How to dynamic malloc memory in function in c? - c

Question

I want to call a function like this:
char* Seg(char* input, char **segs, int* tags)
in fact input is the real input, segs tags is the return, and now return is the error message.
my program like this:
#include <stdio.h>
char* Seg(char* input, char **segs, int* tags) {
// dynamic malloc the memory here
int count = strlen(input); // this count is according to input
for (int i = 0; i < count; i++) {
segs[i] = "abc";
}
for (int i = 0; i < count; i++) {
tags[i] = i;
}
return NULL;
}
int main(int argc, char *argv[])
{
char** segs = NULL;
int* tags = NULL;
Seg("input", segs, tags);
return 0;
}
I am asking how can I return the value in segs and tags?
Edit
Now I change code to this:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
/**
* input is input params, segs and tags is results
* return: error msg
*/
int Seg(char* input, char ***segs, int** tags) {
int n = strlen(input);
int *tags_ = malloc(n * sizeof(int));
for (int i = 0; i < n; i++) {
tags_[i] = i;
}
tags = &tags_;
char **segs_ = malloc(sizeof(char *) * n);
for (int i = 0; i < n; i++) {
segs_[i] = "haha";
}
segs = &segs_;
return n;
}
int main(int argc, char *argv[])
{
char** segs = NULL;
int* tags = NULL;
int n = Seg("hahahahah", &segs, &tags);
printf("%p", tags);
free(segs);
free(tags);
return 0;
}
Why tags is still nil?

Answer 1

If I have understood you correctly then you need something like the following.
I used sentinel values for the both dynamically allocated arrays. You can use your own approach instead of using sentinel values.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char * Seg( const char *input, char ***segs, int **tags )
{
// dynamic malloc the memory here
size_t count = strlen( input ); // this count is according to input
*segs = malloc((count + 1) * sizeof(char *));
*tags = malloc((count + 1) * sizeof(int));
for ( size_t i = 0; i < count; i++ )
{
( *segs )[i] = "abc";
}
(*segs)[count] = NULL;
for ( size_t i = 0; i < count; i++ )
{
( *tags )[i] = ( int )i;
}
(*tags)[count] = -1;
return NULL;
}
int main( void )
{
char **segs = NULL;
int *tags = NULL;
Seg( "input", &segs, &tags );
for (char **p = segs; *p; ++p)
{
printf( "%s ", *p );
}
putchar('\n');
for (int *p = tags; *p != -1; ++p)
{
printf("%d ", *p);
}
putchar('\n');
free(segs);
free(tags);
return 0;
}
The program output is
abc abc abc abc abc
0 1 2 3 4
After you updated your post then the function can look also the following way
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
size_t Seg( const char *input, char ***segs, int **tags )
{
// dynamic malloc the memory here
size_t count = strlen( input ); // this count is according to input
*segs = malloc(count * sizeof(char *));
*tags = malloc(count * sizeof(int));
for ( size_t i = 0; i < count; i++ )
{
( *segs )[i] = "abc";
}
for ( size_t i = 0; i < count; i++ )
{
( *tags )[i] = ( int )i;
}
return count;
}
int main( void )
{
char **segs = NULL;
int *tags = NULL;
size_t n = Seg( "input", &segs, &tags );
for (size_t i = 0; i < n; i++)
{
printf( "%s ", segs[i] );
}
putchar('\n');
for (size_t i = 0; i < n; i++)
{
printf("%d ", tags[i]);
}
putchar('\n');
free(segs);
free(tags);
return 0;
}
You can also add code to the function that checks whether the memory was allocated successfully.
As for your additional question then such a code like for example this
int *tags_ = malloc(n * sizeof(int));
tags = &tags_;
changes local variable tags of the type int ** (function parameters are function local variables) instead of changing the original pointer tags of the type int * passed to the function by reference as an argument.