I have code this add_str function :
void add_str(char *str1, char *str2, char **res)
{
int i;
int j = 0;
res[0] = malloc((strlen(str1) + strlen(str2) + 1) * sizeof(char));
if (res[0] == NULL)
return;
for (i = 0; str1[i]; i++)
res[0][j++] = str1[i];
for (i = 0; str2[i]; i++)
res[0][j++] = str2[i];
res[0][j] = '\0';
}
It receives 2 strings, str1 and str2 and a pointer an string **res which is not malloc. My function add str1 and str2 on **res.
My question is : Is there a way to do not write res[0] each time I have to do something with it ?
res is a pointer to a pointer, so you need to dereference it before you use it. You're right that res[0] isn't the right way to do that though in this context. Use (*res) instead.
What you really want is to dereference:
*res
This is not a direct answer to your question, but more of a general coding advice:
In order to properly manage all dynamic memory operations within your program, you should strive (try as much as you can) to have both operations x=malloc(...) and free(x) executed from the same function.
In most cases, if you design your code correctly, then you can achieve this.
In some cases, when the flow is asynchronous, it is impossible to do so. For example, one function allocates the buffer, and after some event occurs in the system, another function (AKA callback) releases the buffer.
In such cases, you should still try to have both operations within the same "scope" (class, file or module).
In your coding example, function add_str allocates the buffer but does not release it, which means some other function will eventually have to do it (or you will have memory leaks). Hence, if possible, then you should try to perform the malloc outside of this function.
Why do you need to pass a pointer pointer anyway? You could just as well change it to this:
char *add_str(char *str1, char *str2)
{
char *res = malloc((strlen(str1) + strlen(str2) + 1));
if (res == NULL)
return NULL;
char *ret = res;
while(*str1 != 0)
*res++ = *str1++;
while(*str2 != 0)
*res++ = *str2++;
*res = '\0';
return ret;
}
It has the same effect and you don't have to deal wit that ugly construct.
I would use another pointer variable to hold the result string, which avoids dereferencing the result variable all the time. Apart from that, of course, you should avoid using the array subscript operator where a simple pointer dereference is what you do. I have also made a few other changes to your example code to make it more concise, feel free to ignore them.
void add_str(char *str1, char *str2, char **res) {
char* result = *res = malloc((strlen(str1) + strlen(str2) + 1) * sizeof(char));
//Dereferencing a NULL pointer will safely crash your program on any sane system.
//if (!result) return;
int j = 0; //Since C99 you are allowed to mix code and variable declarations.
for(int i = 0; str1[i]; i++) result[j++] = str1[i];
for(int i = 0; str2[i]; i++) result[j++] = str2[i];
result[j] = '\0';
}
If pointers are used to their full potential, your code can look like this:
void add_str(char *str1, char *str2, char **res) {
char* result = *res = malloc((strlen(str1) + strlen(str2) + 1) * sizeof(char));
while(*str1) *result++ = *str1++;
while(*str2) *result++ = *str2++;
*result = '\0';
}
Related
I'm trying to add a character at a defined position. I've created a new function, allocate a memory for one more char, save characters after the position then added my character at the defined position, and now I don't know how to erase characters after that position to concatenate the saved string. Any solution?
Here is the beginning of my function:
void appendCharact(char *source, char carac, int position) {
source = realloc(source, strlen(source) * sizeof(char) + 1); //Get enough memory
char *temp = source.substr(position); //Save characters after my position
source[position] = carac; //Add the character
}
EDIT :
I'm trying to implement another "barbarous" solution, in debug mode I can see that I've approximately my new string but it look like I can't erase the older pointer...
void appendCharact(char *source, char carac, int position) {
char *temp = (char *)malloc((strlen(source) + 2) * sizeof(char));
int i;
for(i = 0; i < position; i++) {
temp[i] = source[i];
}
temp[position] = carac;
for (i = position; i < strlen(source); i++) {
temp[i + 1] = source[i];
}
temp[strlen(temp) + 1] = '\0';
free(source);
source = temp;
}
I mentioned that I could see five problems with the code as shown (copied here for reference)
void appendCharact(char * source, char carac , int position)
{
source = realloc(source, strlen(source) * sizeof(char) + 1); //Get enough memory
char * temp = source.substr(position); //Save characters after my position
source[position] = carac; //Add the charactere
}
The problems are (in no specific order):
strlen(source) * sizeof(char) + 1 is equal to (strlen(source) * sizeof(char)) + 1. It should have been (strlen(source) + 1) * sizeof(char). However, this works fine since sizeof(char) is defined in the C++ specification to always be equal to 1.
Related to the above: Simple char strings are really called null-terminated byte strings. As such they must be terminated by a "null" character ('\0'). This null character of course needs space in the allocated string, and is not counted by strlen. Therefore to add a character you need allocate strlen(source) + 2 characters.
Never assign back to the pointer you pass to realloc. If realloc fails, it will return a null pointer, making you lose the original memory, and that is a memory leak.
The realloc function return type is void*. In C++ you need to cast it to the correct pointer type for assignment.
You pass source by value, meaning inside the function you have a local copy of the pointer. When you assign to source you only assign to the local copy, the original pointer used in the call will not be modified.
Here are some other problems with the code, or its possible use:
Regarding the null-terminator, once you allocate enough memory for it you also need to add it to the string.
If the function is called with source being a literal string or an array or anything that wasn't returned by a previous call to malloc, calloc or realloc, then you can't pass that pointer to realloc.
You use source.substr(position) which is not possible since source isn't an object and therefore doesn't have member functions.
Your new solution is much closer to a working function but it still has some problems:
you do not check for malloc() failure.
you should avoid computing the length of the source string multiple times.
temp[strlen(temp) + 1] = '\0'; is incorrect as temp is not yet a proper C string and strlen(temp) + 1 would point beyond the allocated block anyway, you should just write temp[i + 1] = '\0';
the newly allocated string should be returned to the caller, either as the return value or via a char ** argument.
Here is a corrected version:
char *insertCharact(char *source, char carac, size_t position) {
size_t i, len;
char *temp;
len = source ? strlen(source) : 0;
temp = (char *)malloc(len + 2);
if (temp != NULL) {
/* sanitize position */
if (position > len)
position = len;
/* copy initial portion */
for (i = 0; i < position; i++) {
temp[i] = source[i];
}
/* insert new character */
temp[i] = carac;
/* copy remainder of the source string if any */
for (; i < len; i++) {
temp[i + 1] = source[i];
}
/* set the null terminator */
temp[i + 1] = '\0';
free(source);
}
return temp;
}
int pos = 1;
char toInsert = '-';
std::string text = "hallo";
std::stringstream buffer;
buffer << text.substr(0,pos);
buffer << toInsert;
buffer << text.substr(pos);
text = buffer.str();
Try using something like:
#include <string>
void appendCharAt(std::string& src, char c , int pos)
{
std::string front(src.begin(), src.begin() + pos - 1 ); // use iterators
std::string back(src.begin() + pos, src.end() );
src = front + c + back; // concat together +-operator is overloaded for strings
}
Not 100% sure weather the positions are right. Maybe front hast to be src.begin() + pos and back src.begin() + pos + 1. Just try it out.
The C version of this will have to take care of the situation where realloc fails, in which case the original string is preserved. You should only overwrite the old pointer with the one returned from realloc upon success.
It might look something like this:
bool append_ch (char** str, char ch, size_t pos)
{
size_t prev_size = strlen(*str) + 1;
char* tmp = realloc(*str, prev_size+1);
if(tmp == NULL)
{
return false;
}
memmove(&tmp[pos+1], &tmp[pos], prev_size-pos);
tmp[pos] = ch;
*str = tmp;
return true;
}
Usage:
const char test[] = "hello word";
char* str = malloc(sizeof test);
memcpy(str, test, sizeof test);
puts(str);
bool ok = append_ch(&str, 'l', 9);
if(!ok)
asm ("HCF"); // error handling here
puts(str);
free(str);
I have trouble with my code and I need your help! What I need to do is to write a function that will extract the web address that starts from www. and ends with .edu from an inputted string. The inputted string will have no spaces in it so scanf() should work well here.
For example:
http://www.school.edu/admission. The extracted address should be www.school.edu.
This is what I came up with so far, it obviously didn't work, and I can't think of anything else unfortunately.
void extract(char *s1, char *s2) {
int size = 0;
char *p, *j;
p = s1;
j = s2;
size = strlen(s1);
for(p = s1; p < (s1 + size); p++) {
if(*p == 'w' && *(p+1) == 'w' && *(p+2) == 'w' && *(p+3) == '.'){
for(p; p < (p+4); p++)
strcat(*j, *p);
}
else if(*p=='.' && *(p+1)=='e' && *(p+2)=='d' && *(p+3)=='u'){
for(p; (p+1) < (p+4); p++)
strcat(*j, *p);
}
}
size = strlen(j);
*(j+size+1) = '\0';
}
The function has to use pointer arithmetic. The errors I get have something to do with incompatible types and casting. Thanks ahead!
So the most trivial approach might be:
#include <stdio.h>
int main(void)
{
char str[1000];
sscanf("http://www.school.edu/admission", "%*[^/]%*c%*c%[^/]", str);
puts(str);
}
Now, here goes the fixed code:
#include <stdio.h>
#include <string.h>
void extract(char *s1, char *s2) {
size_t size = strlen(s1), i = 0;
while(memcmp(s1 + i, "www.", 4)){
i++;
}
while(memcmp(s1 + i, ".edu", 4)){
*s2++ = *(s1 + i);
i++;
}
*s2 = '\0';
strcat(s2, ".edu");
}
int main(void)
{
char str1[1000] = "http://www.school.edu/admission", str2[1000];
extract(str1, str2);
puts(str2);
}
Note that s2 must be large enough to contain the extracted web address, or you may get a segfault.
This is an easy solution for your problem:
char* extract(char *s1) {
char* ptr_www;
char* ptr_edu;
int len ;
char* s2;
ptr_www = strstr(s1,"www");
ptr_edu = strstr(s1,".edu");
len = ptr_edu -ptr_www + 4;
s2 = malloc (sizeof(char)*len+1);
strncpy(s2,ptr_www,len);
s2[len] = '\0';
printf ("%s",s2);
return s2;
}
There is a lot wrong unfortunately. Your compilation is failing because you pass a char to strcat when it expects a char*. Even if it did compile though it would crash.
for(p = s1; p < (s1 + size); p++) {
// This if statement will reference beyond s1+size when p=s1+size-2. Consequently it may segfault
if(*p=='w' && *(p+1)=='w' && *(p+2)=='w' && *(p+3)=='.') {
for(p; p < (p+4); p++) // This is an infinite loop
// strcat concatenates one string onto another.
// Dereferencing the pointer makes no sense.
// This is the likely causing your compilation error.
// If this compiled it would almost certainly segfault.
strcat(*j, *p);
}
// This will also reference beyond s1+size. Consequently it may segfault
else if(*p=='.' && *(p+1)=='e' && *(p+2)=='d' && *(p+3)=='u') {
for(p; (p+1) < (p+4); p++) // This is also an infinite loop
// Again strcat expects 2x char* (aka. strings) not 2x char
// This will also almost certainly segfault.
strcat(*j, *p);
}
}
// strlen() counts the number of chars until the first '\0' occurrence
// It is never correct to call strlen() to determine where to add a '\0' string termination character.
// If the character were actually absent this would almost certainly result in a segfault.
// As it is strcat() (when called correctly) will add the terminator anyway.
size = strlen(j);
*(j+size+1) = '\0';
EDIT: This seems like a homework question, so I thought it would be more constructive to mention where your current code is going wrong, so you can recheck your knowledge in those areas.
The answer to your exact question is it doesn't compile because you dereference the string and hence pass 2x char instead of char* to strcat().
I try to code my own concatenation function in C without library, but I have issue and I don't know where it comes from.
To do my function I use pointers of char.
This is my Code :
#include <stdio.h>
#include <stdlib.h>
int longueur(char *str)
{
int i =0;
while(str[i] != '\0')
{
i++;
}
return i;
}
void concat(char* source, char* dest)
{
int longStr1 = (longueur(source));
int longStr2 = (longueur(dest));
int i=0, j=0;
char* temp = dest;
free(dest);
dest = (char*) realloc(dest, ((longStr1 + longStr2)* sizeof(char)));
/*dest[0] = temp[0]; <------ If I do this it will generate issue, so the bellow code too*/
while(temp[i] != '\0')
{
dest[i] = temp[i];
i++;
}
while(source[j] != '\0')
{
dest[i] = source[j];
i++;
j++;
}
dest[i] = '\0';
}
int main()
{
char *str1 = "World";
char *str2 = "Hello";
concat(str1, str2);
printf("-------------\n%s", str2);
return 0;
}
EDIT
I read all your answer, so I changed my concat function to :
void concat(char* source, char* dest)
{
int longStr1 = (longueur(source));
int longStr2 = (longueur(dest));
int i=0, j=0;
dest = (char*) malloc((longStr1 + longStr2)* sizeof(char) + sizeof(char));
while(dest[i] != '\0')
{
dest[i] = dest[i];
i++;
}
while(source[j] != '\0')
{
dest[i] = source[j];
i++;
j++;
}
dest[i] = '\0';
}
Now I don't have issue but my code only display "Hello"
In addition to all the good comments and solutions: realloc can give you a different pointer and you must return that pointer. So your function signature should be:
void concat(char* source, char** dest)
{
int longStr1 = (longueur(source));
int longStr2 = (longueur(dest));
int i=0, j=0;
char* temp = *dest, *temp2;
if ((temp2 = realloc(dest, ((longStr1 + longStr2)+1))==NULL) return;
*dest= temp2;
while(temp[i] != '\0')
{
*dest[i] = temp[i];
i++;
}
while(source[j] != '\0')
{
*dest[i] = source[j];
i++;
j++;
}
*dest[i] = '\0';
}
..and this assumes the function will only be called with a dest that was allocated with malloc. And sizeof(char) is always 1. (This resulting function is not optimal.)
--EDIT--
Below the correct, optimized version:
void concat(char* source, char** dest)
{
int longSrc = longueur(source);
int longDst = longueur(dest);
char *pDst, *pSrc;
if ((pDst = realloc(*dest, longSrc + longDst + 1))==NULL) return;
if (pDst != *dest) *dest= pDst;
pDst += longSrc;
pSrc= source;
while(pSrc)
*pDst++ = *pSrc++;
*pDst = '\0';
}
In your code
free(dest);
and
dest = (char*) realloc(dest, ((longStr1 + longStr2)* sizeof(char)));
invokes undefined behavior as none of them use a pointer previously allocated by malloc() or family.
Mostly aligned with your approach, you need to make use of another pointer, allocate dynamic memory and return that pointer. Do not try to alter the pointers received as parameters as you've passed string literals.
That said, you need to have some basic concepts clear first.
You need not free() a memory unless it is allocated through malloc() family.
You need to have a char extra allocated to hold the terminating null.
Please see this discussion on why not to cast the return value of malloc() and family in C..
If your concatenation function allocates memory, then, the caller needs to take care of free()-ing the memory, otherwise it will result in memory leak.
After you have freed dest here:
free(dest);
You cannot use this pointer in following call to realloc:
dest = (char*) realloc(dest, ((longStr1 + longStr2)* sizeof(char)));
/*dest[0] = temp[0]; <------ If I do this it will generate issue, so the bellow code too*/
man realloc
void *realloc(void *ptr, size_t size);
The realloc() function changes the size of the memory block
pointed to by ptr to size bytes. (...)
But this pointer is invalid now and you cannot use it anymore. When you call free(dest), the memory dest points to is being freed, but the value of dest stays untouched, making the dest a dangling pointer. Accessing the memory that has already been freed produces undefined behavior.
NOTE:
Even if free(dest) is technically valid when called on pointer to memory allocated by malloc (it is not an error in your function to call free(dest) then), it is incorrect to use this on pointer to literal string as you do in your example (because str2 points to string literal it is an error to pass this pointer to function calling free on it).
Given your original use, perhaps you would find a variant like this useful
#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>
size_t longueur ( const char * str ) { /* correct type for string lengths */
size_t len = 0;
while (*str++ != '\0') ++len;
return len;
}
char * concat ( const char * first, const char * second ) {
const char * s1 = first ? first : ""; /* allow NULL input(s) to be */
const char * s2 = second ? second : ""; /* treated as empty strings */
size_t ls1 = longueur(s1);
size_t ls2 = longueur(s2);
char * result = malloc( ls1 + ls2 + 1 ); /* +1 for NUL at the end */
char * dst = result;
if (dst != NULL) {
while ((*dst = *s1++) != '\0') ++dst; /* copy s1\0 */
while ((*dst = *s2++) != '\0') ++dst; /* copy s2\0 starting on s1's \0 */
}
return result;
}
int main ( void ) {
const char *str1 = "Hello";
const char *str2 = " World";
char * greeting = concat(str1, str2);
printf("-------------\n%s\n-------------\n", greeting);
free(greeting);
return 0;
}
In this variant, the two inputs are concatenated and the result of the concatenation is returned. The two inputs are left untouched.
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.
What would be an efficient way of converting a delimited string into an array of strings in C (not C++)? For example, I might have:
char *input = "valgrind --leak-check=yes --track-origins=yes ./a.out"
The source string will always have only a single space as the delimiter. And I would like a malloc'ed array of malloc'ed strings char *myarray[] such that:
myarray[0]=="valgrind"
myarray[1]=="--leak-check=yes"
...
Edit I have to assume that there are an arbitrary number of tokens in the inputString so I can't just limit it to 10 or something.
I've attempted a messy solution with strtok and a linked list I've implemented, but valgrind complained so much that I gave up.
(If you're wondering, this is for a basic Unix shell I'm trying to write.)
What's about something like:
char* string = "valgrind --leak-check=yes --track-origins=yes ./a.out";
char** args = (char**)malloc(MAX_ARGS*sizeof(char*));
memset(args, 0, sizeof(char*)*MAX_ARGS);
char* curToken = strtok(string, " \t");
for (int i = 0; curToken != NULL; ++i)
{
args[i] = strdup(curToken);
curToken = strtok(NULL, " \t");
}
if you have all of the input in input to begin with then you can never have more tokens than strlen(input). If you don't allow "" as a token, then you can never have more than strlen(input)/2 tokens. So unless input is huge you can safely write.
char ** myarray = malloc( (strlen(input)/2) * sizeof(char*) );
int NumActualTokens = 0;
while (char * pToken = get_token_copy(input))
{
myarray[++NumActualTokens] = pToken;
input = skip_token(input);
}
char ** myarray = (char**) realloc(myarray, NumActualTokens * sizeof(char*));
As a further optimization, you can keep input around and just replace spaces with \0 and put pointers into the input buffer into myarray[]. No need for a separate malloc for each token unless for some reason you need to free them individually.
Were you remembering to malloc an extra byte for the terminating null that marks the end of string?
From the strsep(3) manpage on OSX:
char **ap, *argv[10], *inputstring;
for (ap = argv; (*ap = strsep(&inputstring, " \t")) != NULL;)
if (**ap != '\0')
if (++ap >= &argv[10])
break;
Edited for arbitrary # of tokens:
char **ap, **argv, *inputstring;
int arglen = 10;
argv = calloc(arglen, sizeof(char*));
for (ap = argv; (*ap = strsep(&inputstring, " \t")) != NULL;)
if (**ap != '\0')
if (++ap >= &argv[arglen])
{
arglen += 10;
argv = realloc(argv, arglen);
ap = &argv[arglen-10];
}
Or something close to that. The above may not work, but if not it's not far off. Building a linked list would be more efficient than continually calling realloc, but that's really besides the point - the point is how best to make use of strsep.
Looking at the other answers, for a beginner in C, it would look complex due to the tight size of code, I thought I would put this in for a beginner, it might be easier to actually parse the string instead of using strtok...something like this:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
char **parseInput(const char *str, int *nLen);
void resizeptr(char ***, int nLen);
int main(int argc, char **argv){
int maxLen = 0;
int i = 0;
char **ptr = NULL;
char *str = "valgrind --leak-check=yes --track-origins=yes ./a.out";
ptr = parseInput(str, &maxLen);
if (!ptr) printf("Error!\n");
else{
for (i = 0; i < maxLen; i++) printf("%s\n", ptr[i]);
}
for (i = 0; i < maxLen; i++) free(ptr[i]);
free(ptr);
return 0;
}
char **parseInput(const char *str, int *Index){
char **pStr = NULL;
char *ptr = (char *)str;
int charPos = 0, indx = 0;
while (ptr++ && *ptr){
if (!isspace(*ptr) && *ptr) charPos++;
else{
resizeptr(&ptr, ++indx);
pStr[indx-1] = (char *)malloc(((charPos+1) * sizeof(char))+1);
if (!pStr[indx-1]) return NULL;
strncpy(pStr[indx-1], ptr - (charPos+1), charPos+1);
pStr[indx-1][charPos+1]='\0';
charPos = 0;
}
}
if (charPos > 0){
resizeptr(&pStr, ++indx);
pStr[indx-1] = (char *)malloc(((charPos+1) * sizeof(char))+1);
if (!pStr[indx-1]) return NULL;
strncpy(pStr[indx-1], ptr - (charPos+1), charPos+1);
pStr[indx-1][charPos+1]='\0';
}
*Index = indx;
return (char **)pStr;
}
void resizeptr(char ***ptr, int nLen){
if (*(ptr) == (char **)NULL){
*(ptr) = (char **)malloc(nLen * sizeof(char*));
if (!*(ptr)) perror("error!");
}else{
char **tmp = (char **)realloc(*(ptr),nLen);
if (!tmp) perror("error!");
*(ptr) = tmp;
}
}
I slightly modified the code to make it easier. The only string function that I used was strncpy..sure it is a bit long-winded but it does reallocate the array of strings dynamically instead of using a hard-coded MAX_ARGS, which means that the double pointer is already hogging up memory when only 3 or 4 would do, also which would make the memory usage efficient and tiny, by using realloc, the simple parsing is covered by employing isspace, as it iterates using the pointer. When a space is encountered, it reallocates the double pointer, and malloc the offset to hold the string.
Notice how the triple pointers are used in the resizeptr function.. in fact, I thought this would serve an excellent example of a simple C program, pointers, realloc, malloc, passing-by-reference, basic element of parsing a string...
Hope this helps,
Best regards,
Tom.