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Home/ Questions/Q 329769
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Editorial Team
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Editorial Team
Asked: May 12, 20262026-05-12T09:37:08+00:00 2026-05-12T09:37:08+00:00

The GNU C++ (g++ -pedantic -Wall) accepts this: typedef int MyInt; class Test {

  • 0

The GNU C++ (g++ -pedantic -Wall) accepts this:

typedef int MyInt;

class Test
{
public:
    MyInt foo();
    void bar(MyInt baz);
}; 

int Test::foo()
{
    return 10;
}

void Test::bar(int baz)
{
}

int main(void)
{
    Test t;
    t.bar(t.foo());
    return 0;
}

Is it legal C++? Are other compilers likely to accept it?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-12T09:37:09+00:00Added an answer on May 12, 2026 at 9:37 am

    Yes it is legal:

    7.1.3 The typedef specifier

    A name declared with the typedef
    specifier becomes a typedef-name.
    Within the scope of its declaration, a
    typedef-name is syntactically
    equivalent to a keyword and names the
    type associated with the identifier in
    the way described in clause 8. A
    typedef-name is thus a synonym for
    another type. A typedef-name does not
    introduce a new type the way a class
    declaration (9.1) or enum declaration
    does.

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