what could be wrong with: char* param= new char[200];

Question

Hi,

I'm writing a program in C++ and for some reason I'm getting a segmentation error at the following line:

char* param= new char[200];

I've tried different variations and even tried putting before it

int* param= new int;//for no reason

and the same error occurs. What might I have done to cause this problem?

What could possibly cause a simple memory allocation like this to give problems.

I would include the rest of the program, but it's over 1000 lines. But feel free to ask for more info.

Answer 1

Do you do any memory allocating & freeing before this point? If so, you have probably corrupted the heap in some way. Impossible to diagnose further without seeing more code.

Answer 2

I'd say Neil's on the right track - it's probably something you trampled earlier on that's only being caught there.

Have you made sure that:

1. All previous allocations succeeded.
2. You've not written past the end or beginnings of any arrays (there's a plethora of information and tools for bounds checking out there).

[Edit] In response to your comment about having 4GB of RAM, suppose you code effectively does the following:

unsigned int amountToAllocate = 0;
amountToAllocate -= 1;

int* someStorage = new int[amountToAllocate];

someStorage[0] = 5;

...because amountToAllocate is an unsigned int, it will wrap round to it's maximum value (4294967295 if unsigned ints are 32 bit on your hardware). So you'd be trying to allocate 4294967295*4 bytes (again, assuming ints are 32 bit for you)... which is ~4*4GB and will fail and die a nasty death.

Answer 3

This is how you can use it, I've replaced char with int to store some values in for example but works the same with char.

#include <iostream>

using namespace std;

int main()
{
    int * param = new int[200];

    for (int i = 0; i < 200; ++i) param[i] = i;
    for (int i = 0; i < 200; ++i) cout << param[i] << endl;

    delete[] param;

    return 0;
}

Answer 4

First, verify that this line is the problem. What happens if you comment it out? (or replace it with char *param = NULL)

Second, the only way this can cause a segfault is if something has already gone wrong previously. What happens before this?

If you can't show us the relevant snippets of code, then that is your problem. Start commenting out bits of code or otherwise isolate potential troublespots. Sooner or later, you'll be able to boil the problem down to a small sample that reproduces the error. At that point, the problem (and solution) may be obvious, and if it isn't, at least you've got a reasonably sized code sample to show us.

Answer 5

I second what @jalf is saying. Are you REALLY sure this is the line that is seg faulting? It's very unlikely.

A seg fault happens when you're trying to access memory that you don't have permission to. Since you're not asking for any specific memory but rather letting the "new" memory allocation get it for you, the problem must be elsewhere.

What you should really do is put a bunch of print statements in your code. (I'm assuming you're not comfortable using a debugger?) This will let you see how far the program gets before the seg fault happens. Good luck!