Malloc of pointer to an array- C++

Question

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.

Please try to create an minimal reproducible example to show us of your attempt. — Some programmer dude
– Some programmer dude, Commented Dec 10, 2018 at 20:22
Did you really mean C++? Because this sounds like C code. There is zero reason to write code like this in C++. — MSalters
– MSalters, Commented Dec 10, 2018 at 20:23
I put a value into *segs[0] - be care of operator precedence. Either disambiguate, e.g (*segs)[0] = ... or use a local var and save to your target ptr-to-ptr after everything else is done. — WhozCraig
– WhozCraig, Commented Dec 10, 2018 at 20:23

R Sahu · Accepted Answer · 2018-12-10 20:50:04Z

4

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

Pass the pointer by reference.
```
void foo(int*& segs) { ... }
```
Use new instead of malloc to allocate memory.
```
segs = new int[100];
```
Better yet, use std::vector insteady raw arrays.
```
void foo(std::vector<int>& segs) { ... }
```

edited Dec 10, 2018 at 20:50

answered Dec 10, 2018 at 20:34

R Sahu

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