Passing NULL pointer to pointer and expect allocating for main - c

Question

I was checking pointer to pointer, passing it without allocating. I want to allocate a[0]="something str" a[1]="some thing str" a[2]="something str" a[3]="something str" in pp function. Can I do this (allocating and filling with strcpy in pp function) and get it returned back to main?
This is my attempt:
#include <stdio.h>
#include <malloc.h>
#include <string.h>
#define MAX_LEN 100
// float 3
void pp(char *arr, char *delimiter, char **a)
{
int i = 0;
*a = malloc(sizeof(char) * 10);
}
int main(int argc, void **argv)
{
char *arr = "1.77 1.65 1.56 5.555 6.1";
char **f;
pp(arr, " ", &f[0]);
}
I thought I could just allocate individual char * and then populate as strcpy(*(a+x),"something") but it causes segfaults at *a = malloc(sizeof(char) * 10);.

Answer 1

You have multiple bugs in your code.
You do not initialized f but access f[0].
You allocate memory for 10 single char but not for pointers in your function.
Also the overall approach is broken.
You could try like this:
#include <stdio.h>
#include <malloc.h>
#include <string.h>
#define MAX_LEN 100
void pp(char *arr,char *delimiter,char ***a)
{
// TODO: Handle NULL pointers.
int i=0;
// TODO: Calculate number of strings using arr and delimiter...
int num = 10;
*a=malloc((num+1) * sizeof(**a));
for (int k = 0; k < num; k++)
{
(*a)[k] = malloc( 1+ length of string to copy) ;
strcpy((*a)[k], <string to copy>);
}
(*a)[num] = NULL; // indicate end of array.
}
int main(int argc,void **argv)
{
char *arr="1.77 1.65 1.56 5.555 6.1";
char **f;
pp(arr, " ", &f);
int i = 0;
while (f[i] != NULL)
{
printf("string #%d: %s\n", i, f[i]);
i++;
}
}
You should also think about a way how to report the number of found substrings back to the caller. In the example I added an extra pointer holding NULL to terminate the array.

Answer 2

For starters the header <malloc.h> is not a standard C header. Instead use header <stdlib.h>.
Aa for your task
yes that what I will eventually do. Basically array that holds float
then you can use the approach shown in the demonstration program below
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
size_t split( const char *s, const char *delim, double **a )
{
*a = NULL;
size_t n = 0;
for ( const char *p = s; p += strspn( p, delim ), *p != '\0'; )
{
++n;
p += strcspn( p, delim );
}
if ( n != 0 && ( *a = malloc( n * sizeof( double ) ) ) != NULL )
{
char *endptr;
for ( size_t i = 0; i < n; i++ )
{
( *a )[i] = strtod( s, &endptr );
s = endptr;
}
}
return n;
}
int main( void )
{
char *s= "1.77 1.65 1.56 5.555 6.1";
double *a;
size_t n = split( s, " \t", &a );
if ( a != NULL )
{
for ( size_t i = 0; i < n; i++ )
{
printf( "%.3f ", a[i] );
}
putchar( '\n' );
}
free( a );
}
The program output is
1.770 1.650 1.560 5.555 6.100
Actually the function can be more complicated because you will need to check that conversion to double were suvvessful.