C++ Heap corruption after simple action

Question

I can't get why simple actions with my array corrupt the heap.

For example, this code works fine:

double *matrix = (double*)(malloc(50));

free(matrix);

Then, I try to modify some of the elements of this array like this:

double *matrix = (double*)(malloc(50));

for (int i = 0; i < 20; matrix++, i++) {
    *matrix = (double)i;
}

free(matrix);

The problem is that even this code is not working at all. (VS tells me that the heap was corrupted). Probably it is a stupid mistake, but I am new to all this actions with pointers.

UPDATE

Now my code looks like this:

double *matrix = (double*)malloc(sizeof(double) * 50);

But I got this error message:

correct, and you should check the return code from malloc also — amdixon, Nov 22 '15 at 06:11
you only need the (double*) cast because c++ cant handle the types properly. see **[why c++ requires malloc cast](http://stackoverflow.com/questions/3477741/why-does-c-require-a-cast-for-malloc-but-c-doesnt)** — amdixon, Nov 22 '15 at 06:15

score 2 · Answer 1 · answered Nov 22 '15 at 06:11

2

What your malloc do is allocated 50 bytes, not 50 double. Here are example of how to use malloc:

myVar = (double*)malloc (sizeof(double));

or if you want to use multiple (array) double then you can do it like this:

myVar = (double*)malloc (sizeof(double) * 50);

answered Nov 22 '15 at 06:11

Hernantas

341
1
6

Thank you, but i got another error after changing the code. – CMTV Nov 22 '15 at 06:20

score 2 · Accepted Answer · answered Nov 22 '15 at 06:21

2

Your code modified the pointer "matrix" before you free it, which I believe is an undefined behavior.

No offense, but I suggest you get a good book about c++. "malloc" is not the preferred way for dynamic memory allocation in c++.

Anyway, if you would like to access the memory block as an array, the preferable way to do it is to use operator[]:

for (int i = 0; i < 20; i++) {
    matrix[i] = (double)i;
}

And yes, as @Hernantas mentioned, if you want to allocate a space for 50 double objects, use

myVar = malloc (sizeof(double) * 50);

answered Nov 22 '15 at 06:21

Lucien

134
5

Thank you. I fixed this problem. I am reading "The C programming language". I modified it because, as I understand, it do not change when modified inside other functions. For example, in function doStuff() I can write something like this: matrix+=1000; But in main function matrix still points at zero array element. Did I get it right? – CMTV Nov 22 '15 at 06:28
@CMTV weather the original "matrix" is modified depends on how the argument is passed. If the pointer is pass by value then the change will not affect the original variable. – Lucien Nov 22 '15 at 06:33
Does it mean, that by passing pointer this way: doStuff(int *p) I pass only the value of "p" (and that is why chaning it actually changes nothing in other functions) but *p still is a link to a specific array element? – CMTV Nov 22 '15 at 06:52
@CMTV `doStuff(int *p)` is actually passed-by-value: the value of the pointer itself is copied into the function. If you change the value of p inside doStuff (not changing what it points to), say `p = someOtherPointer`, p will point to a new position while the original pointer will remain unchanged. – Lucien Nov 22 '15 at 06:58

C++ Heap corruption after simple action

2 Answers2