CODEWITHSUNDEEP
c
Bernaldo Hills
Bernaldo Hills

Reputation: 33

Variable declaration not properly checked by compiler

When declaring a variable or pointer, the compiler assumes the variable or pointer itself is already declared when being assigned as a value during declaration.

I have tried both gcc and clang and they compile the "faulty" code without complaining.

CASE 1: This will not compile since "a" is not declared:

void main()
{
int b=sizeof(a);
}

CASE 2: This compiles without a problem:

void main()
{
int a=sizeof(a);
}

Shouldn't the compiler generate the "a is undeclared" error instead, just like in case 1?

Upvotes: 2

Views: 69

Answers (2)

StoryTeller - Unslander Monica
StoryTeller - Unslander Monica

Reputation: 170044

Shouldn't the compiler generate the "a is undeclared" error instead, just like in case 1?

Why? It just saw you declare a.

int a = sizeof(a);
 // ^--- here it is, before its first use

The declaration of a variable begins after its declarator is seen, right before its (optional) initializer. So you can even write the truly faulty

int a = a;

Note however that in your case there is nothing faulty being done. The result of sizeof depends only on the type of a, and the type is known. This is a well defined initializtion (with a conversion from size_t to int, but not one to be worried about).

Upvotes: 4

Tordek
Tordek

Reputation: 10872

sizeof is not a function depending on the value of a; it is a builtin that is evaluated at compile time, so it becomes equivalent to

int a = 4;

Upvotes: 4

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