I begin with C and I have to write a method that returns each word in a sentence separated by a whitespace.
So far I have this:
#include<stdio.h>
#include<string.h>
#define SIZE 100
int nextToken(char const * str, int* from, int* len);
int main(){
char str[SIZE+1];
char mot[SIZE+1];
if(fgets(str, SIZE, stdin) == NULL){
puts("Error");
return -1;
}
str[SIZE] = '\0';
int from = 0;
int len = 0;
while(nextToken(str, &from, &len)){
strncpy(mot, &(str[from]), len);
mot[SIZE] = '\0';
printf("'%s'\n", mot);
return 0;
}
return 0;
}
int nextToken(char const * str, int* from, int* len){
*from = 1;
*len = 2;
return 1;
}
I know the return 0 in the while but it's just to run it one time.
So in the nextToken function, when changing the value pointed by len to 1, I get the expected output for the input hello, i.e 'e'.
But when changing the value pointed by len to 2, I was expecting to get 'el' but I get 'el¿■('.
Why do I have this output, why it doesn't print 'el'?
Sorry if it's stupid but I don't know why this is happening as I begin. I'm compiling in C89.
strncpy does not copy a null-terminator in case it copies len characters. In that eventuality you need to ensure that the string is null-terminated explicitly.
Frankly, I don't think that strncpy is really the right function for you to use here.
I'd write it like this:
memcpy(mot, str+from, len);
mot[len] = '\0';
And you'd want to include a check that this did not result in a buffer overrun.
Related
I have a char array s[11]="0123456789"; and I want to be able to take each digit s[i](using a for loop) and cast it somehow to a char*(I need a char* specifically because I need to use strncat on some other string)
I've been trying to do this for the past 4 hours and I couldn't get anything done.
Unless you're willing to temporarily modify s, or copy the character somewhere else... You're going to have a hard time.
Modify s temporarily.
int main() {
char s[11] = "0123456789";
for (int i = 0; i < strlen(s); i++) {
// Modify s in place.
const char tmp = s[i+1];
s[i+1] = '\0';
char *substr = &s[i];
// Do something with substr.
printf("%s\n", substr);
// Fix s
s[i+1] = tmp;
}
}
Or copy s[i] to a new string.
int main() {
char s[11] = "0123456789";
for (int i = 0; i < strlen(s); i++) {
// Create a temporary string
char substr[2];
substr[0] = s[i];
substr[1] = '\0';
// Do something with substr.
printf("%s\n", substr);
}
}
Or maybe just don't use strncat?
Assuming you're actually using strncat or similar... Those functions are pretty trivial, especially if you are appending only a single character. You might just create your own version.
void append_character(char *buffer, int buffer_size, char new_character) {
int length = strlen(buffer);
if (length + 2 < buffer_size) {
buffer[length] = new_character;
buffer[length+1] = '\0';
} else {
// No space for an additional character, drop it like strncat would.
}
}
Or you could do it as a simple wrapper around strncat:
void append_character(char *buffer, int buffer_size, char new_character) {
char new_string[2] = { new_character, '\0' };
strncat(buffer, buffer_size, new_string);
}
What you are requesting does not make any sense. A char is a small number. A pointer is the address of some memory. Converting a char to a pointer won't give you a valid pointer, but a pointer that will crash as soon as it is used.
For strncat, you need an array of characters containing a C string. You could create a string by writing char array[2]; (now you have an array with space for two characters), then array[0] = whateverchar; array[1] = 0; and now you have a C string with space for exactly one char and one trailing zero byte.
Yet another idea:
#include <stdio.h>
#include <string.h>
int main(void)
{
char src[11]="0123456789";
char dest[50]="abc";
char* p = src;
int i = 0;
int len = strlen(src);
for (i = 0; i < len; i++)
{
strncat(dest,p,1);
p++;
}
printf("src: %s\n", src);
printf("dest: %s\n", dest);
return 0;
}
Compiled with gcc under Ubuntu:
$ gcc hello_str.c -o hello_str
Output:
$ ./hello_str
src: 0123456789
dest: abc0123456789
My str_split function returns (or at least I think it does) a char** - so a list of strings essentially. It takes a string parameter, a char delimiter to split the string on, and a pointer to an int to place the number of strings detected.
The way I did it, which may be highly inefficient, is to make a buffer of x length (x = length of string), then copy element of string until we reach delimiter, or '\0' character. Then it copies the buffer to the char**, which is what we are returning (and has been malloced earlier, and can be freed from main()), then clears the buffer and repeats.
Although the algorithm may be iffy, the logic is definitely sound as my debug code (the _D) shows it's being copied correctly. The part I'm stuck on is when I make a char** in main, set it equal to my function. It doesn't return null, crash the program, or throw any errors, but it doesn't quite seem to work either. I'm assuming this is what is meant be the term Undefined Behavior.
Anyhow, after a lot of thinking (I'm new to all this) I tried something else, which you will see in the code, currently commented out. When I use malloc to copy the buffer to a new string, and pass that copy to aforementioned char**, it seems to work perfectly. HOWEVER, this creates an obvious memory leak as I can't free it later... so I'm lost.
When I did some research I found this post, which follows the idea of my code almost exactly and works, meaning there isn't an inherent problem with the format (return value, parameters, etc) of my str_split function. YET his only has 1 malloc, for the char**, and works just fine.
Below is my code. I've been trying to figure this out and it's scrambling my brain, so I'd really appreciate help!! Sorry in advance for the 'i', 'b', 'c' it's a bit convoluted I know.
Edit: should mention that with the following code,
ret[c] = buffer;
printf("Content of ret[%i] = \"%s\" \n", c, ret[c]);
it does indeed print correctly. It's only when I call the function from main that it gets weird. I'm guessing it's because it's out of scope ?
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#define DEBUG
#ifdef DEBUG
#define _D if (1)
#else
#define _D if (0)
#endif
char **str_split(char[], char, int*);
int count_char(char[], char);
int main(void) {
int num_strings = 0;
char **result = str_split("Helo_World_poopy_pants", '_', &num_strings);
if (result == NULL) {
printf("result is NULL\n");
return 0;
}
if (num_strings > 0) {
for (int i = 0; i < num_strings; i++) {
printf("\"%s\" \n", result[i]);
}
}
free(result);
return 0;
}
char **str_split(char string[], char delim, int *num_strings) {
int num_delim = count_char(string, delim);
*num_strings = num_delim + 1;
if (*num_strings < 2) {
return NULL;
}
//return value
char **ret = malloc((*num_strings) * sizeof(char*));
if (ret == NULL) {
_D printf("ret is null.\n");
return NULL;
}
int slen = strlen(string);
char buffer[slen];
/* b is the buffer index, c is the index for **ret */
int b = 0, c = 0;
for (int i = 0; i < slen + 1; i++) {
char cur = string[i];
if (cur == delim || cur == '\0') {
_D printf("Copying content of buffer to ret[%i]\n", c);
//char *tmp = malloc(sizeof(char) * slen + 1);
//strcpy(tmp, buffer);
//ret[c] = tmp;
ret[c] = buffer;
_D printf("Content of ret[%i] = \"%s\" \n", c, ret[c]);
//free(tmp);
c++;
b = 0;
continue;
}
//otherwise
_D printf("{%i} Copying char[%c] to index [%i] of buffer\n", c, cur, b);
buffer[b] = cur;
buffer[b+1] = '\0'; /* extend the null char */
b++;
_D printf("Buffer is now equal to: \"%s\"\n", buffer);
}
return ret;
}
int count_char(char base[], char c) {
int count = 0;
int i = 0;
while (base[i] != '\0') {
if (base[i++] == c) {
count++;
}
}
_D printf("Found %i occurence(s) of '%c'\n", count, c);
return count;
}
You are storing pointers to a buffer that exists on the stack. Using those pointers after returning from the function results in undefined behavior.
To get around this requires one of the following:
Allow the function to modify the input string (i.e. replace delimiters with null-terminator characters) and return pointers into it. The caller must be aware that this can happen. Note that supplying a string literal as you are doing here is illegal in C, so you would instead need to do:
char my_string[] = "Helo_World_poopy_pants";
char **result = str_split(my_string, '_', &num_strings);
In this case, the function should also make it clear that a string literal is not acceptable input, and define its first parameter as const char* string (instead of char string[]).
Allow the function to make a copy of the string and then modify the copy. You have expressed concerns about leaking this memory, but that concern is mostly to do with your program's design rather than a necessity.
It's perfectly valid to duplicate each string individually and then clean them all up later. The main issue is that it's inconvenient, and also slightly pointless.
Let's address the second point. You have several options, but if you insist that the result be easily cleaned-up with a call to free, then try this strategy:
When you allocate the pointer array, also make it large enough to hold a copy of the string:
// Allocate storage for `num_strings` pointers, plus a copy of the original string,
// then copy the string into memory immediately following the pointer storage.
char **ret = malloc((*num_strings) * sizeof(char*) + strlen(string) + 1);
char *buffer = (char*)&ret[*num_strings];
strcpy(buffer, string);
Now, do all your string operations on buffer. For example:
// Extract all delimited substrings. Here, buffer will always point at the
// current substring, and p will search for the delimiter. Once found,
// the substring is terminated, its pointer appended to the substring array,
// and then buffer is pointed at the next substring, if any.
int c = 0;
for(char *p = buffer; *buffer; ++p)
{
if (*p == delim || !*p) {
char *next = p;
if (*p) {
*p = '\0';
++next;
}
ret[c++] = buffer;
buffer = next;
}
}
When you need to clean up, it's just a single call to free, because everything was stored together.
The string pointers you store into the res with ret[c] = buffer; array point to an automatic array that goes out of scope when the function returns. The code subsequently has undefined behavior. You should allocate these strings with strdup().
Note also that it might not be appropriate to return NULL when the string does not contain a separator. Why not return an array with a single string?
Here is a simpler implementation:
#include <stdlib.h>
char **str_split(const char *string, char delim, int *num_strings) {
int i, n, from, to;
char **res;
for (n = 1, i = 0; string[i]; i++)
n += (string[i] == delim);
*num_strings = 0;
res = malloc(sizeof(*res) * n);
if (res == NULL)
return NULL;
for (i = from = to = 0;; from = to + 1) {
for (to = from; string[to] != delim && string[to] != '\0'; to++)
continue;
res[i] = malloc(to - from + 1);
if (res[i] == NULL) {
/* allocation failure: free memory allocated so far */
while (i > 0)
free(res[--i]);
free(res);
return NULL;
}
memcpy(res[i], string + from, to - from);
res[i][to - from] = '\0';
i++;
if (string[to] == '\0')
break;
}
*num_strings = n;
return res;
}
I want to create a program in C that takes an arbitrary number of lines of arbitrary length as input and then prints to console the last line that was inputted. For example:
input:
hi
my name is
david
output: david
I figured the best way to do this would be to have a loop that takes each line as input and stores it in a char array, so at the end of the loop the last line ends up being what is stored in the char array and we can just print that.
I have only had one lecture in C so far so I think I just keep setting things up wrong with my Java/C++ mindset since I have more experience in those languages.
Here is what I have so far but I know that it's nowhere near correct:
#include <stdio.h>
int main()
{
printf("Enter some lines of strings: \n");
char line[50];
for(int i = 0; i < 10; i++){
line = getline(); //I know this is inproper syntax but I want to do something like this
}
printf("%s",line);
}
I also have i < 10 in the loop because I don't know how to find the total number of lines in the input which, would be the proper amount of times to loop this. Also, the input is being put in all at once from the
./program < test.txt
command in Unix shell, where test.txt has the input.
Use fgets():
while (fgets(line, sizeof line, stdin)) {
// don't need to do anything here
}
printf("%s", line);
You don't need a limit on the number of iterations. At the end of the file, fgets() returns NULL and doesn't modify the buffer, so line will still hold the last line that was read.
I'm assuming you know the maximum length of the input line.
This one here will surely do the job for you
static char *getLine( char * const b , size_t bsz ) {
return fgets(b, bsz, stdin) );
}
But remember fgets also puts a '\n' character at the end of buffer so perhaps something like this
static char *getLine( char * const b , size_t bsz ) {
if( fgets(b, bsz, stdin) ){
/* Optional code to strip NextLine */
size_t size = strlen(b);
if( size > 0 && b[size-1] == '\n' ) {
b[--size] = '\0';
}
/* End of Optional Code */
return b;
}
return NULL;
}
and your code needs to be altered a bit while calling the getline
#define BUF_SIZE 256
char line[BUF_SIZE];
for(int i = 0; i < 10; i++){
if( getLine(line, BUF_SIZE ) ) {
fprintf(stdout, "line : '%s'\n", line);
}
}
Now it is how ever quite possible to create function like
char *getLine();
but then one needs to define the behavior of that function for instance if the function getLine() allocates memory dynamically then you probably need use a free to de-allocate the pointer returned by getLine()
in which case the function may look like
char *getLine( size_t bsz ) {
char *b = malloc( bsz );
if( b && fgets(b, bsz, stdin) ){
return b;
}
return NULL;
}
depending on how small your function is you can entertain thoughts about making it inline perhaps that's a little off topic for now.
In order to have dynamic number of input of dynamic length, you have to keep on reallocating your buffer when the input is of greater length. In order to store the last line, you have to take another pointer to keep track of it and to stop the input from the terminal you have to press EOF key(ctrl+k). This should do your job.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char *get_last_line(FILE* fp, size_t size){
//The size is extended by the input with the value of the provisional
char *str, *last_str = NULL;
int ch;
size_t len = 0, last_len = 0;
str = realloc(NULL, sizeof(char)*size);//size is start size
if(!str)return str;
while(ch=fgetc(fp)){
if(ch == EOF){
break;
}
if(ch == '\n'){
str[len]='\0';
last_len = len;
last_str = realloc(last_str,sizeof(char)*last_len);
last_str[last_len]='\0';
//storing the last line
memcpy(last_str,str,sizeof(char)*last_len);
str = realloc(NULL, sizeof(char)*size);//size is start size
len = 0;
}
else {
str[len++]=ch;
if(len==size){
str = realloc(str, sizeof(char)*(size+=16));
if(!str)return str;
}
}
}
free(str);
return last_str;
}
int main(void){
char *m;
printf("input strings : ");
m = get_last_line(stdin, 10);
printf("last string :");
printf("%s\n", m);
free(m);
return 0;
}
I have a program that reverses a string from an input of a variable length character array. The function returns a variable length character array and is printed. When I print the output, I do get the reversed string, but there are garbage characters appended to it in my console print.
Is this a "legal" operation in terms of returning to buffers? Can someone please critique my code and suggest a better alternative if it is not the right approach?
Thanks.
#include <stdio.h>
#include <stdlib.h>
char *reverse_string(char *input_string);
char *reverse_string(char *input_string)
{
int i=0;
int j=0;
char *return_string;
char filled_buffer[16];
while (input_string[i]!='\0')
i++;
while (i!=0)
{
filled_buffer[j]=input_string[i-1];
i--;
j++;
}
return_string=filled_buffer;
printf("%s", return_string);
return return_string;
}
int main (void)
{
char *returned_string;
returned_string=reverse_string("tasdflkj");
printf("%s", returned_string);
return 1;
}
This is my output from Xcode - jklfdsat\347\322̲\227\377\231\235
No, it isn't safe to return a pointer to a local string in a function. C won't stop you doing it (though sometimes the compiler will warn you if you ask it to; in this case, the local variable return_string prevents it giving the warning unless you change the code to return filled_buffer;). But it is not safe. Basically, the space gets reused by other functions, and so they merrily trample on what was once a neatly formatted string.
Can you explain this comment in more detail — "No, it isn't safe..."
The local variables (as opposed to string constants) go out of scope when the function returns. Returning a pointer to an out-of-scope variable is undefined behaviour, which is something to be avoided at all costs. When you invoke undefined behaviour, anything can happen — including the program appearing to work — and there are no grounds for complaint, even if the program reformats your hard drive. Further, it is not guaranteed that the same thing will happen on different machines, or even with different versions of the same compiler on your current machine.
Either pass the output buffer to the function, or have the function use malloc() to allocate memory which can be returned to and freed by the calling function.
Pass output buffer to function
#include <stdio.h>
#include <string.h>
int reverse_string(char *input_string, char *buffer, size_t bufsiz);
int reverse_string(char *input_string, char *buffer, size_t bufsiz)
{
size_t j = 0;
size_t i = strlen(input_string);
if (i >= bufsiz)
return -1;
buffer[i] = '\0';
while (i != 0)
{
buffer[j] = input_string[i-1];
i--;
j++;
}
printf("%s\n", buffer);
return 0;
}
int main (void)
{
char buffer[16];
if (reverse_string("tasdflkj", buffer, sizeof(buffer)) == 0)
printf("%s\n", buffer);
return 0;
}
Memory allocation
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char *reverse_string(char *input_string);
char *reverse_string(char *input_string)
{
size_t j = 0;
size_t i = strlen(input_string) + 1;
char *string = malloc(i);
if (string != 0)
{
string[--i] = '\0';
while (i != 0)
{
string[j] = input_string[i-1];
i--;
j++;
}
printf("%s\n", string);
}
return string;
}
int main (void)
{
char *buffer = reverse_string("tasdflkj");
if (buffer != 0)
{
printf("%s\n", buffer);
free(buffer);
}
return 0;
}
Note that the sample code includes a newline at the end of each format string; it makes it easier to tell where the ends of the strings are.
This is an alternative main() which shows that the allocated memory returned is OK even after multiple calls to the reverse_string() function (which was modified to take a const char * instead of a plain char * argument, but was otherwise unchanged).
int main (void)
{
const char *strings[4] =
{
"tasdflkj",
"amanaplanacanalpanama",
"tajikistan",
"ablewasiereisawelba",
};
char *reverse[4];
for (int i = 0; i < 4; i++)
{
reverse[i] = reverse_string(strings[i]);
if (reverse[i] != 0)
printf("[%s] reversed [%s]\n", strings[i], reverse[i]);
}
for (int i = 0; i < 4; i++)
{
printf("Still valid: %s\n", reverse[i]);
free(reverse[i]);
}
return 0;
}
Also (as pwny pointed out in his answer before I added this note to mine), you need to make sure your string is null terminated. It still isn't safe to return a pointer to the local string, even though you might not immediately spot the problem with your sample code. This accounts for the garbage at the end of your output.
First, returning a pointer to a local like that isn't safe. The idiom is to receive a pointer to a large enough buffer as a parameter to the function and fill it with the result.
The garbage is probably because you're not null-terminating your result string. Make sure you append '\0' at the end.
EDIT: This is one way you could write your function using idiomatic C.
//buffer must be >= string_length + 1
void reverse_string(char *input_string, char* buffer, size_t string_length)
{
int i = string_length;
int j = 0;
while (i != 0)
{
buffer[j] = input_string[i-1];
i--;
j++;
}
buffer[j] = '\0'; //null-terminate the string
printf("%s", buffer);
}
Then, you call it somewhat like:
#define MAX_LENGTH 16
int main()
{
char* foo = "foo";
size_t length = strlen(foo);
char buffer[MAX_LENGTH];
if(length < MAX_LENGTH)
{
reverse_string(foo, buffer, length);
printf("%s", buffer);
}
else
{
printf("Error, string to reverse is too long");
}
}
I'm stuck at yet another C problem. How can I concatenate two strings with the second string being inserted before the first string?
This is what I came up with. Unfortunately I'm stuck at all these pointer to chars, char arrays et cetera.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(int argc, char* argv[] )
{
char* output;
int i;
for(i = 9; i > 0; i--)
{
unsigned int value = (unsigned int)i;
char buffer[20];
sprintf(buffer, "%u", value);
// strcat(ouput, buffer); // append before the string.
// first loop: 9
// second loop: 89
// third loop: 789
}
printf("%s", output);
}
How must I correct my code to make it work? I guess I have to somehow set the output variable to empty. When do I need fixed widths for the char array or the pointer? 20 was just a random guess from me.
I'm very confused, as your posted code has absolutely nothing to do with the problem you state. (Well, they both use strings, but that's about it)
char* src = "Hello, ";
char* dest = "World!";
char* temp;
temp = malloc(strlen(src) +strlen(dest) + 1);
strcpy(temp, src);
strcat(temp, dest);
dest = temp;
Unless dest is a fixed buffer of adequate size for the combined string. If so, then replace the last line with:
strcpy(dest, temp);
free(temp);
Now, if you want to specifically build the list of digits backwards, let's try a different tack:
char buffer[10];
buffer[9] = '\0'; // null terminate our string.
char* output;
int i;
for(i = 9; i > 0; i--)
{
// this is a fast way of saying, sprintf("%u", i);
// works only for single digits
char d = (char)('0' + i);
buffer[i-1] = d;
output = &buffer[i-1];
printf("%s", output);
}
Usually, you should just avoid the situation to start with. The most obvious solution for your example would be to simply count upward to start with. When that's not suitable, a recursive solution to reverse the order in which the string is built can still allow you to generate the string from beginning to end:
int build_string(int value, char *string) {
char temp[10];
if (value > -1)
build_string(value-1, string);
sprintf(temp, "%d", value); // use snprintf if available.
strcat(string, temp);
return string;
}
int main() {
char result[20] = {0};
build_string(9, result);
printf("%s", result);
return 0;
}
You can append the integer at the end of the string as:
int i;
char buffer[20];
for(i = 0; i < 10; i++) {
sprintf(buffer+i, "%u", i);
}
printf("%s", buffer); // prints 0123456789
For your stated problem (insert one string in front of another), this code will do the job - but has no error checking. It assumes there is enough space in the target buffer for the existing string and the new prefix:
/* Insert string t in front of string s in string s */
char *strinsert(char *s, const char *t)
{
char *p = s + strlen(s);
char *q = p + strlen(t);
char *r = s;
while (p >= s)
*q-- = *p--;
while (*t)
*s++ = *t++;
return(r);
}
What it does is copy the existing string up by the correct number of places so that there is space for the new string at the beginning.
Assuming that the destination buffer is big enough and that the source and destination do not overlap:
// not sure what order to put the params - the usual C way is destination
// followed by source, but it's also potentially confusing that the result of
// prepend(foo,bar) is "<bar><foo>".
char* prepend(char *restrict dest, const char *restrict src) {
size_t len = strlen(src);
memmove(dest + len, dest, strlen(dest));
return memcpy(dest, src, len);
}
If the buffers may overlap (for example, if src is the second half of dest), this approach doesn't work.
If the destination buffer is not big enough, then someone has to allocate new memory for the result, in which case the question of which is the "source" and which the "destination" disappears - they're both "source" and neither is "destination".