2

My code:

#include "pch.h"
#include <iostream>

using namespace std;

   int main()
   {
      char** pptr = new char*[5];

      for (int i = 0; i < 5; i++)
      pptr[i] = new char[5];
   }

What I want to happen is that pptr now points to the beginning of an array of 5 pointers that each point to the beginning of an array of 5 characters.

I put a breakpoint at the end of the main function and added pptr to watch, and it only stores one pointer. Why does this happen and how do I do it correctly?

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2
  • 1
    You said you wanted pptr to point to the beginning of an array of pointers... the beginning of an array is indeed one pointer - the first one in the array. The other pointers come later in the same array. Commented Jan 4, 2019 at 14:39
  • 1
    Please next time reduce the picture to its interesting part to not have that large white margin ;-) Commented Jan 4, 2019 at 14:44

3 Answers 3

Reset to default
6

This is the default knowledge of your pointer type in Visual Studio. You indicate in the code that char** pptr is a pointer, but it cannot know how big.

To fix this, you can add a watch on pptr[0], and then you can specify that it has a "size" of 5 by changing it to pptr[0],5. Also, if the size is variable you can do "ptr[0],[size]" where size is an expression that evaluates to the number of elements to show.

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6 Comments

That's an interesting solution to an annoying problem, I'll try it out in the future. Though I'm surprised, since pptr[0],5 has a distinct meaning in c++. It seems inconsistent for the debugger to arbitrarily interpret certain operators differently when it treats most operators the same way the language does.
Yes, the , is specific to the debugger watch variable. I'm not sure it works with a comma operator as the expression. Probably triggeres a bunch of errors somewhere!
yes is strange, is it not possible to change the type of the element to display to be char*[5] being c++ as I proposed ? (I do not use Visual Studio)
No, it only displays the type as VS understands it. But then, you can always have a reinterprect_cast in the expression as well (I think).
Yes, you should be able to dereference the pointer and cast it to a reference to an array of 5 char* which I think is written char*(&)[5]. Edit : The watch expression (char*(&)[5])(*pptr) will give you pptr as an array of 5 char*. Edit 2 : Though pptr,5 is still a way nicer way of doing it.
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2

Your program does what you want, but the debugger cannot know the number of elements, it just know it is a pointer, so it write the contains of that pointer.

I don't know what debugger you use, but probably when you display the values you can modify char** by char*[5]` to see all

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Comments

-2 
// #include "pch.h"
#include <iostream>

using namespace std;

int main(){
    char** pptr = new char*[5];

    for (int i = 0; i < 5; i++)
        pptr[i] = new char[5];

    for(int i=0;i<5;i++){
        char ch='A';
        pptr[0][i]=ch;
    }

    for(int i=0;i<5;i++){
        cout<<pptr[0][i]<<" ";
    }
}

Now the pptr[0] that is a pointer is pointing to an array of characters. Hope that it helps.

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1 Comment

the problem doesn't concern the code but what the debugger shows by default

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