How can pointers be allocated in main function?

Question

I just started learning the programming language C and am having a problem while solving 1 exercise below.

Write a program that dynamically allocates 10 bytes of memory for a pointer, with the provided function template as follows: bool allocate10Bytes(uint8_t *outPtr);

outPtr: output pointer
return: true: If the allocation is successful. false: If the allocation failed.

I tried with the following code but the pointer in main function is still not allocated.

bool allocate10Bytes(uint8_t* outPtr)
{
    outPtr = (uint8_t*)malloc(10 * sizeof(uint8_t));

    if (outPtr == NULL) {
        return false;
    }

    return true;
}

int main(void)
{
    uint8_t* p = NULL;

    bool success = allocate10Bytes(p);
    
    if (success) free(p);
    return 0;
}

Please help me. Thank you very much!

There’s no way to get the value of the pointer to the allocated memory back to `main` with the function signature that’s given. — Pete Becker, Jul 27 '21 at 03:09
[This](https://stackoverflow.com/a/5580790) and [this](https://stackoverflow.com/a/26004257) will help you. — kakkoko, Jul 27 '21 at 03:09
Impossible. `allocate10Bytes` either needs to return a `uint8_t*` or accept a `unit8_t**` as an argument. — yano, Jul 27 '21 at 03:10
[Is this you?](https://stackoverflow.com/questions/68537895/write-a-program-that-dynamically-allocates-10-bytes-of-memory-for-a-pointer) (10K only) — dbush, Jul 27 '21 at 03:16

prehistoricpenguin · Accepted Answer · 2021-07-27T03:47:15.230

6

Your code aimed to change a pointer argument's value since arguments are passed by value, the code have changed the local value outPtr's value, not the one in main

To pass a pointer as an argument and change the pointer's value in callee, we have only one choice in C : to pass a pointer to the pointer


bool allocate10Bytes(uint8_t** outPtr) {
  *outPtr = malloc(10 * sizeof(uint8_t));

  if (*outPtr == NULL) {
    return false;
  }

  return true;
}

int main(void) {
  uint8_t* p = NULL;

  bool success = allocate10Bytes(&p);

  if (success) free(p);
  return 0;
}

For code smells: it's bad to use a magic number 10 in the code, it would be much better to add a macro or constant here.

If the function prototype must be bool allocate10Bytes(uint8_t* outPtr), then we need some ugly casting here. I don't think it's a good exercise problem, maybe your teacher is doing something wrong.

#include <stdbool.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
bool allocate10Bytes(uint8_t* outPtr) {
  uint8_t** outPtr1 = (uint8_t**)(outPtr);
  *outPtr1 = malloc(10 * sizeof(uint8_t));

  if (*outPtr1 == NULL) {
    return false;
  }

  return true;
}

int main(void) {
  uint8_t* p = NULL;

  bool success = allocate10Bytes((uint8_t*)&p);

  if (success) free(p);
  return 0;
}

edited Jul 27 '21 at 03:47

answered Jul 27 '21 at 03:11

prehistoricpenguin

6,130
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2

No need to cast the pointer – smac89 Jul 27 '21 at 03:13
2

`*outPtr = malloc(10 * sizeof(uint8_t));` could be cleaned up a bit. Use the type of `outPtr` in `sizeof` – Ted Lyngmo Jul 27 '21 at 03:15
1

@prehistoricpenguin This exercise does not allow changing the input parameters of the function. The function parameter must be uint8_t* outPtr. – Justin Jul 27 '21 at 03:16
1

@Justin If the function prototype must be `bool allocate10Bytes(uint8_t* outPtr)`, then we need some ugly casting here. I don't think it's a good exercise problem, maybe your teacher is doing something wrong. – prehistoricpenguin Jul 27 '21 at 03:19
1

@prehistoricpenguin I think so, too. Thanks for the great advice! – Justin Jul 27 '21 at 03:22
1

@Justin the requirement is broken, your teacher should be fired. – Jabberwocky Jul 27 '21 at 07:11

How can pointers be allocated in main function?

1 Answers1