Function that reads a string from a file of unknown length - c

I've been trying to piece together a function that allows me to create a string out of a given file of unknown length.
What it's supposed to do, is set the size of the output string to a single character, then for each character besides EOF, increment the size of the string by 1 and add the newly read character to it.
void readstring(FILE *f, char *s[])
{
int size = 1;
int c = 0, i = 0;
s = malloc(size*sizeof(char));
while(c != -1)
{
c = fgetc(f);
s[i] = (char)c;
i++;
if(i == size)
{
size++;
s = realloc(s, size*sizeof(char));
}
}
s[i] = '\0';
}
int main()
{
char *in = malloc(2*sizeof(char));
FILE *IN;
IN = fopen("in.txt", "r");
readstring(IN, in);
printf("%s",&in);
fclose(IN);
free(in);
return 0;
}

If you are on a POSIX compliant system (any modern Linux), don't try to reinvent the wheel. Just use getline(). It will do the tricky stuff of managing a growing buffer for you, and return a suitably malloc()'ed string.
You are assigning the result of malloc()/realloc() to s instead of *s. s is purely local to readstring(), *s is the pointer variable that s points to. Pass in the adress of the pointer in main. Correct code looks like this:
void foo(char** out_string) {
*out_string = malloc(...);
}
int main() {
char* string;
foo(&string);
}
Without the address taking & and pointer dereferenciation *, your main() cannot know where readstring() has stored the characters.
You also need to dereference the double pointer when you set the characters of the string: Use (*s)[i] instead of s[i], as s[i] denotes a pointer, not a character.
Also, try running your program with valgrind, and see if you can learn from the errors that it will spew out at you. It's a great tool for debugging memory related problems.

Related

Return a string made with a line read from input

i am trying to code a C function which returns a line read from the input as a char* . I am on Windows and i test my program in the command line by giving files as input and output of my program like this:
cl program.c
program < test_in.txt > test_out.txt
This is my (not working) function:
char* getLine(void)
{
char* result = "";
int i, c;
i = 1;
while((c = getchar()) != EOF)
{
*result++ = c;
i++;
if(c == '\n')
return result - i;
}
return result - i;
}
I was expecting it to work because i previously wrote:
char* getString(char* string)
{
//char* result = string; // the following code achieve this.
char* result = "";
int i;
for(i = 1; *result++ = *string++; i++);
return result - i;
}
And these lines of code have a correct behaviour.
Even if every answers will be appreciated, i would be really thankfull
if any of you could explain me why my getString() function works while my getLine() function doesn't.
Your function does not allocate enough space for the string being read. The variable char* result = "" defines a char pointer to a string literal ("", empty string), and you store some arbitrary number of characters into the location pointed to by result.
char* getLine(void)
{
char* result = ""; //you need space to store input
int i, c;
i = 1;
while((c = getchar()) != EOF)
{
*result++ = c; //you should check space
i++;
if(c == '\n')
return result - i; //you should null-terminate
}
return result - i; //you should null-terminate
}
You need to allocate space for your string, which is challenging because you don't know how much space you are going to need a priori. So you need to decide whether to limit how much you read (ala fgets), or dynamically reallocate space as you read more. Also, how to you indicate that you have finished input (reached EOF)?
The following alternative assumes dynamic reallocation is your chosen strategy.
char* getLine(void)
{
int ch; int size=100; size_t pos=0;
char* result = malloc(size*sizeof(char*));
while( (ch=getchar()) != EOF )
{
*result++ = ch;
if( ++pos >= size ) {
realloc(result,size+=100);
//or,realloc(result,size*=2);
if(!result) exit(1); //realloc failed
}
if( c=='\n' ) break;
}
*result = '\0'; //null-terminate
return result - pos;
}
When you are done with the string returned from the above function, please remember to free() the allocated space.
This alternative assumes you provide a buffer to store the string (and specifies the size of the buffer).
char* getLine(char* buffer, size_t size)
{
int ch;
char* result = buffer;
size_t pos=0;
while( (ch=getchar()) != EOF )
{
*result++ = ch;
if( ++pos >= size ) break; //full
if( c=='\n' ) break;
}
*result = '\0'; //null-terminate
return buffer;
}
Both avoid the subtle interaction between detecting EOF, and having enough space to store a character read. The solution is to buffer a character if you read and there is not enough room, and then inject that on a subsequent read. You will also need to null-ter
Both functions have undefined behaviour since you are modifying string literals. It just seems to work in one case. Basically, result needs to point to memory that can be legally accessed, which is not the case in either of the snippets.
On the same subject, you might find this useful: What Every C Programmer Should Know About Undefined Behavior.
Think of it this way.
When you say
char* result = "";
you are setting up a pointer 'result' to point to a 1-byte null terminated string (just the null). Since it is a local variable it will be allocated on the stack.
Then when you say
*result++ = c;
you are storing that value 'c' in to that address + 1.
So, where are you putting it?
Well, most stacks are to-down; so they grow toward lower addresses; so, you are probably writing over what is already on the stack (the return address for whatever called this, all the registers it needs restore and all sorts of important stuff).
That is why you have to be very careful with pointers.
When you expect to return a string from a function, you have two options (1) provide a string to the function with adequate space to hold the string (including the null-terminating character), or (2) dynamically allocate memory for the string within the function and return a pointer. Within your function you must also have a way to insure your are not writing beyond the end of the space available and you are leaving room for the null-terminating character. That requires passing a maximum size if you are providing the array to the function, and keeping count of the characters read.
Putting that together, you could do something similar to:
#include <stdio.h>
#define MAXC 256
char* getLine (char *s, int max)
{
int i = 0, c = 0;
char *p = s;
while (i + 1 < max && (c = getchar()) != '\n' && c != EOF) {
*p++ = c;
i++;
}
*p = 0;
return s;
}
int main (void) {
char buf[MAXC] = {0};
printf ("\ninput : ");
getLine (buf, MAXC);
printf ("output: %s\n\n", buf);
return 0;
}
Example/Output
$ ./bin/getLine
input : A quick brown fox jumps over the lazy dog.
output: A quick brown fox jumps over the lazy dog.

Returning a string [char pointer] from a function [duplicate]

This question already has answers here:
Return char[]/string from a function [duplicate]
(5 answers)
Closed 8 years ago.
I am writing a program that returns a string from stdin, but i am getting warning that it returns an adress of local wariable. How can i return the string?
thanks in advance
#include <stdio.h>
char* readLine()
{
int i;
char input[1024];
for(i=0;i<1024;i++)
{
input[i]=fgetc(stdin);
if(input[i]=='\n')
{
break;
}
}
return input;
}
int main()
{
printf("%s",readLine());
return 0;
}
This should work for you:
You can pass input from main as reference:
#include <stdio.h>
char * readLine(char * input, int length) {
int i;
for(i = 0; i < length; i++) {
input[i] = fgetc(stdin);
input[length] = '\0';
if(input[i] == '\n')
break;
}
return input;
}
int main() {
int length = 1024;
char input[length+1];
printf("%s", readLine(input, length));
return 0;
}
Try to do something like that instead:
#include <stdio.h>
char* readLine()
{
int i;
char *input;
if ((input = malloc(sizeof(char) * 1024)) == NULL)
return (NULL);
for(i=0;i<1024;i++)
{
input[i]=fgetc(stdin);
if(input[i]=='\n')
{
input[i] = '\0';
break;
}
}
return input;
}
int main()
{
char *str;
if (str = readLine()) != NULL) {
printf("%s\n", str);
free(str);
}
return 0;
}
}
There is nothing wrong here - that is just a WARNING because usually it is a common mistake of new programmers. I used to run into problems with this usage all the time.
The first thing... this "string" is not null-terminated. You'll want to put at the end of that function something like *(input + i) = '\0'; and make either the array size 1025 or the condition i < 1023 (so that the null character isn't assigned beyond the end of the buffer), because at the moment using this array in a function that expects null termination will cause it to possibly continue past the end of the array, resulting in a memory access violation. Alternately, you could use memset(input,0,1024);, just still make sure that the condition is something like i < 1023 so that the standard input you receive doesn't end up writing all the way to the last null character in the array.
The other problem is that this memory is local, as in it "belongs" to this function. And for the usage you have here, it is probably just fine to use the same memory... if you plan to call the function, do something with the result, and then call the function again, do something with the result... But if you want to keep what's given to you by it, you'll have to either (1) copy the string to another buffer that isn't going to be written to again when the function is called in the future, or (2) make the function allocate a new buffer each time it runs, and then be sure to delete that memory when you're done with it. For example, instead of char input [1024]; (which by the way would have the same pointer for the life of the program, so it's not really necessary to return it each time) you could write char* input = malloc(1024); and later, when the caller is done with the string, you should free(input);. (Of course, the name might not be input in this case since you would probably not want to free the memory in the function whose purpose is to allocate it.)
I will edit this later with code showing changes.

append a character from an array to a char pointer

Ok, so I'm a person who usually writes Java/C++, and I've just started getting into writing C. I'm currently writing a lexical analyser, and I can't stand how strings work in C, since I can't perform string arithmetic. So here's my question:
char* buffer = "";
char* line = "hello, world";
int i;
for (i = 0; i < strlen(line); i++) {
buffer += line[i];
}
How can I do that in C? Since the code above isn't valid C, how can I do something like that?
Basically I'm looping though a string line, and I'm trying to append each character to the buffer string.
string literals are immutable in C. Modifying one causes Undefined Behavior.
If you use a char array (your buffer) big enough to hold your characters, you can still modify its content :
#include <stdio.h>
int main(void) {
char * line = "hello, world";
char buffer[32]; // ok, this array is big enough for our operation
int i;
for (i = 0; i < strlen(line) + 1; i++)
{
buffer[i] = line[i];
}
printf("buffer : %s", buffer);
return 0;
}
First off the buffer needs to have or exceed the length of the data being copied to it.
char a[length], b[] = "string";
Then the characters are copied to the buffer.
int i = 0;
while (i < length && b[i] != '\0') { a[i] = b[i]; i++; }
a[i] = '\0';
You can reverse the order if you need to, just start i at the smallest length value among the two strings, and decrement the value instead of increment. You can also use the heap, as others have suggested, ordinate towards an arbitrary or changing value of length. Furthermore, you can change up the snippet with pointers (and to give you a better idea of what is happening):
int i = 0;
char *j = a, *k = b;
while (j - a < length && *k) { *(j++) = *(k++); }
*j = '\0';
Make sure to look up memcpy; and don't forget null terminators (oops).
#include <string.h>
//...
char *line = "hello, world";
char *buffer = ( char * ) malloc( strlen( line ) + 1 );
strcpy( buffer, line );
Though in C string literals have types of non-const arrays it is better to declare pointers initialized by string literals with qualifier const:
const char *line = "hello, world";
String literals in C/C++ are immutable.
If you want to append characters then the code can look the following way (each character of line is appended to buffer in a loop)
#include <string.h>
//...
char *line = "hello, world";
char *buffer = ( char * ) malloc( strlen( line ) + 1 );
buffer[0] = '\0';
char *p = Buffer;
for ( size_t i = 0; i < strlen( line ); i++ )
{
*p++ = line[i];
*p = '\0';
}
The general approach is that you find the pointer to the terminating zero substitute it for the target character advance the pointer and appenf the new terminating zero. The source buffer shall be large enough to accomodate one more character.
If you want to append a single character to a string allocated on the heap, here's one way to do it:
size_t length = strlen(buffer);
char *newbuffer = realloc(buffer, length + 2);
if (newbuffer) { // realloc succeeded
buffer = newbuffer;
buffer[length] = newcharacter;
buffer[length + 1] = '\0';
}
else { // realloc failed
// TODO handle error...
free(buffer); // for example
}
However, this is inefficient to do repeatedly in a loop, because you'll be repeatedly calling strlen() on (essentially) the same string, and reallocating the buffer to fit one more character each time.
If you want to be smarter about your reallocations, keep track of the buffer's current allocated capacity separately from the length of the string within it — if you know C++, think of the difference between a std::string object's "size" and its "capacity" — and when it's necessary to reallocate, multiply the buffer's size by a scaling factor (e.g. double it) instead of adding 1, so that the number of reallocations is O(log n) instead of O(n).
This is the sort of thing that a good string class would do in C++. In C, you'll probably want to move this buffer-management stuff into its own module.
The simplest solution, lacking any context, is to do:
char buffer[ strlen(line) + 1 ];
strcpy(buffer, line);
You may be used to using pointers for everything in Java (since non-primitive types in Java are actually more like shared pointers than anything else). However you don't necessarily have to do this in C and it can be a pain if you do.
Maybe a good idea given your background would be to use a counted string object in C, where the string object owns its data. Write struct my_string { char *data; size_t length; } . Write functions for creating, destroying, duplicating, and any other operation you need such as appending a character, or checking the length. (Separate interface from implementation!) A useful addition to this would be to make it allocate 1 more byte than length, so that you can have a function which null-terminates and allows it to be passed to a function that expects a read-only C-style string.
The only real pitfall here is to remember to call a function when you are doing a copy operation, instead of allowing structure assignment to happen. (You can use structure assignment for a move operation of course!)
The asprintf function is very useful for building strings, and is available on GNU-based systems (Linux), or most *BSD based systems. You can do things like:
char *buffer;
if (asprintf(&buffer, "%s: adding some stuff %d - %s", str1, number, str2) < 0) {
fprintf(stderr, "Oops -- out of memory\n");
exit(1); }
printf("created the string \"%s\"\n", buffer);
free(buffer); /* done with it */
Appending is best done with snprintf
Include the stdio.h header
#include <stdio.h>
then
char* buffer;
char line[] = "hello, world";
// Initialise the pointer to an empty string
snprintf(buffer, 1, "%s", "");
for (i = 0; i < strlen(line); ++i) {
snprintf(buffer, sizeof line[i], "%s%s", buffer, line[i]);
}
As you have started the code you have there is different from the question you are asking.
You could have split the line with strtok though.
But I hope my answer clarifies it.

from static array assignment to array from file

I have this piece of code outside the main function
mystr * arrstr[] = {
"rec",
"cent",
"ece",
"ce",
"recent",
"nt",
};
I modified it so that it can read the values from a text file. for this purpose i modified this working code to read line from file into array named string.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void) {
int i=0,j;
char* string[100];
char line[100];
FILE *file;
file = fopen("patt", "r");
while(fgets(line, sizeof(line), file)!=NULL) {
printf("%s", line);
string[i] = (char*)malloc(strlen(line));
strcpy(string[i], line);
i++;
}
fclose(file);
return 0;
}
so the final code is now something like this.
..
..
char *getpatterns(const char *filename) {
int i=0;
char* string[100];
char line[100];
FILE *file;
file = fopen(filename, "r");
while(fgets(line, sizeof(line), file)!=NULL) {
//printf("%s", line);
string[i] = (char*)malloc(strlen(line));
strcpy(string[i], line);
i++;
}
fclose(file);
return(string);
}
mystr * arrstr[] = getpatterns("patt");/*{
"rec",
"cent",
"ece",
"ce",
"recent",
"nt",
};*/
..
..
But i get errors like this.
example1.c: In function ‘getpatterns’:
example1.c:43:2: warning: return from incompatible pointer type [enabled by default]
example1.c:43:2: warning: function returns address of local variable [enabled by default]
example1.c: At top level:
example1.c:45:1: error: invalid initializer
make: *** [example1.o] Error 1
Here line 45 is this line
mystr * arrstr[] = getpatterns("patt");/*{
Please suggest corrective action.
The first warnings are that you are trying to return a char ** as a char * (which is not a good idea), and that you are returning a local variable which is deallocated when the function returns (also not a good idea). The last is telling you that you can't use function calls in initializers of global variables in C (you can do some of that in C++, though I'm not convinced you can do this one).
Fixing it will take some rethinking. You need the function to return allocated memory, or you need to pass the memory to the function. And you'll have to change the type of the global variable. And you'll need to know how many entries there are in the array, somehow.
mystr **arrstr = 0; // Either
mystr *arrstr[100]; // Or
On the whole, I'd probably go with memory allocation and the 'either' declaration:
mystr **arrstr = 0;
char **getpatterns(const char *file)
{
char **array = 0;
...code similar to yours that allocates entries in the array...
...include space for a null pointer to mark the end of the list of strings...
return(array);
}
int main(void)
{
arrstr = getpatterns("patt");
...
}
(Another 'cheat' mechanism would use static char *string[100]; in getpatterns(); you still have to fix the return type and the type of the global variable.)
I tried these but, errors were not resolved: ...
It's impossible to tell exactly what was wrong without your code. However, the code below works for me. The source code was in a file gp.c; the source code prints itself, and releases the memory. Checked under valgrind with a clean bill of health.
Note that your original code did not allocate enough space for the strings it was copying (because you retained the newline read by fgets() — but you were at least using fgets() and not gets(), which is very important). This code uses memmove() — it could use memcpy() instead since there's guaranteed to be no overlap, but memmove() always works and memcpy() doesn't necessarily work when the source data overlaps the target data. It knows how long the string is, so the copy function doesn't need to test for whether the character being copied is a NUL '\0'. The code carefully ensures that there's a null pointer at the end of the list of pointers; that's how you know when you've reached the end of the list of strings. The code also works when gp.c is an empty file.
The algorithm using three items num_xxx, max_xxx, and xxx is a typical way to handle incremental allocation. It typically over-allocates slightly; if you're concerned about the space, you could use strings = realloc(strings, (num_strings+1) * sizeof(*strings)); max_strings = num_strings + 1; at the end of the loop to release the extra space. The + 1 is to allow for the null pointer. By roughly doubling the size allocated each time you allocate, you avoid quadratic behaviour compared with incrementing by one each time.
Notice too that the code carefully avoids losing the allocated space if the realloc() fails. You should 'never' use space = realloc(space, new_size); to avoid losing your pointer. The code carefully avoids dereferencing null pointers, and simply stops reading when there is a memory shortage.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
extern char **getpatterns(const char *filename);
char **getpatterns(const char *filename)
{
size_t num_strings = 0;
size_t max_strings = 0;
char **strings = 0;
FILE *file = fopen(filename, "r");
if (file != 0)
{
char line[4096];
while (fgets(line, sizeof(line), file) != NULL)
{
if (max_strings == 0 || num_strings >= max_strings - 1)
{
size_t new_num = max_strings * 2 + 2;
char **new_space = realloc(strings, new_num * sizeof(*new_space));
if (new_space == 0)
break;
strings = new_space;
max_strings = new_num;
}
size_t len = strlen(line); /* Includes '\n' at end */
strings[num_strings] = (char*)malloc(len);
memmove(strings[num_strings], line, len - 1);
strings[num_strings][len] = '\0';
strings[++num_strings] = 0; /* Null terminate list of strings */
}
fclose(file);
}
return(strings);
}
int main(void)
{
char **data = getpatterns("gp.c");
char **argp = data;
if (argp != 0)
{
/* Print data */
while (*argp != 0)
puts(*argp++);
/* Free space */
argp = data;
while (*argp != 0)
free(*argp++);
free(data);
}
return(0);
}

Why am I getting a segmentation fault?

I'm trying to write a program that takes in a plaintext file as it's argument and parses through it, adding all the numbers together and then print out the sum. The following is my code:
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
static int sumNumbers(char filename[])
{
int sum = 0;
FILE *file = fopen(filename, "r");
char *str;
while (fgets(str, sizeof BUFSIZ, file))
{
while (*str != '\0')
{
if (isdigit(*str))
{
sum += atoi(str);
str++;
while (isdigit(*str))
str++;
continue;
}
str++;
}
}
fclose(file);
return sum;
}
int main(int argc, char *argv[])
{
if (argc != 2)
{
fprintf(stderr, "Please enter the filename as the argument.\n");
exit(EXIT_FAILURE);
}
else
{
printf("The sum of all the numbers in the file is : %d\n", sumNumbers(argv[1]));
exit(EXIT_SUCCESS);
}
return 0;
}
And the text file I'm using is:
This a rather boring text file with
some random numbers scattered
throughout it.
Here is one: 87 and here is another: 3
and finally two last numbers: 12
19381. Done. Phew.
When I compile and try to run it, I get a segmentation fault.
You've not allocated space for the buffer.The pointer str is just a dangling pointer. So your program effectively dumps the data read from the file into memory location which you don't own, leading to the segmentation fault.
You need:
char *str;
str = malloc(BUFSIZ); // this is missing..also free() the mem once done using it.
or just:
char str[BUFSIZ]; // but then you can't do str++, you'll have to use another
// pointer say char *ptr = str; and use it in place of str.
EDIT:
There is another bug in:
while (fgets(str, sizeof BUFSIZ, file))
The 2nd argument should be BUFSIZ not sizeof BUFSIZ.
Why?
Because the 2nd argument is the maximum number of characters to be read into the buffer including the null-character. Since sizeof BUFSIZ is 4 you can read max upto 3 char into the buffer. That is reason why 19381 was being read as 193 and then 81<space>.
You haven't allocated any memory to populate str. fgets takes as its first argument a buffer, not an unassigned pointer.
Instead of char *str; you need to define a reasonably sized buffer, say, char str[BUFSIZ];
Because you've not allocated space for your buffer.
A number of people have already addressed the problem you asked about, but I've got a question in return. What exactly do you think this accomplishes:
if (isdigit(*str))
{
if (isdigit(*str))
{
sum += atoi(str);
str++;
while (isdigit(*str))
str++;
continue;
}
}
What's supposed to be the point of two successive if statements with the exact same condition? (Note for the record: neither one has an else clause).
You have declared char* str, but you have not set aside memory for it just yet. You will need to malloc memory for it.
Many memory related errors such as this one can be easily found with valgrind. I'd highly recommend using it as a debugging tool.
char *str;
str has no memory allocated for it. Either use malloc() to allocate some memory for it, or declared it with a predefined size.
char str[MAX_SIZE];
Your program has several bugs:
It does not handle long lines correctly. When you read a buffer of some size it may happen that some number starts at the end of the buffer and continues at the beginning of the next buffer. For example, if you have a buffer of size 4, there might be the input The |numb|er 1|2345| is |larg|e., where the vertical lines indicate the buffer's contents. You would then count the 1 and the 2345 separately.
It calls isdigit with a char as argument. As soon as you read any "large" character (greater than SCHAR_MAX) the behavior is undefined. Your program might crash or produce incorrect results or do whatever it wants to do. To fix this, you must first cast the value to an unsigned char, for example isdigit((unsigned char) *str). Or, as in my code, you can feed it the value from the fgetc function, which is guaranteed to be a valid argument for isdigit.
You use a function that requires a buffer (fgets) but you fail to allocate the buffer. As others noted, the easiest way to get a buffer is to declare a local variable char buffer[BUFSIZ].
You use the str variable for two purposes: To hold the address of the buffer (which should remain constant over the whole execution time) and the pointer for analyzing the text (which changes during the execution). Make these two variables. I would call them buffer and p (short for pointer).
Here is my code:
#include <ctype.h>
#include <stdio.h>
static int sumNumbers(const char *filename)
{
int sum, num, c;
FILE *f;
if ((f = fopen(filename, "r")) == NULL) {
/* TODO: insert error handling here. */
}
sum = 0;
num = 0;
while ((c = fgetc(f)) != EOF) {
if (isdigit(c)) {
num = 10 * num + (c - '0');
} else if (num != 0) {
sum += num;
num = 0;
}
}
if (fclose(f) != 0) {
/* TODO: insert error handling here. */
}
return sum;
}
int main(int argc, char **argv) {
int i;
for (i = 1; i < argc; i++)
printf("%d\t%s\n", sumNumbers(argv[i]), argv[i]);
return 0;
}
Here is a function, that does your job:
static int sumNumbers(char* filename) {
int sum = 0;
FILE *file = fopen(filename, "r");
char buf[BUFSIZ], *str;
while (fgets(buf, BUFSIZ, file))
{
str=buf;
while (*str)
{
if (isdigit(*str))
{
sum += strtol(str, &str, 10);
}
str++;
}
}
fclose(file);
return sum;
}
This doesn't includes error handling, but works quite well. For your file, output will be
The sum of all the numbers in the file is : 19483

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