CODEWITHSUNDEEP
c++cpointers
SaurabhS
SaurabhS

Reputation: 663

Passing a char pointer array to a function

I have written following sample code to demonstrate my problem

#include <iostream>
#include <string.h>

using namespace std;

void f (char*** a)
{
    *a = new (char*[2]);
    *a[0] = new char[10];
    *a[1] = new char[10];
    strcpy(*a[0], "abcd");
    strcpy(*a[1],"efgh");
}

int main ()
{
    char** b = NULL;
    f(&b);
    cout<<b[0]<<"\n";
    cout<<b[1]<<"\n";
    return 1;
}

In this code i found that a = new (char[2]); not allocating memory of *a[1].

In gdb i am getting following seg fault. 

Program received signal SIGSEGV, Segmentation fault.
0x0000000000400926 in f (a=0x7fffffffdfe8) at array1.cpp:10
10          *a[1] = new char[10];
(gdb) p *a[1]
Cannot access memory at address 0x0

This is really confusing me. Can someone explain where i am going wrong.

I know i can pass argument like void f (char**& a) and by calling function f(b) and this works . But i want to know want happening if i use char*** a. It should also work. Sorry if this is an stupid question.
Any tutorial or reference on above problem will be appreciated.
Thanks in advance.

Upvotes: 3

Views: 371

Answers (2)

Dr. Sahib
Dr. Sahib

Reputation: 166

a[0], a[1] are char, and they have a value in them, dereferencing that value will obviously cause a segmentation fault. What you may want to do is:

(*a)[...]

dereference 'a' which is a pointer, this will give u an array.

Upvotes: 0

Some programmer dude
Some programmer dude

Reputation: 409442

It's because of the operator precedence, where the array-indexing operator [] have higher precedence than the dereference operator *.

So the expression *a[0] is really, from the compilers point of view, the same as *(a[0]), which is not what you want.

You have to explicitly add parentheses to change the precedence:

(*a)[0] = ...

Upvotes: 8

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