CODEWITHSUNDEEP
c++deque
manuzhang
manuzhang

Reputation: 3025

deque push back error

I am writing a compiler and use deque to store methods labels of a class and here's the sample code:

#include <deque>
#include <iostream>
#include <string>

using std::cout;
using std::deque;
using std::endl;
using std::string;

int main()
{
  deque<const char *> names;

  string prefix = "___";
  const char *classname = "Point";

  const char *methodname[] = {"Init", "PrintBoth", "PrintSelf", "equals"};

  for (int i = 0; i < 4; i++)
    {
      string label = prefix + classname + "." + methodname[i];
      names.push_back(label.c_str());
    }

  for (int i = 0; i < 4; i++)
    cout << names[i] << endl;

  return 0;
}

However, the result is not what I've expected:

___Point
___Point.PrintSelf
___Point.PrintSelf
___Point.equals

Also, I noticed if I simply push back the methodname

names.push_back(methodname[i])

I get all the methodnames in order.

What have I done wrong here?

Upvotes: 3

Views: 2049

Answers (2)

Sarfaraz Nawaz
Sarfaraz Nawaz

Reputation: 361442

for (int i = 0; i < 4; i++)
{
  string label = prefix + classname + "." + methodname[i];
  names.push_back(label.c_str()); //what you're pushing? a temporary!

} //<--- `label` is destroyed here and it's memory is freed.

Here label is a variable which gets destroyed at the closing brace and gets created again, in each iteration.

That means, what you're pushing to names is a temporary value. That is causing the problem.

I would suggest you to use this:

std::deque<std::string> names;

then do this:

names.push_back(label); //a copy is pushed to the deque!

Upvotes: 9

Israel Unterman
Israel Unterman

Reputation: 13510

this is because the string label is a temporary, and its chhar pointer isn't valid once it exit the scope - the for loop, in this case.

I suggest to use deque<string> instead. this way you can push label itself, and then a real copy of label will be created inside the deque.

Upvotes: 0

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