I am writing my own trim() in C. There is a structure which contains all string values, the structure is getting populated from the data coming from a file which contains spaces before and after the beginning of a word.
char *trim(char *string)
{
int stPos,endPos;
int len=strlen(string);
for(stPos=0;stPos<len && string[stPos]==' ';++stPos);
for(endPos=len-1;endPos>=0 && string[endPos]==' ';--endPos);
char *trimmedStr = (char*)malloc(len*sizeof(char));
strncpy(trimmedStr,string+stPos,endPos+1);
return trimmedStr;
}
int main()
{
char string1[]=" a sdf ie ";
char *string =trim(string1);
printf("%s",string);
return 0;
}
Above code is working fine, but i don't want to declare new variable that stores the trimmed word. As the structure contains around 100 variables.
Is there any way to do somthing like below where I dont need any second variable to print the trimmed string.
printf("%s",trim(string1));
I believe above print can create dangling pointer situation.
Also, is there any way where I don't have to charge original string as well, like if I print trim(string) it will print trimmed string and when i print only string, it will print original string
elcuco was faster. but it's done so here we go:
char *trim(char *string)
{
char *ptr = NULL;
while (*string == ' ') string++; // chomp away space at the start
ptr = string + strlen(string) - 1; // jump to the last char (-1 because '\0')
while (*ptr == ' '){ *ptr = '\0' ; ptr--; } ; // overwrite with end of string
return string; // return pointer to the modified start
}
If you don't want to alter the original string I'd write a special print instead:
void trim_print(char *string)
{
char *ptr = NULL;
while (*string == ' ') string++; // chomp away space at the start
ptr = string + strlen(string) - 1; // jump to the last char (-1 because '\0')
while (*ptr == ' '){ ptr--; } ; // find end of string
while (string <= ptr) { putchar(*string++); } // you get the picture
}
something like that.
You could the original string in order to do this. For trimming the prefix I just advance the pointer, and for the suffix, I actually add \0. If you want to keep the original starting as is, you will have to move memory (which makes this an O(n^2) time complexity solution, from an O(n) I provided).
#include <stdio.h>
char *trim(char *string)
{
// trim prefix
while ((*string) == ' ' ) {
string ++;
}
// find end of original string
char *c = string;
while (*c) {
c ++;
}
c--;
// trim suffix
while ((*c) == ' ' ) {
*c = '\0';
c--;
}
return string;
}
int main()
{
char string1[] = " abcdefg abcdf ";
char *string = trim(string1);
printf("String is [%s]\n",string);
return 0;
}
(re-thinking... is it really O(n^2)? Or is it O(2n) which is a higher O(n)...? I guess depending on implementation)
You can modify the function by giving the output in the same input string
void trim(char *string)
{
int i;
int stPos,endPos;
int len=strlen(string);
for(stPos=0;stPos<len && string[stPos]==' ';++stPos);
for(endPos=len-1;endPos>=0 && string[endPos]==' ';--endPos);
for (i=0; i<=(endPos-stPos); i++)
{
string[i] = string[i+stPos];
}
string[i] = '\0'; // terminate the string and discard the remaining spaces.
}
...is there any way where i don't have to charge original string as well, like if i do trim(string) it will print trimmed string and when i print only string, it will print original string – avinashse 8 mins ago
Yes, though it gets silly.
You could modify the original string.
trim(string);
printf("trimmed: %s\n", string);
The advantage is you have the option of duplicating the string if you want to retain the original.
char *original = strdup(string);
trim(string);
printf("trimmed: %s\n", string);
If you don't want to modify the original string, that means you need to allocate memory for the modified string. That memory then must be freed. That means a new variable to hold the pointer so you can free it.
char *trimmed = trim(original);
printf("trimmed: %s\n", trimmed);
free(trimmed);
You can get around this by passing a function pointer into trim and having trim manage all the memory for you.
#include <string.h>
#include <stdlib.h>
#include <stdio.h>
void trim(char *string, void(*func)(char *) )
{
// Advance the pointer to the first non-space char
while( *string == ' ' ) {
string++;
}
// Shrink the length to the last non-space char.
size_t len = strlen(string);
while(string[len-1]==' ') {
len--;
}
// Copy the string to stack memory
char trimmedStr[len + 1];
strncpy(trimmedStr,string, len);
// strncpy does not add a null byte, add it ourselves.
trimmedStr[len] = '\0';
// pass the trimmed string into the user function.
func(trimmedStr);
}
void print_string(char *str) {
printf("'%s'\n", str);
}
int main()
{
char string[]=" a sdf ie ";
trim(string, print_string);
printf("original: '%s'\n", string);
return 0;
}
Ta da! One variable, the original is left unmodified, no memory leaks.
While function pointers have their uses, this is a bit silly.
It's C. Get used to managing memory. ¯\_(ツ)_/¯
Also, is there any way where I don't have to charge original string as
well, like if I print trim(string) it will print trimmed string and
when i print only string, it will print original string
Yes you can, but you cannot allocate new memory in the trim function as you will not be holding the return memory.
You can have a static char buffer in the trim function and operate on it.
Updated version of #elcuco answer.
#include <stdio.h>
char *trim(char *string)
{
static char buff[some max length];
// trim prefix
while ((*string) == ' ' ) {
string++;
}
// find end of original string
int i = 0;
while (*string) {
buff[i++] = *string;
string++;
}
// trim suffix
while ((buff[i]) == ' ' ) {
buff[i] = '\0';
i--;
}
return buff;
}
int main()
{
char string1[] = " abcdefg abcdf ";
char *string = trim(string1);
printf("String is [%s]\n",string);
return 0;
}
With this you don't need to worry about holding reference to trim function return.
Note: Previous values of buff will be overwritten with new call to trim function.
If you don't want to change the original, then you will need to make a copy, or pass a second array of sufficient size as a parameter to your function for filling. Otherwise a simple in-place trmming is fine -- so long as the original string is mutable.
An easy way to approach trimming on leading and trailing whitespace is to determine the number of leading whitespace characters to remove. Then simply use memmove to move from the first non-whitespace character back to the beginning of the string (don't forget to move the nul-character with the right portion of the string).
That leaves only removing trailing whitespace. An easy approach there is to loop from the end of the string toward the beginning, overwriting each character of trailing whitespace with a nul-character until your first non-whitespace character denoting the new end of string is found.
A simple implementation for that could be:
#include <stdio.h>
#include <string.h>
#include <ctype.h>
#define DELIM " \t\n" /* whitespace constant delimiters for strspn */
/** trim leading and trailing whitespace from s, (s must be mutable) */
char *trim (char *s)
{
size_t beg = strspn (s, DELIM), /* no of chars of leading whitespace */
len = strlen (s); /* length of s */
if (beg == len) { /* string is all whitespace */
*s = 0; /* make s the empty-string */
return s;
}
memmove (s, s + beg, len - beg + 1); /* shift string to beginning */
for (int i = (int)(len - beg - 1); i >= 0; i--) { /* loop from end */
if (isspace(s[i])) /* checking if char is whitespace */
s[i] = 0; /* overwrite with nul-character */
else
break; /* otherwise - done */
}
return s; /* Return s */
}
int main (void) {
char string1[] = " a sdf ie ";
printf ("original: '%s'\n", string1);
printf ("trimmed : '%s'\n", trim(string1));
}
(note: additional intervening whitespace was added to your initial string to show that multiple intervening whitespace is left unchanged, the output is single-quoted to show the remaining text boundaries)
Example Use/Output
$ ./bin/strtrim
original: ' a sdf ie '
trimmed : 'a sdf ie'
Look things over and let me know if you have further questions.
Related
I just want the string without underscore. I tried below few codes all doesn't work:
string is char pointer from another function, it looks like this: " "_I_have_1_dog.dat)" "
void func1(char *string)
{
char buffer[256]="";
unsigned long count = 0;
count = sscanf_s(string, " \"%*c%255[^\"]\"", buffer, _countof(buffer));
output:
_I_have_1_dog.dat
count = sscanf_s(string, " \"%*[^_]_%255[^\"]\"", buffer, _countof(buffer));
output:
_I_have_1_dog.dat
count = sscanf_s(string, " \"[^_]_%255[^\"]\"", buffer, _countof(buffer));
output:
_I_have_1_dog.dat
count = sscanf_s(string, " \"[^_]_%255[^\"]\"", buffer, _countof(buffer));
output:
_I_have_1_dog.dat
Edit Based on Removing 1st char '_' Instead of All '_'
The easiest approach to remove a leading '_' is simply to shift all characters down by 1 in string if the first character is an '_'.
You can use the functions like memmove from string.h to do the same thing. (in a single function call) However, looping is just as easy.
A simple function using the loop method could be:
void rm_1st_underscore (char *string)
{
int i = 1; /* set index to 1 (2nd char in string) */
if (*string != '_') /* if 1st char not '_', just return */
return;
do /* loop over each char in string */
string[i-1] = string[i]; /* shift chars back by 1 in string */
while (string[i++] != 0); /* (note: causes '\0' to copy) */
}
A short example would be:
#include <stdio.h>
void rm_1st_underscore (char *string)
{
int i = 1; /* set index to 1 (2nd char in string) */
if (*string != '_') /* if 1st char not '_', just return */
return;
do /* loop over each char in string */
string[i-1] = string[i]; /* shift chars back by 1 in string */
while (string[i++] != 0); /* (note: causes '\0' to copy) */
}
int main (void) {
char str[] = "_I_have_1_dog.dat";
rm_1st_underscore (str);
puts (str);
}
Example Use/Output
$ ./bin/rmunderscore
I_have_1_dog.dat
Look things over and let me know if you still have questions. If you need help trying it with memmove let me know and I'll drop another example.
Using memmove
Since we are just removing the first '_' instead of all of them, using memmove makes it trivial. Simply include string.h and get the length of string and then call memmove copying from the second char in string back to the first, e.g.
...
#include <string.h>
void rm_1st_underscore (char *string)
{
size_t len = strlen (string);
memmove (string, string + 1, len);
}
...
(the output is the same)
I am trying to write a program that reads the stdin stream looking for words (consecutive alphabetic characters) and for each word rotates it left to the first vowel (e.g. "friend" rotates to "iendfr") and writes this sequence out in place of the original word. All other characters are written to stdout unchanged.
So far, I have managed to reverse the letters, but have been unable to do much more. Any suggestions?
#include <stdio.h>
#include <ctype.h>
#include <string.h>
#define MAX_STK_SIZE 256
char stk[MAX_STK_SIZE];
int tos = 0; // next available place to put char
void push(int c) {
if (tos >= MAX_STK_SIZE) return;
stk[tos++] = c;
}
void putStk() {
while (tos >= 0) {
putchar(stk[--tos]);
}
}
int main (int charc, char * argv[]) {
int c;
do {
c = getchar();
if (isalpha(c) && (c == 'a' || c == 'A' || c == 'e' || c == 'E' || c == 'i' || c == 'o' || c == 'O' || c == 'u' || c == 'U')) {
push(c);
} else if (isalpha(c)) {
push(c);
} else {
putStk();
putchar(c);
}
} while (c != EOF);
}
-Soul
I am not going to write the whole program for you, but this example shows how to rotate a word from the first vowel (if any). The function strcspn returns the index of the first character matching any in the set passed, or the length of the string if no matches are found.
#include <stdio.h>
#include <string.h>
void vowelword(const char *word)
{
size_t len = strlen(word);
size_t index = strcspn(word, "aeiou");
size_t i;
for(i = 0; i < len; i++) {
printf("%c", word[(index + i) % len]);
}
printf("\n");
}
int main(void)
{
vowelword("friend");
vowelword("vwxyz");
vowelword("aeiou");
return 0;
}
Program output:
iendfr
vwxyz
aeiou
There are a number of ways your can approach the problem. You can use a stack, but that just adds handling the additional stack operations. You can use a mathematical reindexing, or you can use a copy and fill solution where you copy from the first vowel to a new string and then simply add the initial characters to the end of the string.
While you can read/write a character at a time, you are probably better served by creating the rotated string in a buffer to allow use of the string within your code. Regardless which method you use, you need to validate all string operations to prevent reading/writing beyond the end of your input and/or rotated strings. An example of a copy/fill approach to rotating to the first vowel in your input could be something like the following:
/* rotate 's' from first vowel with results to 'rs'.
* if 's' contains a vowel, 'rs' contains the rotated string,
* otherwise, 'rs' contais 's'. a pointer to 'rs' is returned
* on success, NULL otherwise and 'rs' is an empty-string.
*/
char *rot2vowel (char *rs, const char *s, size_t max)
{
if (!rs || !s || !max) /* validate params */
return NULL;
char *p = strpbrk (s, "aeiou");
size_t i, idx, len = strlen (s);
if (len > max - 1) { /* validate length */
fprintf (stderr, "error: insuffieient storage (len > max - 1).\n");
return NULL;
}
if (!p) { /* if no vowel, copy s to rs, return rs */
strcpy (rs, s);
return rs;
}
idx = p - s; /* set index offset */
strcpy (rs, p); /* copy from 1st vowel */
for (i = 0; i < idx; i++) /* rotate beginning to end */
rs[i+len-idx] = s[i];
rs[len] = 0; /* nul-terminate */
return rs;
}
Above, strpbrk is used to return a pointer to the first occurrence of a vowel in string 's'. The function takes as parameters a pointer to a adequately sized string to hold the rotated string 'rs', the input string 's' and the allocated size of 'rs' in 'max'. The parameters are validated and s is checked for a vowel with strpbrk which returns a pointer to the first vowel in s (if it exists), NULL otherwise. The length is checked against max to insure adequate storage.
If no vowels are present, s is copied to rs and a pointer to rs returned, otherwise the pointer difference is used to set the offset index to the first vowel, the segment of the string from the first vowel-to-end is copied to rs and then the preceding characters are copied to the end of rs with the loop. rs is nul-terminated and a pointer is returned.
While I rarely recommend the use of scanf for input, (a fgets followed by sscanf or strtok is preferable), for purposes of a short example, it can be used to read individual strings from stdin. Note: responding to upper/lower case vowels is left to you. A short example setting the max word size to 32-chars (31-chars + the nul-terminating char) will work for all known words in the unabridged dictionary (longest word is 28-chars):
#include <stdio.h>
#include <string.h>
enum { BUFSZ = 32 };
char *rot2vowel (char *rs, const char *s, size_t max);
int main (void)
{
char str[BUFSZ] = {0};
char rstr[BUFSZ] = {0};
while (scanf ("%s", str) == 1)
printf (" %-8s => %s\n", str, rot2vowel (rstr, str, sizeof rstr));
return 0;
}
Example Use/Output
(shamelessly borrowing the example strings from WeatherVane :)
$ echo "friend vwxyz aeiou" | ./bin/str_rot2vowel
friend => iendfr
vwxyz => vwxyz
aeiou => aeiou
Look it over and let me know if you have any questions. Note: you can call the rot2vowel function prior to the printf statement and print the results with rstr, but since the function returns a pointer to the string, it can be used directly in the printf statement. How you use it is up to you.
Is there any efficient (- in terms of performance) way for printing some arbitrary string, but only until the first new line character in it (excluding the new line character) ?
Example:
char *string = "Hello\nWorld\n";
printf(foo(string + 6));
Output:
World
If you are concerned about performance this might help (untested code):
void MyPrint(const char *str)
{
int len = strlen(str) + 1;
char *temp = alloca(len);
int i;
for (i = 0; i < len; i++)
{
char ch = str[i];
if (ch == '\n')
break;
temp[i] = ch;
}
temp[i] = 0;
puts(temp);
}
strlen is fast, alloca is fast, copying the string up to the first \n is fast, puts is faster than printf but is is most likely far slower than all three operations mentioned before together.
size_t writetodelim(char const *in, int delim)
{
char *end = strchr(in, delim);
if (!end)
return 0;
return fwrite(in, 1, end - in, stdout);
}
This can be generalized somewhat (pass the FILE* to the function), but it's already flexible enough to terminate the output on any chosen delimiter, including '\n'.
Warning: Do not use printf without format specifier to print a variable string (or from a variable pointer). Use puts instead or "%s", string.
C strings are terminated by '\0' (NUL), not by newline. So, the functions print until the NUL terminator.
You can, however, use your own loop with putchar. If that is any performance penalty is to be tested. Normally printf does much the same in the library and might be even slower, as it has to care for more additional constraints, so your own loop might very well be even faster.
for ( char *sp = string + 6 ; *sp != '\0'; sp++ ) {
if ( *sp == '\n' ) break; // newline will not be printed
putchar(*sp);
}
(Move the if-line to the end of the loop if you want newline to be printed.)
An alternative would be to limit the length of the string to print, but that would require finding the next newline before calling printf.
I don't know if it is fast enough, but there is a way to build a string containing the source string up to a new line character only involving one standard function.
char *string = "Hello\nWorld\nI love C"; // Example of your string
static char newstr [256]; // String large enough to contain the result string, fulled with \0s or NULL-terimated
sscanf(string + 6, "%s", newstr); // sscanf will ignore whitespaces
sprintf(newstr); // printing the string
I guess there is no more efficient way than simply looping over your string until you find the first \n in it. As Olaf mentioned it, a string in C ends with a terminating \0 so if you want to use printf to print the string you need to make sure it contains the terminating \0 or yu could use putchar to print the string character by character.
If you want to provide a function creating a string up to the first found new line you could do something like that:
#include <stdio.h>
#include <string.h>
#define MAX 256
void foo(const char* string, char *ret)
{
int len = (strlen(string) < MAX) ? (int) strlen(string) : MAX;
int i = 0;
for (i = 0; i < len - 1; i++)
{
if (string[i] == '\n') break;
ret[i] = string[i];
}
ret[i + 1] = '\0';
}
int main()
{
const char* string = "Hello\nWorld\n";
char ret[MAX];
foo(string, ret);
printf("%s\n", ret);
foo(string+6, ret);
printf("%s\n", ret);
}
This will print
Hello
World
Another fast way (if the new line character is truly unwanted)
Simply:
*strchr(string, '\n') = '\0';
As you can see below I have created a little program to concatenate 2 strings using C, as you may imagine this code doesn't work, I have already corrected it myself by using Array notation instead of pointers, and it works just fine, however I'm still not sure why is it that my code fails being almost a replica of my corrected code.
#include <string.h>
#include <stdlib.h>
#include <stdio.h>
void concatena(char *str1, char *str2){
char *strAux;
int mover;
mover = 0;
strAux = (char *)(malloc(strlen(str1) + strlen(str2)+2));
*(strAux) = '\0';
if(str1 == '\0')
*strAux = '\0';
else
while(str1 != '\0'){
*(strAux+mover++)=*(str1++);
}
if(str2 == '\0')
*strAux = '\0';
else
while(str2 != '\0'){
*(strAux+mover++)=*(str2++);
}
strAux='\0';
str1=strAux;
printf("%s", str1);
free(strAux);
}
I´m still a C beginner (And yes, I'm aware that there are libraries like string.h, I'm asking this for academic reasons) and I have been told that char pointers and arrays are the same thing, something that confuses the heck out of me.
Any help is greatly appreciated.
The first problem I see is with this section:
if(str2 == '\0')
*strAux = '\0';
Just before this code, you've filled up strAux with the string from str1.
Then, if str2 is empty, you suddenly put a null-terminator at the beginning of strAux, eliminating all the work you've done so far!
I think what you intend is:
if(*str2 == '\0')
*(strAux+mover) = '\0';
Its the same thing again after your loop for str2, you have the code:
strAux='\0';
Again, this puts a null-terminator at the start of strAux, effectively ending the newly created string before it even gets started.
Here's how I'd re-write your code:
void concatena(char *str1, char *str2){
char *strAux;
int mover = 0;
strAux = (char *)(malloc(strlen(str1) + strlen(str2)+1)); // Changed to +1, NOT +2
*(strAux) = '\0'; // Start the string as (empty)
while(*str1 != '\0'){ // Copy the first string over.
*(strAux+mover++)=*(str1++);
}
while(*str2 != '\0'){ // Copy the second string over.
*(strAux+mover++)=*(str2++);
}
*(strAux+mover)='\0'; // End the new, combined string.
printf("%s", strAux); // Show the results.
free(strAux);
}
Accepting the same constraints, here is how I would (re)write your code. Unfortunately there is a specification shortcoming: should the concatenation occur to the first string passed? Or should a new string be created? Here are both methods:
#include <string.h>
#include <stdlib.h>
#include <stdio.h>
char *concatena (const char *str1, const char *str2)
{
char *op, *newStr = (char*)malloc (strlen (str1) + strlen (str2) + 1);
if (!newStr)
{
fprintf (stderr, "concatena: error allocating\n");
return;
}
op = newStr; // set up output pointer
while (str1 && *str1) // copy first string
*op++ = *str1++;
while (str2 && *str2) // concatenate second string
*op++ = *str2++;
*op = '\000'; // add conventional NUL termination
return newStr;
}
void concatenb (char *str1, const char *str2)
{
char *op;
if (!str1)
{
fprintf (stderr, "concatenb: NULL string 1\n");
return;
}
op = &str1 [strlen (str1)]; // set output pointer at trailing NUL
while (str2 && *str2) // concatenate second string
*op++ = *str2++;
*op = '\000'; // add conventional NUL termination
}
strAux = (char *)(malloc(strlen(str1) + strlen(str2)+2));
2 is not required, just 1 is sufficient for the termination character.
*(strAux) = '\0';
This should be happening only at the end of all your computation. Not in between the concatenation i.e.,
while(*str1 != '\0'){ // This loops copies the first string
// ^ Notice that you need to dereference to check for the termination character.
*(strAux+mover++)=*(str1++);
}
while(*str2 != '\0'){ // This loop copies the second string
*(strAux+mover++)=*(str2++);
}
// Finally adding termination character
*(strAux+mover) = '\0'; // since with mover you are keeping track of locations.
The amount of errors in your code is disheartening. You should probably pick up a good C book and start over.
First off, there's a library function that you can use to concatenate strings:
const unsigned int len = strlen(str1) + strlen(str2) + 1;
char * dst = malloc(len);
strncat(dst, str1, len);
strncat(dst, str2, len);
Now, if you insist on doing it manually, you have to get pointers and dereferencing right:
char * d = dst;
while (*str1 != 0) *dst++ = *str1++;
while (*str2 != 0) *dst++ = *str2++;
*dst = 0;
// d now points to the beginning of the concatenated string
The two loops check if the current character in the input string is nonzero, and if so, then they copy that character to the current character in the output string, and then both input and output pointer are advanced. (This is all done in one wash by use of the postfix ++ operator.) Finally, the last character is set to zero to create a new null-terminator.
In the process we modified all three pointers dst, str1 and str2. The latter two came in as input function arguments by copy, so that's fine. For returning the concatenated string we made a copy of dst before the loop, which we can return in the end.
I'm trying to parse the string below in a good way so I can get the sub-string stringI-wantToGet:
const char *str = "Hello \"FOO stringI-wantToGet BAR some other extra text";
str will vary in length but always same pattern - FOO and BAR
What I had in mind was something like:
const char *str = "Hello \"FOO stringI-wantToGet BAR some other extra text";
char *probe, *pointer;
probe = str;
while(probe != '\n'){
if(probe = strstr(probe, "\"FOO")!=NULL) probe++;
else probe = "";
// Nulterm part
if(pointer = strchr(probe, ' ')!=NULL) pointer = '\0';
// not sure here, I was planning to separate it with \0's
}
Any help will be appreciate it.
I had some time on my hands, so there you are.
#include <string.h>
#include <stddef.h>
#include <stdio.h>
#include <stdlib.h>
int getStringBetweenDelimiters(const char* string, const char* leftDelimiter, const char* rightDelimiter, char** out)
{
// find the left delimiter and use it as the beginning of the substring
const char* beginning = strstr(string, leftDelimiter);
if(beginning == NULL)
return 1; // left delimiter not found
// find the right delimiter
const char* end = strstr(string, rightDelimiter);
if(end == NULL)
return 2; // right delimiter not found
// offset the beginning by the length of the left delimiter, so beginning points _after_ the left delimiter
beginning += strlen(leftDelimiter);
// get the length of the substring
ptrdiff_t segmentLength = end - beginning;
// allocate memory and copy the substring there
*out = malloc(segmentLength + 1);
strncpy(*out, beginning, segmentLength);
(*out)[segmentLength] = 0;
return 0; // success!
}
int main()
{
char* output;
if(getStringBetweenDelimiters("foo FOO bar baz quaz I want this string BAR baz", "FOO", "BAR", &output) == 0)
{
printf("'%s' was between 'FOO' and 'BAR'\n", output);
// Don't forget to free() 'out'!
free(output);
}
}
In first loop, scan until to find your first delimiter string. Set an anchor pointer there.
if found, from the anchor ptr, in a second loop, scan until you find your 2nd delimiter string or you encounter end of the string
If not at end of string, copy characters between the anchor ptr and the 2nd ptr (plus adjustments for spaces, etc that you need)