Strange outputs from appended char* in C++

Question

I was just writing simple code and then I saw something strange. The code is supposed to append a string to another string. The output from the new appended string outputs not only the correct appended string, but it also adds every time four time the number 2 and I don't understand why. I thought it was some memory leak I overlooked or something like that, but there it outputs every time the same four numbers.

Code:

#include <iostream>

using namespace std;


unsigned int getStrSize(const char* string)
{
    unsigned int size = 0;
    while (string[size] != '\0')
    {
        size++;
    }
    return size;
}

int main()
{
    const char* bla1 = "hello";
    const char* bla2 = " blaah";

    int size1 = getStrSize(bla1);
    int size2 = getStrSize(bla2);
    int size12 = size1 + size2;

    char* bla12 = new char[size12];

    for (int i = 0; i < size1; i++)
    {
        bla12[i] = bla1[i];
    }
    for (int i = 0; i < size2; i++)
    {
        bla12[i + size1] = bla2[i];
    }
    char* blaNew = bla12;

    cout << bla1 << "\n";
    cout << bla2 << "\n";
    cout << bla12 << "\n";
    cout << blaNew << "\n";
}

Outputs:

hello
 blaah
hello blaah²²²²
hello blaah²²²²

Aykhan Hagverdili · Accepted Answer · 2020-01-05T13:36:53.830

You're missing the nil:

#include <iostream>

using namespace std;
unsigned int getStrSize(const char* string)
{
    unsigned int size = 0;
    while (string[size] != '\0')
    {
        size++;
    }
    return size;
}

int main()
{
    const char* bla1 = "hello";
    const char* bla2 = " blaah";

    int size1 = getStrSize(bla1);
    int size2 = getStrSize(bla2);
    int size12 = size1 + size2 + 1; // notice +1

    char* bla12 = new char[size12];

    for (int i = 0; i < size1; i++)
    {
        bla12[i] = bla1[i];
    }
    for (int i = 0; i < size2; i++)
    {
        bla12[i + size1] = bla2[i];
    }
    bla12[size12 - 1] = '\0'; // terminate with nil
    char* blaNew = bla12;

    cout << bla1 << "\n";
    cout << bla2 << "\n";
    cout << bla12 << "\n";
    cout << blaNew << "\n";

    delete[] bla12;  // Don't leak memory
    delete[] blaNew; // 
}

Better yet consider using standard library functions:

#include <cstring>
#include <iostream>

int main() {
  const char* bla1 = "hello";
  const char* bla2 = " blaah";

  auto const size1 = std::strlen(bla1);
  auto const size2 = std::strlen(bla2);
  auto const size12 = size1 + size2 + 1;

  char* bla12 = new char[size12];

  std::strcpy(bla12, bla1);
  std::strcat(bla12, bla2);
  char* blaNew = bla12;

  std::cout << bla1 << "\n";
  std::cout << bla2 << "\n";
  std::cout << bla12 << "\n";
  std::cout << blaNew << "\n";

  delete[] bla12;
  delete[] blaNew;
}

Nah I want to build my own string with no dependences from any standard library functions. — larryStacker, Jan 05 '20 at 13:31

score 1 · Answer 2 · answered Jan 05 '20 at 13:20

The problem is that you don't null terminated your char* buffer. std::cout.operator<<(char*) will try to find \0 as its terminating character. So you just need to append \0 at the end of your buffer.

Note: Pay attention to increase the size of the buffer of 1 or you will access/write a not allocated memory.

Strange outputs from appended char* in C++

2 Answers2