Dynamic array and functionalize malloc [duplicate] - c

Question

This question already has an answer here:
Dynamic memory access only works inside function
(1 answer)
Closed 6 years ago.
I try to malloc an array functionally, and this function also can check whether the memory is enough or not.
The first code block can't work.
When it executes " *pi = 5 ", the compiler shows this error message " Thread 1: EXC_BAD_ACCESS(code=1, address=0x0) " .
#include <stdio.h>
#include <stdlib.h>
void malloc_and_check(int *var)
{
if(!(var = (int*)malloc(sizeof(int))))
{
fprintf(stderr, "Insufficient Memory\n");
exit(EXIT_FAILURE);
}
}
int main(int argc, const char * argv[])
{
int *pi;
malloc_and_check(pi);
*pi = 5;
printf("%d", *pi);
free(pi);
}
However, the codes as below can work normally.
#include <stdio.h>
#include <stdlib.h>
int main(int argc, const char * argv[])
{
int *pi;
if(!(pi = (int*)malloc(sizeof(int))))
{
fprintf(stderr, "Insufficient Memory\n");
exit(EXIT_FAILURE);
}
*pi = 5;
printf("%d", *pi);
free(pi);
}
What are the difference between them? Thanks!

Answer 1

When you pass the int * var into malloc_and_check you are updating a local copy of the pointer so the original is unchanged. You can fix this in two ways:
(a) return the pointer from malloc_and_check
int * malloc_and_check()
{
int * var = 0;
if(!(var = (int*)malloc(sizeof(int))))
{
fprintf(stderr, "Insufficient Memory\n");
return 0;
}
return var;
}
Then you would call it as
pi = malloc_and_check();
(b) pass in a pointer to the pointer
void malloc_and_check(int ** var)
{
if(!( *var = (int*)malloc(sizeof(int))))
{
fprintf(stderr, "Insufficient Memory\n");
exit(EXIT_FAILURE);
}
}
Then you would pass it in as
malloc_and_check(&pi);

Answer 2

The pointer var is allocated memory dynamically but it is local to the function malloc_and_check. The pointer pi in the function main will not get to know the address of the memory allocated to var. It is a bad access and may also result in crash sometimes.
try the code below to get the expected results :
#include <stdio.h>
#include <stdlib.h>
void malloc_and_check(int **var)
{
int *var1;
if(!(var1 = (int*)malloc(sizeof(int))))
{
fprintf(stderr, "Insufficient Memory\n");
exit(EXIT_FAILURE);
}
else
*var = var1;
}
int main(int argc, const char * argv[])
{
int *pi;
malloc_and_check(&pi);
*pi = 5;
printf("%d", *pi);
free(pi);
}

Answer 3

You are trying to pass the pointer by value (pass by value). You need to pass a double pointer to the function malloc_and_check. Due to this the value assigned to the pointer is not reflecting back in the caller.
Please modify your code as below:
#include <stdio.h>
#include <stdlib.h>
void malloc_and_check(int **var)
{
if(!(*var = malloc(sizeof(int))))
{
fprintf(stderr, "Insufficient Memory\n");
exit(EXIT_FAILURE);
}
}
int main(int argc, const char * argv[])
{
int *pi;
malloc_and_check(&pi);
*pi = 5;
printf("%d", *pi);
free(pi);
}

Answer 4

The change you make to var in the function is lost when the function returns. Either you need to pass var as a double pointer or you need to return a pointer from the function.
Example where the function returns a pointer:
#include <stdio.h>
#include <stdlib.h>
int* malloc_and_check()
{
int *var;
if(!(var = malloc(sizeof(int))))
{
fprintf(stderr, "Insufficient Memory\n");
exit(EXIT_FAILURE);
}
return var;
}
int main(int argc, const char * argv[])
{
int *pi;
pi = malloc_and_check();
*pi = 5;
printf("%d", *pi);
free(pi);
}