Appending char to C array - c

I have a string declared as such:
char *mode_s = (char *)calloc(MODE_S_LEN, sizeof(char));
How can I add a char to the end of the array?

Lets assume " first available position " means at index 0.
char *mode_s = (char *)calloc(MODE_S_LEN, sizeof(char));
*mode_s='a';
To store a character at an arbitrary index n
*(mode_s+n)='b';
Use pointer algebra, as demonstrated above, which is equivalent to
mode_s[n]='b';
One sees that the first case simply means that n=0.
If you wish to eliminate incrementing the counter, as specified in the comment bellow, you can write a data structure and a supporting function that fits your needs. A simple one would be
typedef struct modeA{
int size;
int index;
char *mode_s;
}modeA;
The supporting function could be
int add(modeA* a, char toAdd){
if(a->size==a->index) return -1;
a->mode_s[index]=toAdd;
a->index++;
return 0;
}
It returns 0 when the add was successful, and -1 when one runs out of space.
Other functions you might need can be coded in a similar manner. Note that as C is not object oriented, the data structure has to be passed to the function as a parameter.
Finally you code code a function creating an instance
modeA genModeA(int size){
modeA tmp;
tmp.mode_s=(char *)calloc(size, sizeof(char));
tmp.size=size;
tmp.index=0;
return tmp;
}
Thus using it with no need to manually increment the counter
modeA tmp=genModeA(MODE_S_LEN);
add(&tmp,'c');

There is no standard function to concatenate a character to a string in C. You can easily define such a function:
#include <string.h>
char *strcatc(char *str, char c) {
size_t len = strlen(str);
str[len++] = c;
str[len] = '\0';
return str;
}
This function only works if str is allocated or defined with a larger size than its length + 1, ie if there is available space at its end. In your example, mode_s is allocated with a size of MODE_S_LEN, so you can put MODE_S_LEN-1 chars into it:
char *mode_s = calloc(MODE_S_LEN, sizeof(*mode_s));
for (int i = 0; i < MODE_S_LEN - 1; i++) {
strcatc(mode_s, 'X');
}

char newchar = 'a'; //or getch() from keyboard
//realloc memory:
char *mode_sNew = (char *)calloc(MODE_S_LEN + 1, sizeof(char));
//copy the str:
srncpy(mode_sNew, mode_s, MODE_S_LEN);
//put your char:
mode_sNew[MODE_S_LEN] = newchar;
//free old memory:
free(mode_s);
//reassign to the old string:
mode_s = mode_sNew;
//in a loop you can add as many characters as you want. You also can add more than one character at once, but assign only one in a new position

Related

inverse String in c

i've done a function that inverse a String(array of character) given in parameter , but it's not working , any idea why ?
I'm getting something like this : æIGt(Kt$0#
thanks you
#include <stdio.h>
#include <string.h>
char *
inverse(char *s)
{
int i, taille = strlen(s);
char r[taille];
for (i = 0 ; i < taille ; i++)
{
r[i] = s[taille - i - 1];
}
r[i] = '\0';
return r;
}
int
main()
{
char s[] = "kira";
char *r = inverse(s);
printf("%s",r);
return 1;
}
You are returning a pointer to a local variable. That variable gets destroied when the function inverse returns, so accessing the pointer after the function exits will return invalid data.
It's slightly hard to tell from you question, because you haven't given any outputs, but my best guess is that it's because your returning a pointer to an item on the stack, which will get over-written by the next call, in your case printf. You need to pass inverse a place to put its answer. Try this instead:
#include <stdio.h>
#include <string.h>
void inverse(char *s, char *r)
{
int i,taille=strlen(s);
for(i=0;i<taille;i++)
{
r[i]=s[taille-i-1];
}
r[i]='\0';
}
int main()
{
char s[] = "kira";
char r[sizeof(s)];
inverse(s, r);
printf("%s",r);
return 1;
}
Another standard approach to reversing a string uses pointers to work from both the beginning and end of the string swapping two characters with each iteration. It swaps the original string in place (make a copy if you need to preserve the original, or pass a second string and place the reversed string in there)
/** strrevstr - reverse string, swaps 2 chars per-iteration.
* Takes valid string and reverses, original is not preserved.
* If 's' is valid and non-empty, returns pointer to 's',
* returns NULL otherwise.
*/
char *strrevstr (char *s)
{
if (!s || !*s) { /* validate string is not NULL and non-empty */
printf ("strrevstr() error: invalid string\n");
return NULL;
}
char *begin = s; /* pointers to beginning and end, and tmp char */
char *end = begin + strlen (s) - 1;
char tmp;
while (end > begin) /* swap both beginning and end each iteration */
{
tmp = *end;
*end-- = *begin;
*begin++ = tmp;
}
return s;
}
As you can tell, there are a number of ways to approach the problem, with this and the other answers provided, you should be able to tailor a solution to meet your needs.
There are advantages and disadvantages to every approach. There is nothing wrong with dynamically allocating a new block of memory to hold the reversed string, it just adds an additional responsibility to (1) preserve a pointer to the starting address for the new block so (2) it can be freed when no longer needed. If you need to preserve the original string, passing a pointer to an character array of sufficient size to hold the reversed string is another option preserve the original.
Look over all the answers and let me know if you have any questions.

loop to reverse string in C

So I've looked around on SO and can't find code that answers my question. I have written a function that is supposed to reverse a string as input in cmd-line. Here is the function:
void reverse (char string[]) {
int x;
int i = 0;
char line[strlen(string)];
for (x = strlen(string) - 1; x > 0; x--) {
char tmp = string[x];
line[i] = tmp;
i++;
}
string = line;
}
When I call my reverse() function, the string stays the same. i.e., 'abc' remains 'abc'
If more info is needed or question is inappropriate, let me know.
Thanks!!
You're declaring your line array one char shorter remember the null at the end.
Another point, it should be for (x = strlen(string) - 1; x >= 0; x--) since you need to copy the character at 0.
void reverse (char string[]) {
int x;
int i = 0;
char line[strlen(string) + 1];
for (x = strlen(string) - 1; x >= 0; x--) {
char tmp = string[x];
line[i] = tmp;
i++;
}
for(x = 0; x < strlen(string); x++)
{
string[x] = line[x];
}
}
Note that this function will cause an apocalypse when passed an empty string or a string literal (as Bobby Sacamano said).
Suggestion you can probably do: void reverse(char source[], char[] dest) and do checks if the source string is empty.
I think that your answer is almost correct. You don't actually need an extra slot for the null character in line. You just need two minor changes:
Change the assignment statement at the bottom of the procedure to a memcpy.
Change the loop condition to <-
So, your correct code is this:
void reverse (char string[]) {
int x;
int i = 0;
char line[strlen(string)];
for (x = strlen(string) - 1; x >= 0; x--) {
char tmp = string[x];
line[i] = tmp;
i++;
}
memcpy(string, line, sizeof(char) * strlen(line));
}
Since you want to reverse a string, you first must decide whether you want to reverse a copy of the string, or reverse the string in-situ (in place). Since you asked about this in 'C' context, assume you mean to change the existing string (reverse the existing string) and make a copy of the string in the calling function if you want to preserve the original.
You will need the string library
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
Array indexing works, and this version takes that approach,
/* this first version uses array indexing */
char*
streverse_a(char string[])
{
int len; /*how big is your string*/
int ndx; /*because 'i' is hard to search for*/
char tmp; /*hold character to swap*/
if(!string) return(string); /*avoid NULL*/
if( (len=strlen(string)) < 2 ) return(string); /*one and done*/
for( ndx=0; ndx<len/2; ndx++ ) {
tmp=string[ndx];
string[ndx]=string[len-1-ndx];
string[len-1-ndx]=tmp;
}
return(string);
}
But you can do the same with pointers,
/* this is how K&R would write the function with pointers */
char*
streverse(char* sp)
{
int len, ndx; /*how big is your string */
char tmp, *bp, *ep; /*pointers to begin/end, swap temporary*/
if(!sp) return(sp); /*avoid NULL*/
if( (len=strlen(bp=sp)) < 2 ) return(sp); /*one and done*/
for( ep=bp+len-1; bp<ep; bp++, ep-- ) {
tmp=*bp; *bp=*ep; *ep=tmp; /*swap*/
}
return(sp);
}
(No, really, the compiler does not charge less for returning void.)
And because you always test your code,
char s[][100] = {
"", "A", "AB", "ABC", "ABCD", "ABCDE",
"hello, world", "goodbye, cruel world", "pwnz0r3d", "enough"
};
int
main()
{
/* suppose your string is declared as 'a' */
char a[100];
strcpy(a,"reverse string");
/*make a copy of 'a', declared the same as a[]*/
char b[100];
strcpy(b,a);
streverse_a(b);
printf("a:%s, r:%s\n",a,b);
/*duplicate 'a'*/
char *rp = strdup(a);
streverse(rp);
printf("a:%s, r:%s\n",a,rp);
free(rp);
int ndx;
for( ndx=0; ndx<10; ++ndx ) {
/*make a copy of 's', declared the same as s[]*/
char b[100];
strcpy(b,s[ndx]);
streverse_a(b);
printf("s:%s, r:%s\n",s[ndx],b);
/*duplicate 's'*/
char *rp = strdup(s[ndx]);
streverse(rp);
printf("s:%s, r:%s\n",s[ndx],rp);
free(rp);
}
}
The last line in your code does nothing
string = line;
Parameters are passed by value, so if you change their value, that is only local to the function. Pointers are the value of the address of memory they are pointing to. If you want to modify the pointer that the function was passed, you need to take a pointer to that pointer.
Here is a short example of how you could do that.
void reverse (char **string) {
char line = malloc(strlen(*string) + 1);
//automatic arrays are deallocated once the function ends
//so line needs to be dynamically or statically allocated
// do something to line
*string = line;
}
The obvious issue with this is that you can initialize the string with static memory, then this method will replace the static memory with dynamic memory, and then you'll have to free the dynamic memory. There's nothing functionally wrong with that, it's just a bit dangerous, since accidentally freeing the string literal is illegal.
char *test = "hello";
reverse(test);
free(test); //this is pretty scary
Also, if test was allocated as dynamic memory, the pointer to it would be lost and then it would become a memory leak.

Parsing a string with multi-char delimiter and got junk data along with the result in the second call in c

I have writen a code to split the string with multiple char delimiter.
It is working fine for first time of calling to this function
but i calling it second time it retuns the correct word with some unwanted symbol.
I think this problem occurs because of not clearing the buffer.I have tried a lot but cant solve this. please help me to solve this problem.
char **split(char *phrase, char *delimiter) {
int i = 0;
char **arraylist= malloc(10 *sizeof(char *));
char *loc1=NULL;
char *loc=NULL;
loc1 = phrase;
while (loc1 != NULL) {
loc = strstr(loc1, delimiter);
if (loc == NULL) {
arraylist[i]=malloc(sizeof(loc1));
arraylist[i]=loc1;
break;
}
char *buf = malloc(sizeof(char) * 256); // memory for 256 char
int length = strlen(delimiter);
strncpy(buf, loc1, loc-loc1);
arraylist[i]=malloc(sizeof(buf));
arraylist[i]=buf;
i++;
loc = loc+length;
loc1 = loc;
}
return arraylist;
}
called this function first time
char **splitdetails = split("100000000<delimit>0<delimit>hellooo" , "<delimit>");
It gives
splitdetails[0]=100000000
splitdetails[1]=0
splitdetails[2]=hellooo
but i called this second time
char **splitdetails = split("20000000<delimit>10<delimit>testing" , "<delimit>");
splitdetails[0]=20000000��������������������������
splitdetails[1]=10����
splitdetails[2]=testing
Update:-
thanks to #fatelerror. i have change my code as
char** split(char *phrase, char *delimiter) {
int i = 0;
char **arraylist = malloc(10 *sizeof(char *));
char *loc1=NULL;
char *loc=NULL;
loc1 = phrase;
while (loc1 != NULL) {
loc = strstr(loc1, delimiter);
if (loc == NULL) {
arraylist[i]=malloc(strlen(loc1) + 1);
strcpy(arraylist[i], loc1);
break;
}
char *buf = malloc(sizeof(char) * 256); // memory for 256 char
int length = strlen(delimiter);
strncpy(buf, loc1, loc-loc1);
buf[loc - loc1] = '\0';
arraylist[i]=malloc(strlen(buf));
strcpy(arraylist[i], buf);
i++;
loc = loc+length;
loc1 = loc;
}
}
In the caller function, i used it as
char *id
char **splitdetails = split("20000000<delimit>10<delimit>testing" , "<delimit>");
id = splitdetails[0];
//some works done with id
//free the split details with this code.
for(int i=0;i<3;i++) {
free(domaindetails[i]);
}free(domaindetails);
domaindetails=NULL;
then i called the same for the second as,
char **splitdetails1= split("10000000<delimit>1000<delimit>testing1" , "<delimit>");
it makes error and i can't free the function.
thanks in advance.
Your problem boils down to three basic things:
sizeof is not strlen()
Assignment doesn't copy strings in C.
strncpy() doesn't always nul-terminate strings.
So, when you say something like:
arraylist[i]=malloc(sizeof(loc1));
arraylist[i]=loc1;
thisdoes not copy the string. The first one allocates the size of loc1, which is a char *. In other words, you allocated the size of a pointer. You want to allocate storage to store the string, i.e. using strlen():
arraylist[i]=malloc(strlen(loc1) + 1);
Note the + 1 as well, because you also need room for the nul-terminator. Then, to copy the string you want to use strcpy():
strcpy(arraylist[i], loc1);
The way you had it was just assigning a pointer to your old string (and in the process leaing the memory you had just allocated). It's also common to use strdup() which combines both of these steps, i.e.
arraylist[i] = strdup(loc1);
This is convenient but strdup() is not part of the official C library. You need to assess the portability needs of your code before you consider using it.
Additionally, with strncpy(), you should be aware that it does not always nul-terminate:
strncpy(buf, loc1, loc-loc1);
This copies less bytes than were in the original string and doesn't terminate buf. Thus, it's necessary to include a nul terminator yourself:
buf[loc - loc1] = '\0';
This is the root cause of what you are seeing with the garbage. Since you didn't nul terminate, C doesn't know where your string ends and so it keeps on reading whatever happens to be in memory.

Substring in C using pointer

I have a character pointer that points the begining of a string and an index less than the length of the string. Now I want to create a pointer to point a substring of original string from the begining to the index or a substring with above constraints. Please help me to find a way to get it done. Here is a bit of the code:
char* ch="323+465";//this is the original string
int index=2; //this is the index upto which I wish to create a substring,
// in java, it would have been ch.substring(0,3), if ch were a String
Thanks in advance.
You can't do that without creating 3 strings. The char point only marks the beginning of the string, so you would need to combine a pointer and an index into a new type. Remember you don't have strings in C. In languages like Java (and others) will create copies of the sub string anyway.
struct pseudo_string { char *s, int index; } vstring[3];
char* ch="323+465";
vstring[0].s = ch;
vstring[0].index = 2;
vstring[1].s = ch + index + 1; // weird
vstring[1].index = 1;
vstring[2].s = vstring[1].s + 1;
vstring[2].index = 2;
So it is overly complex and useless. In this case index is being used as counter...
If you want to keep the same base pointer, you gonna need 2 indices or 1 index and a len:
struct pseudo_string2 { char *s; int start; int end; };
But that's an overkill for small strings.
If don't want to use malloc, you can try to use a matrix:
char vstring[3][10]={0};
strncpy(vstring[0], ch, 3);
strncpy(vstring[1], ch+3, 1);
strncpy(vstring[2], ch+4, 3);
The advantage of the matrix, even if you waste few bytes, is that you don't need to deallocate it. But if you need to use these values outside this function, than you don't have another scape than to use malloc and free (don't consider globals for that ;-).
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
char * substr(char *s, int start, int end)
{
int size = end - start + 2; // 1 for the inclusive limits and another 1 for the \0
char * r = (char*)malloc(size);
strncpy(r,s+start, size-1);
r[size-1]=0;
return r;
}
int main()
{
char* ch="323+465";
char *parts[3];
parts[0] = substr(ch, 0,2);
parts[1] = substr(ch, 3,3);
parts[2] = substr(ch, 4,6);
printf("%s %s %s\n", parts[0], parts[1], parts[2]);
free(parts[0]);
free(parts[1]);
free(parts[2]);
}
Make a copy of a suitable number of characters:
char * substr = malloc(index + 2);
strncpy(substr, ch, index + 1);
substr[index + 1] = 0;
// ...
free(substr);
If you're happy to mutilate the original string, just insert a null byte:
ch[index + 1] = 0;
The odd + 1 comes from the fact that your index seems to be inclusive, which is generally a bad idea.
You can't, because that would imply modifying the string literal, which is illegal.
Alternative:
char ch[]="323+465";
int index=2;
ch[index] = '\0';
First, make the string writable by using a char array instead of a pointer to a string literal. Then
char ch[] = "323+456";
int idx = 2;
ch[idx] = 0; /* Now ch holds the string "32" */
You should avoid an identifier clash with the classic BSD index function, that's why I used idx instead.
This solution assumes it is okay to modify the original string. If not, you need to allocate a new string first.
the same behavior of java substring (allocates new string)
char* res = (char*)malloc(index+2);
strncpy(res,ch,index+1);
res[index+1]='\0';
I see that you try to delimit by +, so easier is to use strtok :
char ch[] ="323+465";
char * res;
res = strtok (ch,"+");
// res= 323

c splitting a char* into an char**

I'm reading in a line from a file (char by char, using fgetc()), where all fields(firstname, lastname, ...) are seperated by an ;. What I now want to do is create a char**, add all the chars to that and replace the ; by \0 so that I effectively get a list of all the fields.
Is that actually possibly? And when I create a char**, e.g. char ** buf = malloc(80) can I treat it like a one dimensional array? If the memory returned by malloc contiguous?
EDIT
Sry, meant to replace ; by \0, bot \n.
EDIT 2
This code should demonstrate what I intend to do (may clarify things a little):
int length = 80; // initial length char *buf = malloc(length); int num_chars = 0;
// handle malloc returning NULL
// suppose we already have a FILE pointer fp for (int ch = fgetc(fp); ch != EOF && ferror(fp) == 0; ch = fgetc(fp)) {
if (length == size) { // expand array if necessary
length += 80;
buf = realloc(buf, length);
// handle realloc failure
}
if (ch == ';') {
buf[num_chars] = '\0'; // replace delimiter by null-terminator
} else {
buf[num_chars] = ch; // else put char in buf
} }
// resize the buffer to chars read buf
= realloc(buf, num_chars);
// now comes the part where I'm quite unsure if it works / is possible
char **list = (char **)buf; // and now I should be able to handle it like a list of strings?
Yes, it's possible. Yes, you can treat it as a one dimensional array. Yes, the memory is contiguous.
But you probably want:
char ** fields = malloc(sizeof(char *) * numFields);
Then you can do:
// assuming `field` is a `char *`
fields[i++] = field;
It's not exactly possible as you describe it, but something similar is possible.
More specifically, the last line of your example char **list = (char **)buf; won't do what you believe. char **list means items pointed by *list will be of the type char*, but the content of *buf are chars. Hence it won't be possible to change one into another.
You should understand that a char ** is just a pointer, it can hold nothing by itself. But the char ** can be the start address of an array of char *, each char * pointing to a field.
You will have to allocate space for the char * array using a first malloc (or you also can use a static array of char * instead of a char**. You also will have to find space for every individual field, probably performing a malloc for each field and copying it from the initial buffer. If the initial buffer is untouched after reading, it would also be possible to use it to keep the field names, but if you just replace the initial ; by a \n, you'll be missing the trailing 0 string terminator, you can't replace inplace one char by two char.
Below is a simplified example of what can be done (removed malloc part as it does not add much to the example, and makes it uselessly complex, that's another story):
#include <stdio.h>
#include <string.h>
main(){
// let's define a buffer with space enough for 100 chars
// no need to perform dynamic allocation for a small example like this
char buf[100];
const char * data = "one;two;three;four;";
// now we want an array of pointer to fields,
// here again we use a static buffer for simplicity,
// instead of a char** with a malloc
char * fields[10];
int nbfields = 0;
int i = 0;
char * start_of_field;
// let's initialize buf with some data,
// a semi-colon terminated list of strings
strcpy(buf, data);
start_of_field = buf;
// seek end of each field and
// copy addresses of field to fields (array of char*)
for (i = 0; buf[i] != 0; i++){
if (buf[i] == ';'){
buf[i] = 0;
fields[nbfields] = start_of_field;
nbfields++;
start_of_field = buf+i+1;
}
}
// Now I can address fields individually
printf("total number of fields = %d\n"
"third field is = \"%s\"\n",
nbfields, fields[2]);
}
I now want to do is create a char**, add all the chars to that
You would allocate an array of char (char*) to store the characters, not an array of char* (char**).
And when I create a char*, e.g. char * buf = malloc(80) can I treat it like a one dimensional array? If the memory returned by malloc contiguous?
Yes, malloc always returns continguous blocks. Yes, you can treat it as single-dimensional an array of char* (with 80/sizeof char* elements). But you'll need to seperately allocate memory for each string you store in this array, or create another block for storing the string data, then use your first array to store pointers into that block.
(or something else; lots of ways to skin this cat)
It may be that Helper Method wants each of the "fields" written to a nul-terminated string. I cannot tell. You can do that with strtok(). However strtok() has problems - it destroys the original string "fed" to it.
#define SPLIT_ARRSZ 20 /* max possible number of fields */
char **
split( char *result[], char *w, const char *delim)
{
int i=0;
char *p=NULL;
for(i=0, result[0]=NULL, p=strtok(w, delim); p!=NULL; p=strtok(NULL, delim), i++ )
{
result[i]=p;
result[i+1]=NULL;
}
return result;
}
void
usage_for_split(void)
{
char *result[SPLIT_ARRSZ]={NULL};
char str[]="1;2;3;4;5;6;7;8;9;This is the last field\n";
char *w=strdup(str);
int i=0;
split(result, w, ";\n");
for(i=0; result[i]!=NULL; i++)
printf("Field #%d = '%s'\n", i, result[i]);
free(w);
}

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