'+' cannot add two pointers, but just printing an int and an explicit string?

Question

I am trying to use an array to keep track of the totals of different types of items (up to 50 types). When I want to print the totals out, I get an error saying "'+' cannot add two pointers." I'm thinking the problem is with my totals array somehow, but I can't figure it out. Below is a sample of my code:

  string printSolution()
  {
  int totals[50];

  string printableSolution = "";

  for (int k = 0; k < itemTypeCount; k++)
  {
      totals[k] = 0;
  }

  for (int i = 0; i < itemCount; i++)
  {
      totals[items[i].typeCode]++;
  }

  for (int a = 0; a < itemTypeCount; a++)
  {
      printableSolution.append("There are " + totals[a] + " of Item type " + (a + 1) + ". \n");
  }


}

Wrap the `"There are"` with a `std::string` or use a `std::stringstream`. Right now you're adding `char*` with `int`. — Benjamin Gruenbaum, Jun 19 '14 at 00:46
When you're working with C-style strings, you have to use `strcat()` to concatenate them. `+` only works with `std::string`. — Barmar, Jun 19 '14 at 00:47
possible duplicate of [Error: Cannot add two pointers](http://stackoverflow.com/questions/3108965/error-cannot-add-two-pointers) and http://stackoverflow.com/q/16324599/103167 and http://stackoverflow.com/q/18580262/103167 and http://stackoverflow.com/q/2844986/103167 — Ben Voigt, Jun 19 '14 at 02:03

score 1 · Answer 1 · answered Jun 19 '14 at 01:51

The string literals "Foo" are of const char*, i.e. pointer type.

To understand what happens with:

"There are " + totals[a] + " of Item type " + (a + 1) + ". \n"

Let's look at an expression:

"0123456789" + 5

This actually just offsets 5 bytes from the start, so becomes:

"56789"

So an expression:

"0123456789" + 5 + "foo"

becomes:

"56789" + "foo"

as pointers, and this is not defined.

What you really want is string concatenation; this can be achieved using std::string.

We can write:

   std::string("56789") + "foo"

and this generates a std::string with value: "56789foo" as you desire.

But:

 std::string("0123456789") + 5

is also not defined. You need to use:

std::string("0123456789") + std::to_string(5)

So, finally you want:

std::string("There are ") + std::to_string(totals[a]) + " of Item type " + std::to_string(a + 1) + ". \n"

Note now you do not need to explitly convert all the "" to std:string, as once you have one implicit type conversion will take care of the other operand in operator+. However, adding them would do no harm:

std::string("There are ") + std::to_string(totals[a]) + std::string(" of Item type ") + std::to_string(a + 1) + std::string(". \n")

score 0 · Accepted Answer · edited Jun 19 '14 at 01:40

0

The problem is here:

"There are " + totals[a] + " of Item type " + (a + 1) + ". \n"

It means char* + int + char* + int + char*. You need to print them out separately or change the int to a std::string.

edited Jun 19 '14 at 01:40

Remy Lebeau

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answered Jun 19 '14 at 01:32

Fanl

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score 0 · Answer 3 · answered Jun 19 '14 at 01:37

0

Use C++-style formatting instead:

std::ostringstream oss;
oss << "There are " << totals[a] << " of Item type " << (a + 1) << ". \n";
printableSolution += oss.str();

answered Jun 19 '14 at 01:37

Remy Lebeau

555,201
31
458
770

'+' cannot add two pointers, but just printing an int and an explicit string?

3 Answers3