break;
}
// Do what ever you want with `ch`
beginning = ch == '\n';
}
fgets() approach - needs more code to handle lines longer than N
#define N 1024
char buf[N+1];
while (fgets(buf, sizeof buf, stdin) && buf[0] != '\n') {
; // Do something with buf
}

If you need to read one character at a time then you can with either getchar or fgetc depending upon whether or not you're reading from stdin or some other stream.
But you said you were reading strings, so I'm assuming fgets is more appropriate.
There are primarily two considerations:
maximum line length
whether or not to handle Windows versus non-Windows line endings
Even if you are a beginner--and I won't go into #2 here--you should know you can defend against it. I will at least say that if you compile on one platform and read from stdin from a redirected file from another platform, then you might have to write a defense.
#include <stdio.h>
#include <string.h>
#include <errno.h>
int main (int argc, char *argv[]) {
char buf[32]; // relatively small buf makes testing easier
int lineContinuation = 0;
// If no characters are read, then fgets returns NULL.
while (fgets(buf, sizeof(buf), stdin) != NULL) {
int l = strlen(buf); // No newline in buf if line len + newline exceeds sizeof(buf)
if (buf[l-1] == '\n') {
if (l == 1 && !lineContinuation) {
break; // errno should indicate no error.
}
printf("send line ending (len=%d) to the parser\n", l);
lineContinuation = 0;
} else {
lineContinuation = 1;
printf("send line part (len=%d) to the parser\n", l);
}
}
printf("check errno (%d) if you must handle unexpected end of input use cases\n", errno);
}

Related

How to accept string input only if it of certain length in C else ask user to input the string again

How to accept set of strings as input in C and prompt the user again to re-enter the string if it exceeds certain length. I tried as below
#include<stdio.h>
int main()
{
char arr[10][25]; //maximum 10 strings can be taken as input of max length 25
for(int i=0;i<10;i=i+1)
{
printf("Enter string %d:",i+1);
fgets(arr[i],25,stdin);
}
}
But here fgets accepts the strings greater than that length too.
If the user hits return, the second string must be taken as input. I'm new to C
How to accept string input only if it of certain length
Form a helper function to handle the various edge cases.
Use fgets(), then drop the potential '\n' (which fgets() retains) and detect long inputs.
Some untested code to give OP an idea:
#include <assert.h>
#include <stdio.h>
// Pass in the max string _size_.
// Return NULL on end-of-file without input.
// Return NULL on input error.
// Otherwise return the buffer pointer.
char* getsizedline(size_t sz, char *buf, const char *reprompt) {
assert(sz > 0 && sz <= INT_MAX && buf != NULL); // #1
while (fgets(buf, (int) sz, stdin)) {
size_t len = strlen(buf);
// Lop off potential \n
if (len > 0 && buf[--len] == '\n') { // #2
buf[len] = '\0';
return buf;
}
// OK if next ends the line
int ch = fgetc(stdin);
if (ch == '\n' || feof(stdin)) { // #3
return buf;
}
// Consume rest of line;
while (ch != '\n' && ch != EOF) { // #4
ch = fgetc(stdin);
}
if (ch == EOF) { // #5
return NULL;
}
if (reprompt) {
fputs(reprompt, stdout);
}
}
return NULL;
}
Uncommon: reading null characters remains a TBD issue.
Details for OP who is a learner.
Some tests for sane input parameters. A size of zero does not allow for any input saved as a null character terminated string. Buffers could be larger than INT_MAX, but fgets() cannot directly handle that. Code could be amended to handle 0 and huge buffers, yet leave that for another day.
fgets() does not always read a '\n'. The buffer might get full first or the last line before end-of-file might lack a '\n'. Uncommonly a null character might be read - even the first character hence the len > 0 test, rendering strlen() insufficient to determine length of characters read. Code would need significant changes to accommodate determining the size if null character input needs detailed support.
If the prior fgets() filled its buffer and the next read character attempt resulted in an end-of-file or '\n', this test is true and is OK, so return success.
If the prior fgetc() resulted in an input error, this loops exits immediately. Otherwise, we need to consume the rest of the line looking for a '\n' or EOF (which might be due to an end-of-file or input error.)
If EOF returned (due to an end-of-file or input error), no reason to continue. Return NULL.
Usage
// fgets(arr[i],25,stdin);
if (getsizedline(arr[i], sizeof(arr[i]), "Too long, try again.\n") == NULL) {
break;
}
This code uses a buffer slightly larger than the required max length. If a text line and the newline can't be read into the buffer, it reads the rest of the line and discards it. If it can, it again discards if too long (or too short).
#include <stdio.h>
#include <string.h>
#include <stdbool.h>
#define INPUTS 10
#define STRMAX 25
int main(void) {
char arr[INPUTS][STRMAX+1];
char buf[STRMAX+4];
for(int i = 0; i < INPUTS; i++) {
bool success = false;
while(!success) {
printf("Enter string %d: ", i + 1);
if(fgets(buf, sizeof buf, stdin) == NULL) {
exit(1); // or sth better
}
size_t index = strcspn(buf, "\n");
if(buf[index] == '\0') { // no newline found
// keep reading until end of line
while(fgets(buf, sizeof buf, stdin) != NULL) {
if(strchr(buf, '\n') != NULL) {
break;
}
}
if(feof(stdin)) {
exit(1); // or sth better
}
continue;
}
if(index < 1 || index > STRMAX) {
continue; // string is empty or too long
}
buf[index] = '\0'; // truncate newline
strcpy(arr[i], buf); // keep this OK string
success = true;
}
}
printf("Results:\n");
for(int i = 0; i < INPUTS; i++) {
printf("%s\n", arr[i]);
}
return 0;
}
The nice thing about fgets() is that it will place the line-terminating newline character ('\n') in the input buffer. All you have to do is look for it. If it is there, you got an entire line of input. If not, there is more to read.
The strategy then, is:
fgets( s, size_of_s, stdin );
char * p = strpbrk( s, "\r\n" );
if (p)
{
// end of line was found.
*p = '\0';
return s; (the complete line of input)
}
If p is NULL, then there is more work to do. Since you wish to simply ignore lines that are too long, that is the same as throwing away input. Do so with a simple loop:
int c;
do c = getchar(); while ((c != EOF) && (c != '\n'));
Streams are typically buffered behind the scenes, either by the C Library or by the OS (or both), but even if they aren’t this is not that much of an overhead. (Use a profiler before playing “I’m an optimizing compiler”. Don’t assume bad things about the C Library.)
Once you have tossed everything you didn’t want (to EOL), make sure your input isn’t at EOF and loop to ask the user to try again.
Putting it all together
char * prompt( const char * message, char * s, size_t n )
{
while (!feof( stdin ))
{
// Ask for input
printf( "%s", message );
fflush( stdout ); // This line _may_ be necessary.
// Attempt to get an entire line of input
if (!fgets( s, n, stdin )) break;
char * p = strpbrk( s, "\r\n" );
// Success: return that line (sans newline character(s)) to the user
if (p)
{
*p = '\0';
return s;
}
// Failure: discard the remainder of the line before trying again
int c;
do c = getchar(); while ((c != EOF) && (c != '\n'));
}
// If we get this far it is because we have
// reached EOF or some other input error occurred.
return NULL;
}
Now you can use this utility function easily enough:
char user_name[20]; // artificially small
if (!prompt( "What is your name (maximum 19 characters)? ", user_name, sizeof(user_name) ))
{
complain_and_quit();
// ...because input is dead in a way you likely cannot fix.
// Feel free to check ferror(stdin) and feof(stdin) for more info.
}
This little prompt function is just an example of the kinds of helper utility functions you can write. You can do things like have an additional prompt for when the user does not obey you:
What is your name? John Jacob Jingleheimer Schmidt
Alas, I am limited to 19 characters. Please try again:
What is your name? John Schmidt
Hello John Schmidt.

Use and explanation of getchar() function

I am writing a program to read a user input statement and extract all integers from the input. For example, if I enter "h3ll0", the program will output "30". I have used the fgets function to read the user input.
However, I am currently reading about getchar() and would like to know what would be the best way to use getchar() in my program to read user input instead of fgets. I am not really clear on how getchar() works and what situations it can be useful in.
This question is related to a project that specifically asks for getchar() as the method of reading user input. As I was unclear on how getchar() works, I built the rest of the program using fgets to ensure it was working.
#include <stdio.h>
int main()
{
char user_input[100];
int i;
int j = 0;
printf("Please enter your string: ");
fgets(user_input ,100, stdin);
for(i = 0; user_input[i] ; i++)
{
if(user_input[i] >= '0' && user_input[i] <= '9')
{
user_input[j] = user_input[i];
j++;
}
}
user_input[j] = '\0';
printf("Your output of only integers is: ");
printf("%s\n", user_input);
return 0;
}
OP: unclear on how getchar() works
int fgetc(FILE *stream) typically returns 1 of 257 different values.
"If ... a next character is present, the fgetc function obtains that character as an unsigned char converted to an int C11 §7.21.7.1 2
On end-of-file or input error (rare), EOF, is returned.
OP: to use getchar() in my program to read user input instead of fgets.
Create your own my_fgets() with the same function signature and same function as fgets() and then replace.
char *fgets(char * restrict s, int n, FILE * restrict stream);
The fgets function reads at most one less than the number of characters specified by n from the stream pointed to by stream into the array pointed to by s. No additional characters are read after a new-line character (which is retained) or after end-of-file. A null character is written immediately after the last character read into the array. C11 §7.21.7.2 2
Return the same value
The fgets function returns s if successful. If end-of-file is encountered and no characters have been read into the array, the contents of the array remain unchanged and a null pointer is returned. If a read error occurs during the operation, the array contents are indeterminate and a null pointer is returned. §7.21.7.2 3
Sample untested code
#include <stdbool.h>
#include <stdio.h>
char *my_fgets(char * restrict s, int n, FILE * restrict stream) {
bool something_read = false;
int ch = 0;
char *dest = s;
// Room ("reads at most one less") and EOF not returned?
while (n > 1 && (ch = fgetc(stream)) != EOF) {
n--;
something_read = true;
*dest++ = (char) ch;
if (ch == '\n') {
break; // "No additional characters are read after a new-line character"
}
}
// Did code end the while loop due to EOF?
if (ch == EOF) {
// Was EOF due to end-of-file or rare input error?
if (feof(stream)) {
// "If end-of-file is encountered and no characters ... read into the array ..."
if (!something_read) {
return NULL;
}
} else {
// "If a read error ..."
return NULL; // ** Note 1
}
}
// room for \0?
if (n > 0) {
*dest = '\0'; //" A null character is written immediately after the last character"
}
return s;
}
Perhaps improve fgets() and use size_t for n.
char *my_fgets(char * restrict s, size_t n, FILE * restrict stream);
fgets() with n <= 0 is not clearly defined. Using size_t, an unsigned type, at least eliminates n < 0 concerns.
Note 1: or use s = NULL; instead of return NULL; and let the remaining code null terminate the buffer. We have that option as "array contents are indeterminate".
Something like this should work as a clunky replacement to fgets using only getchar. I don't guarantee the accuracy of the error handling.
I don't think you would ever want to use getchar over fgets in an application. Getchar is more limited and less secure.
#include <stdint.h>
void your_fgets(char *buffer, size_t buffer_size)
{
int i;
size_t j;
if (buffer_size == 0)
return ;
else if (buffer_size == 1)
{
buffer[0] = '\0';
return ;
}
j = 0;
while ((i = getchar()) != EOF)
{
buffer[j++] = i;
if (j == buffer_size - 1 || i == '\n')
{
buffer[j] = '\0';
return ;
}
}
buffer[j] = '\0';
}
I am baffled by the comments on this post suggesting that fgets is easier to use. Using fgets unnecessarily complicates the issue. Just do:
#include <stdio.h>
#include <ctype.h>
#include <stdlib.h>
int
main(int argc, char **argv)
{
int c;
while( ( c = getchar() ) != EOF ) {
if(isdigit(c) && (putchar(c) == EOF)) {
perror("stdout");
return EXIT_FAILURE;
}
}
return ferror(stdin);
}
There is absolutely no reason to use any additional buffering, or read the input one line at a time. Maybe you'll want to output newlines as they come in, but that would be an implementation detail that is left unspecified in the question. Either way, it's utterly trivial (if(( c == '\n' || isdigit(c)) && (putchar(c) == EOF))). Just read a character and decide if you want to output it. The logic is much easier if you don't think about the input as being more complicated than it is. (It's not line-oriented...it's just a stream of bytes.)
If, for some unknown reason you want to make this tool usable only in an interactive setting and load up your output with excess verbosity, you can easily do something like:
#include <stdio.h>
#include <ctype.h>
#include <stdlib.h>
int
main(int argc, char **argv)
{
int c;
do {
int want_header = 1;
printf("Please enter your string: ");
while( ( c = getchar() ) != EOF && c != '\n' ) {
if(! isdigit(c)) {
continue;
}
if(want_header) {
want_header=0;
printf("Your output of only integers is: ");
}
if(putchar(c) == EOF) {
perror("stdout");
return EXIT_FAILURE;
}
}
if( c == '\n')
putchar(c);
want_header = 0;
} while(c == '\n');
return ferror(stdin);
}
but, please, don't do that. (Imagine if grep started by emitting a prompt that said "please enter the regex you would like to search for"!)

How To Read String that contains Spaces, in C language

What is the most accurate way to read strings from the keyboard in C, when the string contains spaces in between words? When I use scanf for that purpose then it doesn't read a string with spaces.The second option is to use gets but it is supposed to be harmful(I also want to know why?).Another thing is that I don't want to use any file handling concept like fgets.
These are 2 ways to read strings containing spaces that don't use gets or fgets
You can use getline (POSIX 2008 may not exist in your system) that conveniently manages allocation of the buffer with adequate size to capture the whole line.
char *line = NULL;
size_t bufsize = 0;
size_t n_read; // number of characters read including delimiter
while ((n_read = getline(&line, &bufsize, stdin)) > 1 && line != NULL) {
// do something with line
}
If you absolutely want scanf, in this example it reads to the end of line unless the line has more than the specified number of chars minus 1 for the delimiter. In the later case the line is truncated and you'll get the remaining chars in the next scanf invocation.
char line[1024];
while (scanf("%1023[^\n]\n", line) == 1) {
// do something with line
}
I should also point out that when you read strings from the keyboard with scanf for example, you are actually reading from a file with file pointer stdin. So you can't really avoid "any file handling concept"
#user3623265,
Please find a sample program which Uses fgets to read string from standard input.
Please refer some sample C documents as to how fgets can be used to get strings from a keyboard and what is the purpose of stdin.
#include <stdio.h>
#include <string.h>
int main(void)
{
char str[80];
int i;
printf("Enter a string: ");
fgets(str, sizeof(str), stdin);
i = strlen(str) - 1;
if (str[i] == '\n')
str[i] = '\0';
printf("This is your string: %s", str);
return 0;
}
There is a third option, you can read the raw data from stdin with the read() call:
#include <unistd.h>
int main(void) {
char buf[1024];
ssize_t n_bytes_read;
n_bytes_read = read(STDIN_FILENO, buf, sizeof(buf) - 1);
if (n_bytes_read < 0) {
// error occured
}
buf[n_bytes_read] = '\0'; // terminte string
printf("\'%s\'", buf);
return 0;
}
Please not that every input is copied raw to buf including the trailing return. That is, if you enter Hello World you will get
'Hello World
'
as output. Try online.
If you insist on not having a FILE * in scope, use getchar().
char buff[1024];
int ch;
int i = 0;
while( (ch = getchar()) != '\n' )
if(i < 1023)
buff[i++] = ch;
buff[i] = 0;
/* now move string into a smaller buffer */
Generally however it's accepted that stdout and stdin and FILE * are available. Your requirement is a bit odd and, since you are obviously not an advanced C programmer who has an unusual need to suppress the FILE * symbol, I suspect your understanding of C IO is shaky.

Extreme troubles with full line input. C Programming Language

I am having the absolute craziest time getting full line input to work. I will explain my problem. I need to get a full line of input, including a space, from the user entered at the keyboard. Simple right? Wrong!
MY GOAL
Store multiple strings, with spaces, into variables. If it makes a difference, I want to make the variables equal to a char pointer. So once I get the input from tempString, I want to set it to a char pointer. Like so:
char *variable1, *variable2;
//get user input
variable1 = tempString;
//get more user input
variable 2 = tempString;
//etc etc etc
Here's what I've tried.
First try
char tempString[100];
scanf("%s", &tempString);
printf("%s", tempString);
Invalid: scanf will stop reading at a white space, so "Example String" would just end up being "Example".
Second try
So I do more research. I thought I found the magic fix.
char tempSTring[100];
fgets(tempString, 100, stdin);
printf("%s", tempString);
Originally this works. However there is a massive problem. I need to get the user to enter about 8 inputs. Meaning I have to use a command like this 8 times. The problem is the program often skips over the fgets command. If I use a scanf previously, somehow the \n character is stuck in the input stream, and automatically feeds into fgets, satisfying its stdin input, and then does not prompt the user for input.
Third try
After thinking fgets was maybe my solution with a work around, I tried some tricks.
char tempSTring[100];
getc(stdin);
fgets(tempString, 100, stdin);
printf("%s", tempString);
I tried adding this getc(stdin) line. It worked for much of my program. It absorbs the \n character left behind in the stream. When it does so, great, it works. But sometimes, for some reason, the \n is NOT left in the stream, and when debugging, it looks like getc(stdin) is requesting input from the user, so it pauses my program to ask for input.
Question
These don't work for me.
How should I be doing this easy task?
To read (up to) 8 lines from a file, you can use either of these solutions. I decline to use variables char *variable1, *variable2, …; — that is an array seeking to escape.
POSIX getline()
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
enum { MAX_LINES = 8 };
char *lines[MAX_LINES];
int index = 0;
char *buffer = 0;
size_t buflen = 0;
while (index < MAX_LINES && getline(&buffer, &buflen, stdin) != -1)
{
lines[index++] = buffer;
buffer = 0;
buflen = 0;
}
free(buffer); // Space may be allocated before EOF is detected
for (int i = 0; i < index; i++)
printf("%d: %s", i, lines[i]);
return 0;
}
If getline() fails to allocate memory, it will report an error, so there is no need to do an explicit error check.
Standard C fgets()
Code using strdup(), another POSIX function. It isn't a part of standard C (though it is widely available). It is trivial to implement.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void)
{
enum { MAX_LINES = 8 };
char *lines[MAX_LINES];
int index = 0;
char buffer[4096];
while (index < MAX_LINES && fgets(buffer, sizeof(buffer), stdin) != 0)
{
if ((lines[index] = strdup(buffer)) == 0)
break;
index++;
}
for (int i = 0; i < index; i++)
printf("%d: %s", i, lines[i]);
return 0;
}
The test in the loop allows for the possibility of strdup() failing to allocate memory.
Notes
Both the solutions above keep the newline at the end of the input string. If you don't want that, you can zap it with:
lines[i][strcspn(lines[i], "\r\n")] = '\0';
This overwrites a carriage return or newline with a null byte, transforming DOS or Unix line endings. You then need to adjust the printing which assumes the string includes a newline. Note that the expression shown works correctly even if there is no carriage return or newline in the string.
The fgets() solution will break lines at 4095 characters, leaving the rest to be read as 'the next line'. If that's not acceptable, you have a variety of strategies open to you.
You can detect whether there is a newline and arrange to allocate more memory and read the next section of the line into the extra memory, repeating until you come across a newline or EOF.
You can read the remaining characters up to the newline or EOF:
int c;
while ((c = getchar()) != EOF && c != '\n')
;
Implementing strdup()
If for some reason your system doesn't have an implementation of strdup(), you can create a surrogate with:
#include <assert.h>
#include <stdlib.h>
#include <string.h>
char *strdup(const char *old_str)
{
assert(old_str != 0);
size_t old_len = strlen(old_str) + 1;
char *new_str = malloc(old_len);
if (new_str != 0)
memmove(new_str, old_str, old_len);
return new_str;
}
Here's how we old fart C programmers would do it:
#include <stdio.h>
#define MAX_LEN 100
int main( )
{
int c;
char input[MAX_LEN+1];
int i = 0;
while ( (c=getchar()) != '\n' && c != EOF && i < MAX_LEN)
input[i++] = c;
if (c == EOF || c =='\n') {
/* received input that terminated within buffer limit */
input[i] = '\0';
printf("read in your input string of: %s\n", input);
}
else {
printf("don't buffer overflow me dude!\n");
return -1;
}
return 0;
}
But nowadays people will tell you to use one of the library functions. I'm still an old fart though.
EDIT: Fixed my embarrassing mistakes pointed out by the helpful comments below.
You can take care of '\n' left by previous scanf by writing it like this -
scanf("%d%*c", &x); //<-- example to take int input
%*c will read from stdin and then discard it, thus '\n' would be removed from stdin.
You can achieve with scanf like this (a way for your previous attempt)-
char tempString[100];
/* As suggested by chqrile it is essential to check return of scanf */
if(scanf("%99[^\n]", tempString)!=1){
// ^^ & not required
tempString[0]='\0';
}
%99[^\n] this will read 99 characters and will stop only after encountering '\n' , thus would read input with spaces.

How to take a line input in C?

I was trying to take a full line input in C. Initially I did,
char line[100] // assume no line is longer than 100 letters.
scanf("%s", line);
Ignoring security flaws and buffer overflows, I knew this could never take more than a word input. I modified it again,
scanf("[^\n]", line);
This, of course, couldn't take more than a line of input. The following code, however was running into infinite loop,
while(fscanf(stdin, "%[^\n]", line) != EOF)
{
printf("%s\n", line);
}
This was because, the \n was never consumed, and would repeatedly stop at the same point and had the same value in line. So I rewrote the code as,
while(fscanf(stdin, "%[^\n]\n", line) != EOF)
{
printf("%s\n", line);
}
This code worked impeccably(or so I thought), for input from a file. But for input from stdin, this produced cryptic, weird, inarticulate behavior. Only after second line was input, the first line would print. I'm unable to understand what is really happening.
All I am doing is this. Note down the string until you encounter a \n, store it in line and then consume the \n from the input buffer. Now print this line and get ready for next line from the input. Or am I being misled?
At the time of posting this question however, I found a better alternative,
while(fscanf(stdin, "%[^\n]%*c", line) != EOF)
{
printf("%s\n", line);
}
This works flawlessly for all cases. But my question still remains. How come this code,
while(fscanf(stdin, "%[^\n]\n", line) != EOF)
{
printf("%s\n", line);
}
worked for inputs from file, but is causing issues for input from standard input?
Use fgets(). #FredK
char buf[N];
while (fgets(buf, sizeof buf, stdin)) {
// crop potential \n if desired.
buf[strcspn(buf, "\n")] = '\0';
...
}
There are to many issues trying to use scanf() for user input that render it prone to mis-use or code attacks.
// Leaves trailing \n in stdin
scanf("%[^\n]", line)
// Does nothing if line begins with \n. \n remains in stdin
// As return value not checked, use of line may be UB.
// If some text read, consumes \n and then all following whitespace: ' ' \n \t etc.
// Then does not return until a non-white-space is entered.
// As stdin is usually buffered, this implies 2 lines of user input.
// Fails to limit input.
scanf("%[^\n]\n", line)
// Does nothing if line begins with \n. \n remains in stdin
// Consumes 1 char after `line`, even if next character is not a \n
scanf("%99[^\n]%*c", line)
Check against EOF is usual the wrong check. #Weather Vane The following, when \n is first entered, returns 0 as line is not populated. As 0 != EOF, code goes on to use an uninitialized line leading to UB.
while(fscanf(stdin, "%[^\n]%*c", line) != EOF)
Consider entering "1234\n" to the following. Likely infinite loop as first fscanf() read "123", tosses the "4" and the next fscanf() call gets stuck on \n.
while(fscanf(stdin, "%3[^\n]%*c", line) != EOF)
When checking the results of *scanf(), check against what you want, not against one of the values you do not want. (But even the following has other troubles)
while(fscanf(stdin, "%[^\n]%*c", line) == 1)
About the closest scanf() to read a line:
char buf[100];
buf[0] = 0;
int cnt = scanf("%99[^\n]", buf);
if (cnt == EOF) Handle_EndOfFile();
// Consume \n if next stdin char is a \n
scanf("%*1[\n]");
// Use buf;
while(fscanf(stdin, "%[^\n]%*c", line) != EOF)
worked for inputs from file, but is causing issues for input from standard input?
Posting sample code and input/data file would be useful. With modest amount of code posted, some potential reasons.
line overrun is UB
Input begins with \n leading to UB
File or stdin not both opened in same mode. \r not translated in one.
Note: The following fails when a line is 100 characters. So meeting the assumption cal still lead to UB.
char line[100] // assume no line is longer than 100 letters.
scanf("%s", line);
Personally, I think fgets() is badly designed. When I read a line, I want to read it in whole regardless of its length (except filling up all RAM). fgets() can't do that in one go. If there is a long line, you have to manually run it multiple times until it reaches the newline. The glibc-specific getline() is more convenient in this regard. Here is a function that mimics GNU's getline():
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
long my_getline(char **buf, long *m_buf, FILE *fp)
{
long tot = 0, max = 0;
char *p;
if (*m_buf == 0) { // empty buffer; allocate
*m_buf = 16; // initial size; could be larger
*buf = (char*)malloc(*m_buf); // FIXME: check NULL
}
for (p = *buf, max = *m_buf;;) {
long l, old_m;
if (fgets(p, max, fp) == NULL)
return tot? tot : EOF; // reach end-of-file
for (l = 0; l < max; ++l)
if (p[l] == '\n') break;
if (l < max) { // a complete line
tot += l, p[l] = 0;
break;
}
old_m = *m_buf;
*m_buf <<= 1; // incomplete line; double the buffer
*buf = (char*)realloc(*buf, *m_buf); // check NULL
max = (*m_buf) - old_m;
p = (*buf) + old_m - 1; // point to the end of partial line
}
return tot;
}
int main(int argc, char *argv[])
{
long l, m_buf = 0;
char *buf = 0;
while ((l = my_getline(&buf, &m_buf, stdin)) != EOF)
puts(buf);
free(buf);
return 0;
}
I usually use my own readline() function. I wrote this my_getline() a moment ago. It has not been thoroughly tested. Please use with caution.

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