CODEWITHSUNDEEP
c++arrayspointersmalloc
Tamir
Tamir

Reputation: 3

Malloc of pointer to an array- C++

I have a function which gets as a parameter a pointer to array, e.g. int** segs. I need to allocate (in the function-body) memory for the array, which has size 100 for example. My attempt was:

*segs=(int*)(malloc(100));

So far so good. I put a value into *segs[0], and still everything is great. But... when I try to reach *segs[1], I get an "invalid write of size 4" error from valgrind, which leads to seg-fault. I have no idea why does that happen. I tried to reach to *segs[2], but then I get even something weirder-

Error of uninitialised value of size 8.

Upvotes: 0

Views: 1184

Answers (1)

R Sahu
R Sahu

Reputation: 206567

Due to operator precedence, *segs[N] is treated as *(segs[N]), which is ok when N is equal to 0. However, when you do the same thing using the index 1, things break since nothing has been allocated for segs[1]

For any index other than zero, you need to use (*segs)[N].

It will be easier to use a temporary pointer in the function.

int* ptr = (int*)(malloc(100));
*segs = ptr;

// and then

ptr[0] = ...;  // Good
ptr[1] = ...;  // Good

...


ptr[99] = ...; // Still good

Upgrading to C++ Way

  1. Pass the pointer by reference.

    void foo(int*& segs) { ... }

  2. Use new instead of malloc to allocate memory.

    segs = new int[100];

  3. Better yet, use std::vector insteady raw arrays.

    void foo(std::vector<int>& segs) { ... }

Upvotes: 4

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