What does {0} mean when initializing an object?

Question

When `{0}` is used to initialize an object, what does it mean? I can't find any references to `{0}` anywhere, and because of the curly braces Google searches are not helpful. Example code: SHELLEXECUTEINFO sexi = {0}; // what does this do?  sexi.cbSize = sizeof(SHELLEXECUTEINFO);  sexi.hwnd = NULL;  sexi.fMask = SEE_MASK_NOCLOSEPROCESS;  sexi.lpFile = lpFile.c_str();  sexi.lpParameters = args;  sexi.nShow = nShow;  if(ShellExecuteEx(&sexi))  {   DWORD wait = WaitForSingleObject(sexi.hProcess, INFINITE);   if(wait == WAIT_OBJECT_0)    GetExitCodeProcess(sexi.hProcess, &returnCode);  } Without it, the above code will crash on runtime.

Answer 1

Using braces {} in this way is part of Uniform Initialization. Let's take a look using the code below:

#include <iostream>

int main(int argc, const char * argv[]) {

     

  int x {100};

  int y = 100;

  int ar[] {1,2,3,4,5};

  int ar2[] = {1,2,3,4,5};

   

  std::cout << "x: " << x << '\n';

  std::cout << "y: " << y << '\n';

   

  for (int i=0; i < 5; ++i) {

    std::cout << " ar[" << i << "]: " << ar[i] << '\n';

    std::cout << "ar2[" << i << "]: " << ar2[i] << '\n';

  }

  return 0;

}

Output

x: 100

y: 100

 ar[0]: 1

ar2[0]: 1

 ar[1]: 2

ar2[1]: 2

 ar[2]: 3

ar2[2]: 3

 ar[3]: 4

ar2[3]: 4

 ar[4]: 5

ar2[4]: 5

Uniform initialization is a feature in C++ 11 that allows the usage of a consistent syntax to initialize variables and objects ranging from primitive type to aggregates. In other words, it introduces brace-initialization that uses braces ({}) to enclose initializer values.

Objects can be initialized with both the equal sign (=) and braces ({}); however, assignment (not initialization) also uses the equal sign (=), therefore, supporting braced initialization maintains a uniformity for initialization as opposed to assignment.

Morale: Even though braces and equal signs are supported for initialization...

1. For initialization, use {}
2. For assignment, use =