I don't understand the behavior of fgets in this example - c

While I could use strings, I would like to understand why this small example I'm working on behaves in this way, and how can I fix it ?
int ReadInput() {
char buffer [5];
printf("Number: ");
fgets(buffer,5,stdin);
return atoi(buffer);
}
void RunClient() {
int number;
int i = 5;
while (i != 0) {
number = ReadInput();
printf("Number is: %d\n",number);
i--;
}
}
This should, in theory or at least in my head, let me read 5 numbers from input (albeit overwriting them).
However this is not the case, it reads 0, no matter what.
I understand printf puts a \0 null terminator ... but I still think I should be able to either read the first number, not just have it by default 0. And I don't understand why the rest of the numbers are OK (not all 0).
CLARIFICATION: I can only read 4/5 numbers, first is always 0.
EDIT:
I've tested and it seems that this was causing the problem:
main.cpp
scanf("%s",&cmd);
if (strcmp(cmd, "client") == 0 || strcmp(cmd, "Client") == 0)
RunClient();
somehow.
EDIT:
Here is the code if someone wishes to compile. I still don't know how to fix
http://pastebin.com/8t8j63vj
FINAL EDIT:
Could not get rid of the error. Decided to simply add #ReadInput
int ReadInput(BOOL check) {
...
if (check)
printf ("Number: ");
...
# RunClient()
void RunClient() {
...
ReadInput(FALSE); // a pseudo - buffer flush. Not really but I ignore
while (...) { // line with garbage data
number = ReadInput(TRUE);
...
}
And call it a day.

fgets reads the input as well as the newline character. So when you input a number, it's like: 123\n.
atoi doesn't report errors when the conversion fails.
Remove the newline character from the buffer:
buf[5];
size_t length = strlen(buffer);
buffer[length - 1]=0;
Then use strtol to convert the string into number which provides better error detection when the conversion fails.

char * fgets ( char * str, int num, FILE * stream );
Get string from stream.
Reads characters from stream and stores them as a C string into str until (num-1) characters have been read or either a newline or the end-of-file is reached, whichever happens first.
A newline character makes fgets stop reading, but it is considered a valid character by the function and included in the string copied to str. (This means that you carry \n)
A terminating null character is automatically appended after the characters copied to str.
Notice that fgets is quite different from gets: not only fgets accepts a stream argument, but also allows to specify the maximum size of str and includes in the string any ending newline character.
PD: Try to have a larger buffer.

Related

C - Using fgets until newline/-1 [closed]

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So I'm trying to make it so that you can write text into a file until you make a newline or type -1. My problem is that when you write, it just keeps going until it crashes and gives the error "Stack around the variable "inputChoice" was corrupted".
I believe the problem is that the program doesn't stop accepting stdin when you want to stop typing (-1, newline) and that causes the error. I've tried with a simple scanf and it works, but you can only write a word. No spaces and it doesn't support multiple lines either. That's why I have to use fgets
Judging from your comments, I assume that there are some basic concepts in C
that you haven't fully understood, yet.
C-Strings
A C-String is a sequence of bytes. This sequence must end with the value 0.
Every value in the sequence represents a character based on the
ASCII encoding, for example the
character 'a' is 97, 'b' is 98, etc. The character '\0' has
the value 0 and it's the character that determines the end of the string.
That's why you hear a lot that C-Strings are '\0'-terminated.
In C you use an array of chars (char string[], char string[SOME VALUE]) to
save a string. For a string of length n, you need an array of dimension n+1, because
you also need one space for the terminating '\0' character.
When dealing with strings, you always have to think about the proper type,
whether your are using an array or a pointer. A pointer
to char doesn't necessarily mean that you are dealing with a C-String!
Why am I telling you this? Because of:
char inputChoice = 0;
printf("Do you wish to save the Input? (Y/N)\n");
scanf("%s", &inputChoice);
I haven't changed much, got very demotivated after trying for a while.
I changed the %s to an %c at scanf(" %c, &inputChoice) and that
seems to have stopped the program from crashing.
which shows that haven't understood the difference between %s and %c.
The %c conversion specifier character tells scanf that it must match a single character and it expects a pointer to char.
man scanf
c
Matches a sequence of characters whose length is specified by the maximum field
width (default 1); the next pointer must be a
pointer to char, and there must be enough room for all the characters
(no terminating null byte is added). The usual skip of
leading white space is suppressed. To skip white space first, use an explicit space in the format.
Forget the bit about the length, it's not important right now.
The important part is in bold. For the format scanf("%c", the function
expects a pointer to char and its not going to write the terminating '\0'
character, it won't be a C-String. If you want to read one letter and one
letter only:
char c;
scanf("%c", &c);
// also possible, but only the first char
// will have a defined value
char c[10];
scanf("%c", c);
The first one is easy to understand. The second one is more interesting: Here
you have an array of char of dimension 10 (i.e it holds 10 chars). scanf
will match a single letter and write it on c[0]. However the result won't be
a C-String, you cannot pass it to puts nor to other functions that expect
C-Strings (like strcpy).
The %s conversion specifier character tells scanf that it must match a sequence of non-white-space characters
man scanf
s
Matches a sequence of non-white-space characters; the next pointer must be a
pointer to the initial element of a character array that is long enough to
hold the input sequence and the terminating null byte ('\0'), which is added
automatically.
Here the result will be that a C-String is saved. You also have to have enough
space to save the string:
char string[10];
scanf("%s", string);
If the strings matches 9 or less characters, everything will be fine, because
for a string of length 9 requires 10 spaces (never forget the terminating
'\0'). If the string matches more than 9 characters, you won't have enough
space in the buffer and a buffer overflow (accessing beyond the size) occurs.
This is an undefined behaviour and anything can happen: your program might
crash, your program might not crash but overwrites another variable and thus
scrwes the flow of your program, it could even kill a kitten somewhere, do
you really want to kill kittens?
So, do you see why your code is wrong?
char inputChoice = 0;
scanf("%s", &inputChoice);
inputChoice is a char variable, it can only hold 1 value.
&inputChoice gives you the address of the inputChoice variable, but the
char after that is out of bound, if you read/write it, you will have an
overflow, thus you kill a kitten. Even if you enter only 1 character, it will
write at least 2 bytes and because you it only has space for one character, a kitten will die.
So, let's talk about your code.
From the perspective of an user: Why would I want to enter lines of text, possibly a lot of lines of text
and then answer "No, I don't want to save the lines". It doesn't make sense to
me.
In my opinion you should first ask the user whether he/she wants to save the
input first, and then ask for the input. If the user doesn't want to save
anything, then there is no point in asking the user to enter anything at
all. But that's just my opinion.
If you really want to stick to your plan, then you have to save every line and
when the user ends entering data, you ask and you save the file.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define BUFFERLEN 1024
void printFile () {
int i;
char openFile[BUFFERLEN];
FILE *file;
printf("What file do you wish to write in?\n");
scanf("%s", openFile);
getchar();
file = fopen(openFile, "w");
if (file == NULL) {
printf("Could not open file.\n");
return;
}
// we save here all lines to be saved
char **lines = NULL;
int num_of_lines = 0;
char buffer[BUFFERLEN];
printf("Enter an empty line of -1 to end input\n");
// for simplicity, we assume that no line will be
// larger than BUFFERLEN - 1 chars
while(fgets(buffer, sizeof buffer, stdin))
{
// we should check if the last character is \n,
// if not, buffer was not large enough for the line
// or the stream closed. For simplicity, I will ignore
// these cases
int len = strlen(buffer);
if(buffer[len - 1] == '\n')
buffer[len - 1] = '\0';
if(strcmp(buffer, "") == 0 || strcmp(buffer, "-1") == 0)
break; // either an empty line or user entered "-1"
char *line = strdup(buffer);
if(line == NULL)
break; // if no more memory
// process all lines that already have been entered
char **tmp = realloc(lines, (num_of_lines+1) * sizeof *tmp);
if(tmp == NULL)
{
free(line);
break; // same reason as for strdup failing
}
lines = tmp;
lines[num_of_lines++] = line; // save the line and increase num_of_lines
}
char inputChoice = 0;
printf("Do you wish to save the Input? (Y/N)\n");
scanf("%c", &inputChoice);
getchar();
if (inputChoice == 'Y' || inputChoice == 'y') {
for(i = 0; i < num_of_lines; ++i)
fprintf(file, "%s\n", lines[i]); // writing every line
printf("Your file has been saved\n");
printf("Please press any key to continue");
getchar();
}
// closing FILE buffer
fclose(file);
// free memory
if(num_of_lines)
{
for(i = 0; i < num_of_lines; ++i)
free(lines[i]);
free(lines);
}
}
int main(void)
{
printFile();
return 0;
}
Remarks on the code
I used the same code as yours as the base for mine, so that you can spot the
differences much quicker.
I use the macro BUFFERLEN for declaring the length of the buffers. That's
my style.
Look at the fgets line:
fgets(buffer, sizeof buffer, stdin)
I use here sizeof buffer instead of 1024 or BUFFERLEN. Again, that's my
style, but I think doing this is better, because even if you change the size
of the buffer by changing the macro, or by using another explicit size, sizeof buffer
will always return the correct size. Be aware that this only works when
buffer is an array.
The function strdup returns a pointer a pointer to a new string that
duplicates the argument. It's used to create a new copy of a string. When
using this function, don't forget that you have to free the memory using
free(). strdup is not part of the standard library, it conforms
to SVr4, 4.3BSD, POSIX.1-2001. If you use Windows (I don't use Windows,
I'm not familiar with the Windows ecosystem), this function might not be
present. In that case you can write your own:
char *strdup(const char *s)
{
char *str = malloc(strlen(s) + 1);
if(str == NULL)
return NULL;
strcpy(str, s);
return str;
}

How to discard the rest of a line in C

I'm trying to write a function that removes the rest of a line in C. I'm passing in a char array and a file pointer (which the char array was read from). The array is only supposed to have 80 chars in it, and if there isn't a newline in the array, read (and discard) characters in the file until you reach it (newline). Here's what I have so far, but it doesn't seem to be working, and I'm not sure what I'm doing wrong. Any help would be greatly appreciated! Here's the given information about what the function should do:
discardRest - if the fgets didn't read a newline than an entire line hasn't been read. This function takes as input the most recently read line and the pointer to the file being read. discardRest looks for the newline character in the input line. If newline character is not in the line, the function reads (and discards) characters from the file until the newline is read. This will cause the file pointer to be positioned to the beginning of the next line in the input file.
And here's the code:
void discardRest(char line[], FILE* file)
{
bool newlineFound = FALSE;
int i;
for(i = 0; i < sizeof(line); i++)
{
if(line[i] == '\n') newlineFound = TRUE;
}
if(!newlineFound)
{
int c = getc(file);
while(c != '\n')
{
c = getc(file);
}
}
}
Your way is much too difficult, besides sizeof always giving the size of its operand, which is a pointer and not the array it points to which you think it is.
fgets has thefollowing contract:
return NULL: Some kind of error, do not use the buffer, its content might be indeterminate.
otherwise the buffer contains a 0-terminated string, with the last non-0 being the retained '\n' if the buffer and the file were both large enough.
Thus, this should work:
So, use strlen() to get the buffer length.
Determine if a whole line was read (length && [length-1] == '\n').
As appropriate:
remove the newline character and return.
discard the rest of the line like you tried.

Why fgets is not inputting first value?

I am writing a program to write my html files rapidly. And when I came to write the content of my page I got a problem.
#include<stdio.h>
int main()
{
int track;
int question_no;
printf("\nHow many questions?\t");
scanf("%d",&question_no);
char question[question_no][100];
for(track=1;track<=question_no;track++)
{
printf("\n<div class=\"question\">%d. ",track);
printf("\nQuestion number %d.\t",track);
fgets(question[track-1],sizeof(question[track-1]),stdin);
printf("\n\n\tQ%d. %s </div>",track,question[track-1]);
}
}
In this program I am writing some questions and their answers (in html file). When I test run this program I input the value of question_no to 3. But when I enter my first question it doesn't go in question[0] and consequently the first question doesn't output. The rest of the questions input without issue.
I searched some questions on stackoverflow and found that fgets() looks for last \0 character and that \0 stops it.
I also found that I should use buffer to input well through fgets() so I used: setvbuf and setbuf but that also didn't work (I may have coded that wrong). I also used fflush(stdin) after my first and last (as well) scanf statement to remove any \0 character from stdin but that also didn't work.
Is there any way to accept the first input by fgets()?
I am using stdin and stdout for now. I am not accessing, reading or writing any file.
Use fgets for the first prompt too. You should also malloc your array as you don't know how long it is going to be at compile time.
#include <stdlib.h>
#include <stdio.h>
#define BUFSIZE 8
int main()
{
int track, i;
int question_no;
char buffer[BUFSIZE], **question;
printf("\nHow many questions?\t");
fgets(buffer, BUFSIZE, stdin);
question_no = strtol(buffer, NULL, 10);
question = malloc(question_no * sizeof (char*));
if (question == NULL) {
return EXIT_FAILURE;
}
for (i = 0; i < question_no; ++i) {
question[i] = malloc(100 * sizeof (char));
if (question[i] == NULL) {
return EXIT_FAILURE;
}
}
for(track=1;track<=question_no;track++)
{
printf("\n<div class=\"question\">%d. ",track);
printf("\nQuestion number %d.\t",track);
fgets(question[track-1],100,stdin);
printf("\n\n\tQ%d. %s </div>",track,question[track-1]);
}
for (i = 0; i < question_no; ++i) free(question[i]);
free(question);
return EXIT_SUCCESS;
}
2D arrays in C
A 2D array of type can be represented by an array of pointers to type, or equivalently type** (pointer to pointer to type). This requires two steps.
Using char **question as an exemplar:
The first step is to allocate an array of char*. malloc returns a pointer to the start of the memory it has allocated, or NULL if it has failed. So check whether question is NULL.
Second is to make each of these char* point to their own array of char. So the for loop allocates an array the size of 100 chars to each element of question. Again, each of these mallocs could return NULL so you should check for that.
Every malloc deserves a free so you should perform the process in reverse when you have finished using the memory you have allocated.
malloc reference
strtol
long int strtol(const char *str, char **endptr, int base);
strtol returns a long int (which in the code above is casted to an int). It splits str into three parts:
Any white-space preceding the numerical content of the string
The part it recognises as numerical, which it will try to convert
The rest of the string
If endptr is not NULL, it will point to the 3rd part, so you know where strtol finished. You could use it like this:
#include <stdio.h>
#include <stdlib.h>
int main()
{
char * endptr = NULL, *str = " 123some more stuff";
int number = strtol(str, &endptr, 10);
printf("number interpreted as %d\n"
"rest of string: %s\n", number, endptr);
return EXIT_SUCCESS;
}
output:
number interpreted as 123
rest of string: some more stuff
strtol reference
This is because the previous newline character left in the input stream by scanf(). Note that fgets() stops if it encounters a newline too.
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
Don't mix fgets() and scanf(). A trivial solution is to use getchar() right after scanf() in order to consume the newline left in the input stream by scanf().
As per the documentation,
The fgets() function shall read bytes from stream into the array
pointed to by s, until n-1 bytes are read, or a < newline > is read and
transferred to s, or an end-of-file condition is encountered
In case of scanf("%d",&question_no); a newline is left in the buffer and that is read by
fgets(question[track-1],sizeof(question[track-1]),stdin);
and it exits.
In order to flush the buffer you should do,
while((c = getchar()) != '\n' && c != EOF)
/* discard */ ;
to clear the extra characters in the buffer

Can't determine value of character at the end of a string

I'm new to C programming. I am trying to make a program that takes some simple input. However I found that on comparison of my input string to what the user "meant" to input, there is an additional character at the end. I thought this might be a '\0' or a '\r' but that seems not to be the case. This is my snippet of code:
char* getUserInput(char* command, char $MYPATH[])
{
printf("myshell$ ");
fgets(command, 200, stdin);
printf("%u\n", (unsigned)strlen(command));
if ((command[(unsigned)strlen(command) - 1] == '\0') || (command[(unsigned)strlen(command) - 1] == '\r'))
{
printf("bye\n");
}
return command;
}
The code shows that when entering, say "exit" that 5 characters are entered. However I can't seem to figure out the identity of this last one. "Bye" never prints. Does anyone know what this mystery character could be?
The magical 5th element most probably is a newline character: \n
From man fgets() (emphasis by me):
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 '\0' is
stored after the last character in the buffer.
To prove this print out each character read by doing so:
char* getUserInput(char* command, char $MYPATH[])
{
printf("myshell$ ");
fgets(command, 200, stdin);
printf("%u\n", (unsigned)strlen(command));
{
size_t i = 0, len = strlen(command);
for (;i < len; ++i)
{
fprintf(stderr, "command[%zu]='%c' (%hhd or 0x%hhx)\n", i, command[i], command[i], command[i]);
}
}
...
assumptions
array indexes in c are started with 0
strlen returns length of string
so, if you have string "exit", this will be 5 symbols in array = e, x, i, t, \0, strlen return 4, but you're trying to decrement it by 1, so you're checking last symbol in string, instead on NULL terminator
to check NULL terminator use command[strlen(command)] - this will give you \0 always, so there is no sense in it
if you want to compare strings use strcmp function
UPDATE: issue with your program is because fgets appends \n symbol at then end of string:
A newline character makes fgets stop reading, but it is considered a
valid character by the function and included in the string copied to
str.
The reason you don't see the last char is because strlen() won't calculate '\0' into the string's length. So testing for '\0' wont succeed.
for instance, const char* a = "abc"; then strlen(a) will be 3. if you want to test it, you need to access it by command[strlen(command)]
The reason for getting strlen equals to 5 on "exit" is because fgets will append the '\n' character at the end of the input. You could test it by command[strlen(command) -1 ] == '\n'

Input/Output scanset in c

#include<stdio.h>
int main()
{
char str[50]={'\0'};
scanf("%[A-Z]s",str);
printf("%s",str);
return 0;
}
1)
Input:
helloWORLD
output:
2)
Input:
HELLoworlD
output:
HELL
In output 1, i expected the output as "WORLD" but it didnt give any outout.
From output 2, i understood that this is working only if the first few characters are in upper case.
Can you please explain how it actually works?
Interpretation of scansets
When it is given helloWORLD, the conversion specification %[A-Z] fails immediately because the h is not an upper-case letter. Therefore, scanf() returns 0, indicating that it did not successfully convert anything. If you tested the return value, you'd know that.
When it is given HELLoworlD, the scanset matches the HELL and stops at the first o. The format string also attempts to match a literal s, but there's no way for scanf() to report that it fails to match that after matching HELL.
Buffer overflow
Note that %[A-Z] is in general dangerous (as is %s) because there is no constraint on the number of characters read. If you have:
char str[50];
then you should use:
if (scanf("%49[A-Z]", str) != 1)
...some problem in the scan...
Also note that there is a 'difference by one' between the declared length of str and the number in the format string. This is awkward; there's no way to provide that number as an argument to scanf() separate from the format string (unlike printf()), so you may end up creating the format string on the fly:
int scan_upper(char *buffer, size_t buflen)
{
char format[16];
if (buflen < 2)
return EOF; // Or other error indication
snprintf(format, sizeof(format), "%%%zu[A-Z]", buflen-1); // Check this too!?
return scanf(format, buffer);
}
When you do
scanf("%[A-Z]s",str);
It takes input as long as you enter upper-case letters.
And since you set all the array to '\0', printf() will stop printing when it meets one.
Therefore, the first input is blank, and the second is printing until the end of the upper-case string.

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