I have to read strings of different length in C.I can do it using an array of pointers and using ,alloc inside the loop.The problem is that the input to the program is a series of sentences like this:
Rene Decartes once said,
"I think, therefore I am."
and there can be many more. How can I check programmaticaly that the user has finished giving the input sentences.
Maybe he's even looking for the end-of-file (end of input, end of stream). You can detect the end of the input stream by checking the return value of various input functions. You did not tell us which function you use for reading input. E.g. with fgets() and input on standard input you would do
#include <stdio.h>
#define BUFFERSIZE 256
char buffer[BUFFERSIZE];
...
while (fgets (buffer, sizeof buffer, stdin) != NULL) {
/* Do something with buffer. */
}
/* No more lines on stdin. */
If I understand the question, you're not looking for the "end of line" signal; you're looking for the "end of sentences" signal.
A common way to do this is to ask the user to enter an empty line when s/he's done. Then (depending on how you're reading the strings), you can check for either an empty string, or a string with only a single (newline) character.
Loop the array keys until you find a "\0" character.
Make use of some delimiters and ask the user to enter it on finishing all the sentences,,
Also in-order to check for the end of each line check with '\n' as the delimiter,,
You can tell the user to type in a special symbol or string which will indicate that the input has ended. For example you can tell the user to enter a . in a new line to terminate the input. Checkout #Adam Liss 's answer also.
You should process them one line at a time using "end of line" as the delimiter. You can store those into a **char -- properly allocated of course. Do be careful of boundaries, and prefer strncpy() to strcpy() -- the later is unsafe.
Related
How to scan total line from user input with c program?
I tried scanf("%99[^\n]",st), but it is not working when I scan something before this scan statment.It worked if this is the first scan statement.
How to scan total line from user input with c program?
There are many ways to read a line of input, and your usage of the word scan suggests you're already focused on the scanf() function for the job. This is unfortunate, because, although you can (to some extent) achieve what you want with scanf(), it's definitely not the best tool for reading a line.
As already stated in the comments, your scanf() format string will stop at a newline, so the next scanf() will first find that newline and it can't match [^\n] (which means anything except newline). As a newline is just another whitespace character, adding a blank in front of your conversion will silently eat it up ;)
But now for the better solution: Assuming you only want to use standard C functions, there's already one function for exactly the job of reading a line: fgets(). The following code snippet should explain its usage:
char line[1024];
char *str = fgets(line, 1024, stdin); // read from the standard input
if (!str)
{
// couldn't read input for some reason, handle error here
exit(1); // <- for example
}
// fgets includes the newline character that ends the line, but if the line
// is longer than 1022 characters, it will stop early here (it will never
// write more bytes than the second parameter you pass). Often you don't
// want that newline character, and the following line overwrites it with
// 0 (which is "end of string") **only** if it was there:
line[strcspn(line, "\n")] = 0;
Note that you might want to check for the newline character with strchr() instead, so you actually know whether you have the whole line or maybe your input buffer was to small. In the latter case, you might want to call fgets() again.
How to scan total line from user input with c program?
scanf("%99[^\n]",st) reads a line, almost.
With the C Standard Library a line is
A text stream is an ordered sequence of characters composed into lines, each line consisting of zero or more characters plus a terminating new-line character. Whether the last line requires a terminating new-line character is implementation-defined. C11dr §7.21.2 2
scanf("%99[^\n]",st) fails to read the end of the line, the '\n'.
That is why on the 2nd call, the '\n' remains in stdin to be read and scanf("%99[^\n]",st) will not read it.
There are ways to use scanf("%99[^\n]",st);, or a variation of it as a step in reading user input, yet they suffer from 1) Not handling a blank line "\n" correctly 2) Missing rare input errors 3) Long line issues and other nuances.
The preferred portable solution is to use fgets(). Loop example:
#define LINE_MAX_LENGTH 200
char buf[LINE_MAX_LENGTH + 1 + 1]; // +1 for long lines detection, +1 for \0
while (fgets(buf, sizeof buf, stdin)) {
size_t eol = strcspn(buf, "\n"); **
buf[eol] = '\0'; // trim potential \n
if (eol >= LINE_MAX_LENGTH) {
// IMO, user input exceeding a sane generous threshold is a potential hack
fprintf(stderr, "Line too long\n");
// TBD : Handle excessive long line
}
// Use `buf[[]`
}
Many platforms support getline() to read a line.
Short-comings: Non C-standard and allow a hacker to overwhelm system resources with insanely long lines.
In C, there is not a great solution. What is best depends on the various coding goals.
** I prefer size_t eol = strcspn(buf, "\n\r"); to read lines in a *nix environment that may end with "\r\n".
scanf() should never be used for user input. The best way to get input from the user is with fgets().
Read more: http://sekrit.de/webdocs/c/beginners-guide-away-from-scanf.html
char str[1024];
char *alline = fgets(str, 1024, stdin);
scanf("%[^'\n']s",alline);
I think the correct solution should be like this. It is worked for me.
Hope it helps.
I have a small program I'm writing to practice programming in C.
I want it to use the getchar(); function to get input from the user.
I use the following function to prompt for user input, then loop using getchar() to store input in an array:
The function is passed a pointer referencing a struct's member.
getInput(p->firstName); //The function is passed an argument like this one
void getInput(char * array)
{
int c;
while((c=getchar()) != '\n')
*array++ = c;
*array = '\0'; //Null terminate
}
This function is called multiple times, as it is a part of a function that creates a structure, and populates it's array members.
However when the program executes, The first two calls to it work fine, but any subsequent calls to this function will cause every-other call to getchar() to not wait for keyboard input.
After some debugging I traced the bug to be that getchar(); was for some reason reading in the '\n' character instead of waiting for input, the while loop test fails, and the function returns essentially an empty string.
I have done some research and keep finding to use
while(getchar() != '\n');
at the end of the function in order to properly flush stdin, however, this produces undesirable results, as the program will prompt again for more input after I type ENTER. Pressing ENTER again continues the program, but every-other subsequent calls continue to read in this mysterious '\n' character right off the bat, causing the test to fail, and resulting in empty strings whenever it comes time to print the contents of the the structure.
Could anyone explain to me what is going on here? Why does getchar() keep fetching a '\n' even though I supposedly cleared the input buffer? I have tried just placing a getchar(); statement at the beginning and end of the function, tried 'do while' loops, and taken other jabs at it, but I can't seem to figure this out.
The code you have written has several drawbacks. I'll try to explain them as it is unclear where your code is failing (probably outside the function you posted)
First of all, you don't check for EOF in getchar() result value. getchar(3) doesn't return a char precisely to allow to return al possible char values plus an extra one, EOF, to mark the end of file (this can be generated from a terminal by input of Ctrl-D in unix, or Ctrl-Z on windows machines) That case must be explicitly contempled in your code, as you'll convert the result to a char and will lose the extra information you received from the function. Read getchar(3) man page to solve this issue.
Second, you don't check for input of enough characters to fill all the array and overflow it. To the function you pass only a pointer to the beginning of the array, but nothing indicates how far it extends, so you can be overfilling past the end of its bounds, just overwritting memory that was not reserved for input purposes. This normally results in something called U.B. in the literature (Undefined Behaviour) and is something you must care of. This can be solved by passing a counter of valid positions to fill in the array and decrementing it for each valid position filled. And not allowing more input once the buffer has filled up.
On other side, you have a standar function that does exactly that, fgets(3) just reads one string array from an input file, and stores it on the pointer (and size) you pass to it:
char *fgets(char *buffer, size_t buffer_size, FILE *file_descriptor);
You can use it as in:
char buffer[80], *line;
...
while (line = fgets(buffer, sizeof buffer, stdin)) {
/* process one full line of input, with the final \n included */
....
}
/* on EOF, fgets(3) returns NULL, so we shall be here after reading the
* full input file */
So I am trying to read a defined number of characters from the input. Let's say that I want to read 30 characters and put them in to a string. I managed to do this with a for loop, and I cleaned the buffer as shown below.
for(i=0;i<30;i++){
string[i]=getchar();
}
string[30]='\0';
while(c!='\n'){
c=getchar(); // c is some defined variable type char
}
And this is working for me, but I was wondering if there is another way to do this. I was researching and some of them are using sprintf() for this problem, but I didn't understand that solution. Then I found that you can use scanf with %s. And some of them use %3s when they want to read 3 characters. I tried this myself, but this command only reads the string till the first empty space. This is the code that I used:
scanf("%30s",string);
And when I run my program with this line, if I for example write: "Today is a beatiful day. It is raining, but it's okay i like rain." I thought that the first 30 characters would be saved in to the string. But when i try to read this string with puts(string); it only shows "Today".
If I use scanf("%s",string) or gets(string) that would rewrite some parts of my memory if the number of characters on input is greater than 30.
You can use scanf("%30[^\n]",s)
Actually, this is how you can set which characters to input. Here, carat sign '^' denotes negation, ie. this will input all characters except \n. %30 asks to input 30 characters. So, there you are.
The API you're looking for is fgets(). The man page describes
char *fgets(char *s, int size, FILE *stream);
fgets() reads in at most one less than size characters from stream and stores them into the buffer pointed to by s. Reading stops after an EOF or a newline. If a newline is read, it is stored into the buffer. A terminating null byte ('\0') is stored after the last character in the buffer.
I am having difficulty with a feature of a segment of code that is designed to illustrate the fgets() function for input. Before I proceed, I would like to make sure that my understanding of I/O and streams is correct and that I'm not completely off base:
Input and Output in C has no specific viable function for working with strings. The one function specific for working with strings is the 'gets()' function, which will accept input beyond the limits of the char array to store the input (thus making it effectively illegal for all but backward compatibility), and create buffer overflows.
This brings up the topic of streams, which to the best of my understanding is a model to explain I/O in a program. A stream is considered 'flowing water' on which the data utilized by programs is conveyed. See links: (also as a conveyor belt)
Can you explain the concept of streams?
What is a stream?
In the C language, there are 3 predefined ANSII streams for standard input and output, and 2 additional streams if using windows or DOS which are as follows:
stdin (keyboard)
stdout (screen)
stderr (screen)
stdprn (printer)
stdaux (serial port)
As I understand, to make things manageable it is okay to think of these as rivers that exist in your operating system, and a program uses I/O functions to put data in them, take data out of them, or change the direction of where the streams are flowing (such as reading or writing a file would require). Never think of the 'beginning' or 'end' of the streams: this is handled by the operating system. What you need to be concerned with is where the water takes your data, and that is mediated by use of specific functions (such as printf(), puts(), gets(), fgets(), etc.).
This is where my questions start to take form. Now I am interested in getting a grasp on the fgets() function and how it ties into streams. fgets() uses the 'stdin' stream (naturally) and has the built in fail safe (see below) that will not allow user input to exceed the array used to store the input. Here is the outline of the fgets() function, rather its prototype (which I don't see why one would ever need to declare it?):
char *fgets(char *str , int n , FILE *fp);
Note the three parameters that the fgets function takes:
p1 is the address of where the input is stored (a pointer, which will likely just be the name of the array you use, e.g., 'buffer')
p2 is the maximum length of characters to be input (I think this is where my question is!)
p3 specifies the input stream, which in this code is 'stdin' (when would it ever be different?)
Now, the code I have below will allow you to type characters until your heart is content. When you hit return, the input is printed on the screen in rows of the length of the second parameter minus 1 (MAXLEN -1). When you enter a return with no other text, the program terminates.
#include <stdio.h>
#define MAXLEN 10
int main(void)
{
char buffer[MAXLEN];
puts("Enter text a line at a time: enter a blank line to exit");
while(1)
{
fgets(buffer, MAXLEN, stdin); //Read comments below. Note 'buffer' is indeed a pointer: just to array's first element.
if(buffer[0] == '\n')
{
break;
}
puts(buffer);
}
return 0;
}
Now, here are my questions:
1) Does this program allow me to input UNLIMITED characters? I fail to see the mechanism that makes fgets() safer than gets(), because my array that I am storing input in is of a limited size (256 in this case). The only thing that I see happening is my long strings of input being parsed into MAXLEN - 1 slices? What am I not seeing with fgets() that stops buffer overflow that gets() does not? I do not see in the parameters of fgets() where that fail-safe exists.
2) Why does the program print out input in rows of MAXLEN-1 instead of MAXLEN?
3) What is the significance of the second parameter of the fgets() function? When I run the program, I am able to type as many characters as I want. What is MAXLEN doing to guard against buffer overflow? From what I can guess, when the user inputs a big long string, once the user hits return, the MAXLEN chops up the string in to MAXLEN sized bites/bytes (both actually work here lol) and sends them to the array. I'm sure I'm missing something important here.
That was a mouthful, but my lack of grasp on this very important subject is making my code weak.
Question 1
You can actually type as much character as your command line tool will allow you per input. However, you call to fgets() will handle only MAXLEN in your example because you tell him to do so.
Moreover, there is no safe check inside fgets(). The second parameter you gave to fgets is the "safety" argument. Try to give to change your call to fgets to fgets(buffer, MAXLEN + 10, stdin); and then type more than MAXLEN characters. Your program will crash because you are accessing unallocated memory.
Question 2
When you make a call to fgets(), it will read MAXLEN - 1 characters because the last one is reserved to the character code \0 which usually means end of string
The second parameter of fgets() is not the number of character you want to store but the maximum capacity of your buffer. And you always have to think about string termination character \0
Question 3
If you undestood the 2 answer before, you will be able to answer to this one by yourself. Try to play with this value. And use a different value than the one used for you buffer size.
Also, you said
p3 specifies the input stream, which in this code is 'stdin' (when would it ever be different?)
You can use fgets to read files stored on your computer. Here is an example :
char buffer[20];
FILE *stream = fopen("myfile.txt", "r"); //Open the file "myfile.txt" in readonly mode
fgets(buffer, 20, stream); //Read the 19 first characters of the file "myfile.txt"
puts(buffer);
When you call fgets(), it lets you type in as much as you want into stdin, so everything stays in stdin. It seems fgets() takes the first 9 characters, attaches a null character, and assigns it to buffer. Then puts() displays buffer then creates a newline.
The key is it's in a while loop -- the code loops again then takes what was remaining in stdin and feeds it into fgets(), which takes the next 9 characters and repeats. Stdin just still had stuff "in queue".
Input and Output in C has no specific viable function for working with strings.
There are several functions for outputting strings, such as printf and puts.
Strings can be input with fgets or scanf; however there is no standard function that both inputs and allocates memory. You need to pre-allocate some memory, and then read some characters into that memory.
Your analogy of a stream as a river is not great. Rivers flow whether or not you are taking items out of them, but streams don't. A better analogy might be a line of people at the gates to a stadium.
C also has the concept of a "line", lines are marked by having a '\n' character at the end. In my analogy let's say the newline character is represented by a short person.
When you do fgets(buf, 20, stdin) it is like "Let the next 19 people in, but if you encounter a short person during this, let him through but not anybody else". Then the fgets function creates a string out of these 0 to 19 characters, by putting the end-of-string marker on the end; and that string is placed in buf.
Note that the second argument to fgets is the buffer size , not the number of characters to read.
When you type in characters, that is like more people joining the queue.
If there were fewer than 19 people and no short people, then fgets waits for more people to arrive. In standard C there's no way to check if people are waiting without blocking to wait for them if they aren't.
By default, C streams are line buffered. In my analogy, this is like there is a "pre-checking" gate earlier on than the main gate, where all people that arrive go into a holding pen until a short person arrives; and then everyone from the holding pen plus that short person get sent onto the main gate. This can be turned off using setvbuf.
Never think of the 'beginning' or 'end' of the streams: this is handled by the operating system.
This is something you do have to worry about. stdin etc. are already begun before you enter main(), but other streams (e.g. if you want to read from a file on your hard drive), you have to begin them.
Streams may end. When a stream is ended, fgets will return NULL. Your program must handle this. In my analogy, the gate is closed.
I'm trying for hours to find the answer for this question i've got in university. I tried running this with writing a file with two lines of :
hello
world
and it reads the file perfectly, So i cant find the answer. I would appreciate your help !
A student wrote the next function for reading a text file and printing it exactly as it is.
void ReadFile(FILE *fIn)
{
char nextLine[MAX_LINE_LENGTH];
while(!feof(fIn))
{
fscanf(fIn,"%s",nextLine);
printf("%s\n",nextLine);
}
}
What are the two errors in this function?
You can assume that each line in the file is not longer than MAX_LINE_LENGTH characters, and that it is a text file that contains only alphabet characters, and that each line is terminated by '\n'.
Thanks.
It discards white space. Try adding multiple spaces and tabs.
It may evaluate a stream more than once, and If there is a read error, the loop never terminates.
See: Why is “while ( !feof (file) )” always wrong?
Reading strings via scanf is dangerous. There is no bounds checking. You may read past you MAX_LINE_LENGTH.(and boom! Segfault)
The main error is that fsacnf( fIn, "%s", nextLine ) doesn't scan a complete line.
From man page:
s
Matches a sequence of non-white-space characters; the next pointer must be a pointer to character array that is long enough to hold the input sequence and the terminating null byte ('\0'), which is added automatically. The input string stops at white space or at the maximum field width, whichever occurs first.
Thus if you have a line "a b" the first fscanf() will scan just "a" and the second one "b" and both are printed in two different lines. You can use fgets() to read a whole line.
The second one is maybe that it's stated "each line in the file is not longer than MAX_LINE_LENGTH characters" but nextLine can contain atmost MAX_LINE_LENGTH-1 characters (+ '\0'). That problem becomes even more important if you replace fscanf() by fgets() because than nextLine must have also capacity to store '\n' or '\r\n' (depending on the platform you're on)
A correct way of doing that is:
void ReadFile(FILE *fIn)
{
char nextLine[MAX_LINE_LENGTH];
while(fgets(nextLine, MAX_LINE_LENGTH, fIn)) {
printf("%s", nextLine);
}
}
As some have posted using feof to control a loop is not a good idea nor using fscanf to read lines.