puts() output is appended "time" string - c

I get very unexpected output from quite simple code
char ch = getchar(), word[100], *p = word;
while (ch != '\n') {
*(p++) = ch;
ch = getchar();
}
puts(word);
output of any 17 character input is appended by "time" like
12345678901234567time
if exceeds "time" is overwritten like
1234567890123456789me
Am I doing something wrong?

puts expects a pointer to string. And a string needs to have a terminating null character - \0 - to signify where the string ends.
But in your case, you did not write the \0 at the end to signify that the string ends there.
You need to do:
char ch = getchar(), word[100], *p = word;
/* Also check that you are not writing more than 100 chars */
int i = 1;
while(ch != '\n' && i++ < 100){
*(p++) = ch;
ch = getchar();
}
*p = '\0'; /* write the terminaring null character */
puts(word);
Before, when you were not writing the terminating null character you could not expect anything determinate to print. It could also have been 12345678901234567AnyOtherWord or something.

There are multiple issues in your code:
You do not null terminate the string you pass to puts(), invoking undefined behavior... in your case, whatever characters happen to be present in word after the last one read from stdin are printed after these and until (hopefully) a '\0' byte is finally found in memory.
You read a byte from stdin into a char variable: this does not allow you to check for EOF, and indeed you do not.
If you read a long line, you will write bytes beyond the end if the word array, invoking undefined behavior. If the end of file is encountered before a '\n' is read from stdin, you will definitely write beyond the end of the buffer... Try for example giving an empty file as input for your program.
Here is a corrected version:
char word[100];
char *p = word;
int ch;
while ((ch = getchar()) != EOF && ch != '\n') {
/* check for long line: in this case, we truncate the line */
if (p < word + sizeof(word) - 1) {
*p++ = ch;
}
}
*p = '\0';
puts(word);

Related

how to stop my program from skipping characters before saving them

I am making a simple program to read from a file character by character, puts them into tmp and then puts tmp in input[i]. However, the program saves a character in tmp and then saves the next character in input[i]. How do I make it not skip that first character?
I've tried to read into input[i] right away but then I wasn't able to check for EOF flag.
FILE * file = fopen("input.txt", "r");
char tmp;
char input[5];
tmp= getc(file);
input[0]= tmp;
int i=0;
while((tmp != ' ') && (tmp != '\n') && (tmp != EOF)){
tmp= getc(file);
input[i]=tmp;
length++;
i++;
}
printf("%s",input);
It's supposed to print "ADD $02", but instead it prints "DD 02".
You are doing things in the wrong order in your code: The way your code is structures, reading and storing the first char is moved out of the loop. In the loop, that char is then overwritten. In that case start with i = 1.
Perhaps you want to read the first character anyway, but I guess you want to read everything up to the first space, which might be the first character. Then do this:
#include <stdio.h>
int main(void)
{
char input[80];
int i = 0;
int c = getchar();
while (c != ' ' && c != '\n' && c != EOF) {
if (i + 1 < sizeof(input)) { // store char if the is room
input[i++] = c;
}
c = getchar();
}
input[i] = '\0'; // null-terminate input
puts(input);
return 0;
}
Things to note:
The first character is read before the loop. the loop condition and the code that stores the char then use that char. Just before the end of the loop body, the next char is read, which will then be processed in the next iteration.
You don't enforce that the char buffer input cannot be overwritten. This is dangerous, especially since your buffer is tiny.
When you construct strings char by char, you should null-terminate it by placing an explicit '\0' at the end. You have to make sure that there is space for that terminator. Nearly all system functions like puts or printf("%s", ...) expect the string to be null-terminated.
Make the result of getchar an int, so that you can distinguish between all valid character codes and the special value EOF.
The code above is useful if the first and subsequent calls to get the next item are different, for example when tokenizing a string with strtok. Here, you can also choose another approach:
while (1) { // "infinite loop"
int c = getchar(); // read a char first thing in a loop
if (c == ' ' || c == '\n' || c == EOF) break;
// explicit break when done
if (i + 1 < sizeof(input)) {
input[i++] = c;
}
}
This approach has the logic of processing the chars in the loop body only, but you must wrap it in an infinite loop and then use the explicit break.

Buffer and its interaction with function?

I am facing some problems while understanding the following code.
It is a program to read Strings from keyboard if the length of the String is lesser than the specified size (i.e 'n' here).
If the length of a string is larger than the specified size, the remaining characters on the line will be discarded.
More specifically, I want to know what is happening inside the buffer and how getchar() is reading the data and not storing it in the buffer.
char * s_gets(char * st, int n)
{
char * ret_val;
int i = 0;
ret_val = fgets(st, n, stdin);
if (ret_val) // i.e., ret_val != NULL
{
while (st[i] != '\n' && st[i] != '\0')
i++;
if (st[i] == '\n')
st[i] = '\0';
else // must have words[i] == '\0'
while (getchar() != '\n')
continue;
}
return ret_val;
}
The code is slightly flawed, but does more or less the job outlined. It uses fgets() to do a lot of the work. That reads up to n - 1 characters from standard input. When it returns, there are a few possibilities:
(EOF) Nothing was available to be read. Nothing has been put in the buffer, but fgets() returned NULL.
(Normal) A line has been read and fitted into the buffer st. The buffer includes the newline.
(Overlong) Part of a line has been read, but the input did not find a newline.
(EOF without newline) Some data was read, but there wasn't a newline before EOF was detected.
Case 1 is simplest: the code returns NULL. Case 2 is handled by scanning the string that's read to find the newline. If the newline is found, it is overwritten with a null byte. Case 3 is handled at the same time; if the value found isn't a newline, it must be the null byte. The code drops into a loop that reads more characters until a newline is read. Case 4 is similar to case 3 in effect, but the code in the loop mishandles this — it doesn't detect and handle EOF, so the code would fall into an indefinite loop. That's a bug that needs to be fixed.
The getchar() loop doesn't assign anything to the buffer st — it makes no changes to st. That continues to contain a null terminated string as read by fgets(). The getchar() loop reads and discards any characters left on the line that was read that did not fit into the buffer.
The code should be:
char *s_gets(char *st, int n)
{
assert(n > 1 && st != NULL);
char *ret_val = fgets(st, n, stdin);
if (ret_val != NULL)
{
int i = 0;
while (st[i] != '\n' && st[i] != '\0')
i++;
if (st[i] == '\n')
st[i] = '\0';
else
{
int c;
while ((c = getchar()) != EOF && c != '\n')
continue;
}
}
return ret_val;
}
The return value of NULL or the original string is the same as fgets() uses, but it isn't the most useful return value. It would be more useful, most often, if the code returned the length of the string that it read, or returned EOF if it encountered EOF (or a read error). The information is readily available — in the variable i.

Is this a valid use of fgetc?

My input stream is from a text file with a list of words separated by the \n character.
The function stringcompare is a function that will compare the equivalence of two strings, case insensitive.
I have two string arrays, word[50] and dict[50]. word is a string that would be given by the user.
Basically what I want to do is pass word[] and each word in the text file as arguments of the stringcompare function.
I've compiled and run this code but it is wrong. Very wrong. What am I doing wrong? Can I even use fgetc() like this? would dict[] even be a string array after the inner loop is done?
char c, r;
while((c = fgetc(in)) != EOF){
while((r = fgetc(in)) != '\n'){
dict[n] = r;
n++;
}
dict[n+1] = '\0'; //is this necessary?
stringcompare(word, dict);
}
It is wrong.
The return value of fgetc() should be stored to int, not char, especially when it will be compared with EOF.
You might forgot to initialize n.
You will miss the first character of each line, which is stored to c.
Use dict[n] = '\0'; instead of dict[n+1] = '\0'; because n is already incremented in the loop.
Possible fix:
int c, r;
while((c = fgetc(in)) != EOF){
ungetc(c, in); // push the read character back to the stream for reading by fgetc later
n = 0;
// add check for EOF and buffer overrun for safety
while((r = fgetc(in)) != '\n' && r != EOF && n + 1 < sizeof(dict) / sizeof(dict[0])){
dict[n] = r;
n++;
}
dict[n] = '\0'; //this is necessary
stringcompare(word, dict);
}

Splitting scanf input into arrays until EOF

Looking to read in using scanf but I want to stop reading if I encounter a ',' '\0' (newline) or EOF
I'm not really sure how to stop achieve this.
I was using
char * aBuff;
char * bBuff;
char * cBuff;
//read in the first three lines and put them into char arrays
//while (scan() != (',' || '\0' || EOF)) //was trying to put it into a while loop, wasn't sure
scanf("%s", aBuff);
scanf("%s", bBuff);
scanf(%s, cBUff);
I plan on taking the input and putting them into separate arrays. Basically take input until a , or new line and place that data into an array and continue this process until the end of file.
scanf() is not a practical method to read until encountering ',', '\0', or EOF. Use fgetc().
The biggest problem is specifying '\0' in the format of scanf(). Example: with format "%[^,\0]", scanf() only reads "%[^," as it stops at the embedded '\0'. So with an invalid format specifier --> undefined behavior.
size_t ReadX(char *dest, size_t size) {
size_t len = 0;
if (size) {
while (--size > 0) {
int ch = fgetc(stdin);
if (ch == 0 || ch == ',' || ch == EOF) break; // maybe add \n too.
*dest[len++] = ch;
}
*dest[len] = '\0';
}
return len; // or maybe return the stopping ch
}
scanf() could be use if code used the ponderous:
scanf("%[\1\2\3...all_char_codes_min_char_to_max_char_except_,_and\0]%*c", &s);
You can try using scansets
scanf() should stop on EOF, but you'd want to maybe do something like this:
scanf("%[^,\0]", &s);

How can know if the end of line in C

If I do :
int main(){
const int LENGTH_LINE = 100;
char line[LENGTH_LINE];
int len;
FILE* fp = fopen(file.txt,"r");
fgets(line,LENGTH_LINE,fp);
len = strlen(line);
if(line[len-1] == '\n')
printf("I've a line");
//This work if the line have \n , but if the end line of the text dont have \n how can do it?
}
I need to know if I take a whole line with fgets because I got a delimiter.
According to http://en.cppreference.com/w/c/io/fgets
Reads at most count - 1 characters from the given file stream and stores them in str.
Parsing stops if end-of-file occurs or a newline character is found, in which case str will contain that newline character.
So, once fgets returns, there are 3 possibilities
LENGTH_LINE was reached
We got a newline
EOF was reached.
I'm assuming you have a line in cases 2 and 3.
In this case the detection condition is :
line[len-1] == '\n' || feof(fp)
Check for the newline character:
size_t len = 0;
// ... your code using fgets
len = strlen(line);
if ((len > 0) && (line[len - 1] == '\n'))
// your input contains the newline
After the fgets call, your line may not have a newline at the end if:
The character limit was reached before a newline was scanned - in your case this is LENGTH_LINE.
The end-of-file (EOF) was reached before a newline.
There was a read error, but in case of an error consider the contents of line unusable.
You should be looking at the return value from fgets so that you'll be able to handle the EOF: fgets returns NULL upon end-of-file or a read error. You can use feof to check for the end-of-file.
If you check feof, and know that you're at the end of your input with no fgets errors, then even without a newline character on the final line you'll know that you've read the entire line.
If for some reason you must have a newline character terminating each line, you can add it yourself:
// you've checked for EOF and know this is your final line:
len = strlen(line);
if (line[len-1] == '\n')
printf("I've a line");
else if ((len + 1) < LENGTH_LINE)
{
line[len] = '\n';
line[len + 1] = '\0';
}
else
// no room in your line buffer for an add'l character
Use like this
while(fgets(line,LENGTH_LINE,fp)!=EOF)
// your code here
Why not just use fgetc instead? That way you can just keep scanning until you get to the end of the line so you don't have to check if you have it.
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
char line[100];
int ch, i = 0;
FILE* fp = fopen(file.txt,"r");
while(ch != '\n' || ch != '\r' || ch != EOF) //or ch != delimiter
{
ch = fgetc(fp);
line[i] = ch;
i++;
}
line[i] = '\n';
line[i+1] = 0x00;
return 0;
}
In that example I just look for a new line, return, or EOF character but you can really make it look for anything you like (e.g. your delimiter). So if your delimiter was q you would just do
while(ch != 'q')...

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