What rules of C++11 standard are used to determine the type of the expression in ({ … })

Question

I hasn't understand what compiler does here and why it's working c++ code

#include <cstdio>
int main()
{
    printf( ({  // (1)
                struct some_noize_struct {
                   // there may be another code
                };
                42;
                "abc";
              }) // (2)
            );

    return 0;
}

Type of expression between (1) and (2) braces is const char*. After some experimens i unrerstood that type of (1)-(2)-expression determined by last part.

Here is a sample code. It works on c++11 and later. http://cpp.sh/5tb47

My question: how it works.

Answer 1

As @HolyBlackCat explains, the code you listed uses a GCC compiler extension to C++ (and to C), while allows for compounds statements to be used as expressions.

In your printf() statement, you need to provide a const char* or const char* & expression as the first argument to the function, eg printf("hello") or printf(getenv("PATH")) . The extension allows the interpretation of a curly-braced block as such an expression, using the last statement in the block. The block in your case is:

{ 
    struct some_noize_struct { 42 };
    42;
    "abc";
}

which has 3 statements. The last statement is the value of the entire expression, which means that what the printf() sees is essentially the same as if you had typed printf("abc") .

This kind of code is not standard C++ (C++11 or any another version), nor is it standard C .

I suggest you write the maintainers of the "C++ Shell" website and ask them to display the exact compilation command-line they use, and/or make sure they use --std=c++11 to compile C++11 code - which it looks like they aren't doing.