This question already has answers here:
strstr not functioning
(2 answers)
Closed 7 years ago.
Basically, I would like to know why this code doesn´t work. It appears that the value of strstr() is always NULL, as all this code ever does is "word not found".
I have already tried if (strstr(retezec,substring)!=NULL), but it doesn't work either.
int main()
{
FILE *files;
files = fopen("Knihovna.txt","rb+");
int i = 0;
while(fgetc(files)!=EOF){
i++;
}
//printf("%d",i);
rewind(files);
char *retezec;
retezec = (char *)malloc(i);
fread(retezec, i, 1, files);
puts("zadejte hledane slovo");
char *substring;
substring = (char *)malloc(50);
fflush(stdin);
fgets(substring,49, stdin);
char *found;
found = strstr(retezec,substring);
if(found){
printf("word found!");
}
else{
puts("word not found");
}
}
It's most likely the result of fgets() reading the trailing newline.
fgets(substring,49, stdin);
This would read the trailing newline if substring has space. So if you input "name". You actually have "name\n".
Remove the trailing newline with:
char *p = strchr(substring, '\n');
if (p) *p = 0; // remove the newline, if present
You have another problem. fread()doesn't NUL terminate. It simply reads the requested bytes. So you would need to check if fread() read i bytes or less and use that number (return value of fread()) to find the actual bytes read. Because it may be less than requested. And then, allocate one extra byte and NUL terminate it if you want to use it as a C-string.
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Closed 5 years ago.
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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;
}
This question already has answers here:
Removing trailing newline character from fgets() input
(14 answers)
Closed 7 years ago.
The task is write a c program to print lines read from an input file (maybe very very large), but without '\n'. Please see the comment in the code below, is it a typical way or good way to do so??
int main() {
const char *input_wavlist_file = "/home/legend0011/Downloads/test-sample-list.txt";
const int BUFFER_SIZE = 100;
FILE *fr = fopen(input_wavlist_file, "r");
if (fr == NULL) {
printf("Error opening input wav list file!\n");
exit(1);
}
char str[BUFFER_SIZE];
while((fgets(str, BUFFER_SIZE, fr)) != NULL) {
char *pch = strchr(str, '\n');
char *filepath = str;
if (pch != NULL) *pch = '\0'; // is this a typical way????????
printf("reading==>%s",filepath);
}
fclose(fr);
}
fgets() comes with a newline character suppress the newline and print it.
size_t n = strlen(str);
if(n>0 && str[n-1] == '\n')
{
str[n-1] = '\0';
}
printf("%s",str);
The issue you're facing here is with fgets() behaviour. As per the man page,
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....
So, it reads and stores the tralining newline into the read buffer. That is the one which is getting printed to the output. Instead of trying n_not to print_ the newline, you can simply replace the trailing \n with \0 after taking the input to get your issue solved.
As this is a simple replacement operation, without having strchr() also you can get the job done in an easy way.
Simple Algo:
Read the line using fgets().
If the return is not NULL, calculate the length of the input using strlen().
Replace the buffer[strlen-1] element (actually \n, you can double-check) with null \0.
note: The proper signature of main() is int main(void)
This answers the question in your comment. There is no string assign operator in c. You have to provide some space for this new string or destructively ammendment the original.
The simplest way to achieve what you want would be use strncpy to copy the first n characters of your source string to a new destination.
This question already has answers here:
strcmp on a line read with fgets
(6 answers)
Closed 3 years ago.
I am reading words from file & need to search some specific words, below is my code
string read = malloc(50 * sizeof(char));
FILE* p = fopen("word","r");
while(fgets(read,50,p))
{
printf("%s\n",read);
if(strcmp(read,"apple") == 0)
{
printf("apple found\n");
break;
}
}
And sample file 'word' is as below,
$: more word
liol
apple
scizzors
big
bang
mentalist
scapegrace
goat
goti
Why is strcmp not working in this case, printf can print string read, so char pointer is working fine.
The fgets() function, in most circumstances, retains the newline at the end of the line. Hence yourtext\n will not compare equal to yourtext. You would have noted this with one of my favorite tricks for checking strings:
printf ("[%s]\n", read);
The presence of a newline before the closing ] would have immediately alerted you to the problem, or at a minimum raised an eyebrow.
If you want to strip a newline off before comparing, you can do something like:
int ln = strlen (read);
if ((ln > 0) && (read[ln-1] == '\n'))
read[ln-1] = '\0';
Alternatively, you could skip that and just do:
if (strcmp (read,"apple\n") == 0)
It's not necessarily elegant but, if you don't need to use the word for anything other than that comparison, it'll be just fine.
However, you may then need to worry about the last line in the file in case it has no newline at the end of it so maybe it's better to use the newline-removing code above.
To search for a specific string in the line, rather than match the entire line, strncmp (which searches for n characters) can be used.
if (strncmp(read,"apple",strlen("apple")) == 0)
{
printf("apple found\n");
break;
}
I am currently learning C, and so I wanted to make a program that asks the user to input a string and to output the number of characters that were entered, the code compiles fine, when I enter just 1 character it does fine, but when I enter 2 or more characters, no matter what number of character I enter, it will always say there is just one character and crashes after that. This is my code and I can't figure out what is wrong.
int main(void)
{
int siz;
char i[] = "";
printf("Enter a string.\n");
scanf("%s", i);
siz = sizeof(i)/sizeof(char);
printf("%d", siz);
getch();
return 0;
}
I am currently learning to program, so if there is a way to do it using the same scanf() function I will appreciate that since I haven't learned how to use any other function and probably won't understand how it works.
Please, FORGET that scanf exists. The problem you are running into, whilst caused mostly by your understandable inexperience, will continue to BITE you even when you have experience - until you stop.
Here is why:
scanf will read the input, and put the result in the char buffer you provided. However, it will make no check to make sure there is enough space. If it needs more space than you provided, it will overwrite other memory locations - often with disastrous consequences.
A safer method uses fgets - this is a function that does broadly the same thing as scanf, but it will only read in as many characters as you created space for (or: as you say you created space for).
Other observation: sizeof can only evaluate the size known at compile time : the number of bytes taken by a primitive type (int, double, etc) or size of a fixed array (like int i[100];). It cannot be used to determine the size during the program (if the "size" is a thing that changes).
Your program would look like this:
#include <stdio.h>
#include <string.h>
#define BUFLEN 100 // your buffer length
int main(void) // <<< for correctness, include 'void'
{
int siz;
char i[BUFLEN]; // <<< now you have space for a 99 character string plus the '\0'
printf("Enter a string.\n");
fgets(i, BUFLEN, stdin); // read the input, copy the first BUFLEN characters to i
siz = sizeof(i)/sizeof(char); // it turns out that this will give you the answer BUFLEN
// probably not what you wanted. 'sizeof' gives size of array in
// this case, not size of string
// also not
siz = strlen(i) - 1; // strlen is a function that is declared in string.h
// it produces the string length
// subtract 1 if you don't want to count \n
printf("The string length is %d\n", siz); // don't just print the number, say what it is
// and end with a newline: \n
printf("hit <return> to exit program\n"); // tell user what to do next!
getc(stdin);
return 0;
}
I hope this helps.
update you asked the reasonable follow-up question: "how do I know the string was too long".
See this code snippet for inspiration:
#include <stdio.h>
#include <string.h>
#define N 50
int main(void) {
char a[N];
char *b;
printf("enter a string:\n");
b = fgets(a, N, stdin);
if(b == NULL) {
printf("an error occurred reading input!\n"); // can't think how this would happen...
return 0;
}
if (strlen(a) == N-1 && a[N-2] != '\n') { // used all space, didn't get to end of line
printf("string is too long!\n");
}
else {
printf("The string is %s which is %d characters long\n", a, strlen(a)-1); // all went according to plan
}
}
Remember that when you have space for N characters, the last character (at location N-1) must be a '\0' and since fgets includes the '\n' the largest string you can input is really N-2 characters long.
This line:
char i[] = "";
is equivalent to:
char i[1] = {'\0'};
The array i has only one element, the program crashes because of buffer overflow.
I suggest you using fgets() to replace scanf() like this:
#include <stdio.h>
#define MAX_LEN 1024
int main(void)
{
char line[MAX_LEN];
if (fgets(line, sizeof(line), stdin) != NULL)
printf("%zu\n", strlen(line) - 1);
return 0;
}
The length is decremented by 1 because fgets() would store the new line character at the end.
The problem is here:
char i[] = "";
You are essentially creating a char array with a size of 1 due to setting it equal to "";
Instead, use a buffer with a larger size:
char i[128]; /* You can also malloc space if you desire. */
scanf("%s", i);
See the link below to a similar question if you want to include spaces in your input string. There is also some good input there regarding scanf alternatives.
How do you allow spaces to be entered using scanf?
That's because char i[] = ""; is actually an one element array.
Strings in C are stored as the text which ends with \0 (char of value 0). You should use bigger buffer as others said, for example:
char i[100];
scanf("%s", i);
Then, when calculating length of this string you need to search for the \0 char.
int length = 0;
while (i[length] != '\0')
{
length++;
}
After running this code length contains length of the specified input.
You need to allocate space where it will put the input data. In your program, you can allocate space like:
char i[] = " ";
Which will be ok. But, using malloc is better. Check out the man pages.
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