Using the sizeof() function correctly

Question

Why is this syntax not correct?

 char* p = new char[dg.sizeof(Payload())];

I want to make a new variable that is a string the size of the Payload portion of dg... I don't understand how to do this? Thank you

 error: expected unqualified-id before 'sizeof'
 error: expected `]' before 'sizeof'
 error: expected ',' or ';' before 'sizeof'

unless dg has a sizeof function of its own, you are using it incorrectly — RisingSun, Aug 27 '15 at 17:22
Is `Payload` actually a function? We have just been told it's a "portion" — doctorlove, Aug 27 '15 at 17:27
What are you trying to do? That's incorrect syntax. sizeof() returns a number, say, 8. dg.8 doesn't makes sense anyway — EvgeniyZh, Aug 27 '15 at 17:31
why is this tagged C? there's no `new` operator in C, and no methods in C either — phuclv, Aug 27 '15 at 17:32
the 'sizeof' is a compile time operator, not a runtime operator. so any activity at runtime is totally unrelated to the 'sizeof' operator. The syntax is not correct for using the 'new' operator. This 'may' be what your looking for: 'char* p = new char[strlen(dg.Payload)];' — user3629249, Aug 27 '15 at 17:42
@user3629249: "*'sizeof' is a compile time operator*" this is not exactly true, since C99 VLAs are around which be nature reveal their size on run-time. `sizeof` can by applied to a VLA. — alk, Aug 27 '15 at 17:49

score 4 · Answer 1 · answered Aug 27 '15 at 17:26

4

sizeof is not a function, it's an operator like + or *, but it operates at compile time. It takes one operand, either a typename in parens, or an expression (which need not be in parens). So if Payload is a member of structure dg, sizeof dg.Payload is the size of the function pointer member, and sizeof dg.Payload() is the size of Payload()'s return type.

answered Aug 27 '15 at 17:26

Lee Daniel Crocker

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[Why is sizeof considered as an operator?](http://stackoverflow.com/q/1393582/995714) – phuclv Aug 27 '15 at 17:31

score 2 · Answer 2 · answered Aug 27 '15 at 17:33

2

sizeof is a c/c++ operator. It cannot be used as a class method. http://en.cppreference.com/w/cpp/language/sizeof

answered Aug 27 '15 at 17:33

Robert Jacobs

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score 1 · Answer 3 · edited Aug 27 '15 at 20:01

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Change

char* p = new char[dg.sizeof(Payload())];

to

char* p = new char[sizeof(dg.Payload())];

edited Aug 27 '15 at 20:01

huysentruitw

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answered Aug 27 '15 at 17:19

JHobern

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I doubt `sizeof` is the right tool at all though. Looks like a function trying to return an array. – Quentin Aug 27 '15 at 17:24
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Note that this will return the sizeof() the return type of Payload, which may or may not be what the OP wants. OP seems to have a fundamental misunderstanding of what sizeof is for. – JPhi1618 Aug 27 '15 at 17:24
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We don't have enough context yet, but I'm already sure this answer is wrong ;) – huysentruitw Aug 27 '15 at 17:25
@WouterHuysentruit: Following up the "missing context" you mention the answer cannot be wrong, nor right ... ;-) - I however agree on that it might not solve the OP's issue ... – alk Aug 27 '15 at 17:44

Using the sizeof() function correctly

3 Answers3