Using sizeof a struct in its constructor

Question

I am trying to initialize a struct in its constructor but I run into a compiler error on the last line when I am initializing the member header.length. Is its size known at the time?

This is the compiler error and the structure:

In constructor ‘stDescriptor::stDescriptor()’:
error: expected primary-expression before ‘)’ token


struct stDescriptor {
    usb_functionfs_descs_head   header;
    stDescriptorBody fs_descs;
    stDescriptorBody hs_descs;

    stDescriptor(){
        header.fs_count = 3;
        header.magic = cpu_to_le32(FUNCTIONFS_DESCRIPTORS_MAGIC);
        header.hs_count = 3;
            header.length = cpu_to_le32(sizeof stDescriptor);
         }
};

score 4 · Answer 1 · edited Jun 20 '20 at 09:12

4

You need

header.length = cpu_to_le32(sizeof(stDescriptor));

because stDescriptor is a type name, not an expression.

5.3.3 Sizeof [expr.sizeof]

1 The sizeof operator yields the number of bytes in the object representation of its operand. The operand is either an expression, which is an unevaluated operand (Clause 5), or a parenthesized type-id. [...]

edited Jun 20 '20 at 09:12

Community

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answered Jan 28 '14 at 16:20

Luchian Grigore

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Question aside: is sizeof(expression) a sizeof expression ? – Jan 28 '14 at 16:26
@DieterLücking if I understand you correctly, yes, this [is a C question](http://stackoverflow.com/questions/18898736/what-does-sizeof-without-do/18898832#18898832) but the logic is the same. – Shafik Yaghmour Jan 28 '14 at 16:35

Using sizeof a struct in its constructor

1 Answers1

5.3.3 Sizeof [expr.sizeof]