Passing a char pointer array to a function

Question

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.

I sure hope this is just a more or less academic question and that you are using std::vector<std::string> et al in production code... :) — Christian Hackl
– Christian Hackl, Commented Jul 6, 2014 at 11:40

Some programmer dude · Accepted Answer · 2014-07-06 08:13:53Z

8

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] = ...

edited Jul 6, 2014 at 8:13

answered Jul 6, 2014 at 7:56

Some programmer dude

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2 Comments

SaurabhS Over a year ago

#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], "Saurab"); strcpy((*a)[1],"srivastav"); } int main () { char* b = NULL; f(&b); cout<<b[0]<<"\n"; cout<<b[1]<<"\n"; return 1; } Now its working.

M.M Over a year ago

Actually *a[0] works because it is correct! *a[0] and (*a)[0] are both a[0][0] - it makes no difference to swap the zeroes around. Cf. *a[1] == a[1][0] but (*a)[1] == a[0][1].

Dr. Sahib · Accepted Answer · 2014-07-07 16:31:08Z

0

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.

edited Jul 7, 2014 at 16:31

answered Jul 7, 2014 at 16:24

Dr. Sahib

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