malloc problem?

Hello I use malloc() to generate a buffer like this where buffer is a char*

buffer = (char*)malloc(chu开发者_运维技巧nksize+1);
  for (k = 0; k < chunksize; k++) {
    buffer[k] = (char) (j+k);
  }

however, in debugger i can see the buffer[3] for example is the char i wrote in, but the buffer buffer is empty(a lot of spaces). But the second time I write stuff in the buffer after free(buffer), it shows the the content I wrote the first time, and overwrite it. Can anyone tell me why? Thank you!!

Generally speaking, malloc implementations work by getting a large chunk of memory from the OS, then sub-allocating bits of it when you ask for buffers. This is because getting memory from the OS is relatively slow.

As a security measure, your OS will give malloc blank memory (full of 0 or some other repeating value). You wouldn't want one process to be able to read the leavings of an unrelated process.

However, malloc has no such problems, because it only serves from that chunk to one process. It doesn't matter if you can see your own leavings (at least not from a security POV).

So what's happened is that you've malloced a buffer, and seen it initially blank. Then you've freed it and asked for "another" buffer, and malloc happens to have handed you the same memory again, this time containing the values you left in it.

This "coincidence" of getting the same memory twice isn't very unlikely, since there's a good reason for malloc to re-use recently-used memory where possible - it's more likely to be in cache.

You might think this doesn't matter to you, because normally you wouldn't read from newly-allocated memory. But it matters to malloc, because typically malloc keeps its own house-keeping data adjacent to the buffer. It also may matter if there's a cost associated with expelling memory from the cache - if you can be "tricked" into re-using the same memory over and over again, then this happens less.

One problem might be that you try to print your char buffer with printf or equivalent? You are missing to assign

buffer[chunksize] = 0;

So your buffer is not well terminated. It may have anything behind that, e.g '\r'.

If on the other hand buffer doesn't represent a string for you, better use a different base type, probably unsigned char.

Your code is correct. The problem could be in some part of the code that you haven't shown if you are concerned that you overwrite some bad memory. For example, if buffer is a pointer to non-char array, in which case pointer arithmetic that you use will result in bad memory access (corruption). You can use the following program as a reference:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main ()
{
  const size_t chunksize = 15;
  size_t k;
  char *buffer;

  buffer = malloc (chunksize + 1);
  memset (buffer, 0, chunksize + 1);

  for (k = 0; k < chunksize; ++k)
    {
      buffer[k] = '0' + (char)k + 49;
    }

  for (k = 0; k < chunksize; ++k)
    {
      printf ("%lu = %c\n", k, buffer[k]);
    }

  free (buffer);

  return 0;
}

Output:

0 = a
1 = b
2 = c
3 = d
4 = e
5 = f
6 = g
7 = h
8 = i
9 = j
10 = k
11 = l
12 = m
13 = n
14 = o