printf("....%d",sizeof((int)(double)(char) i))

[aarklon]

Hi all,

why printf("....%d",sizeof((int)(double)(char) i)) always gives the
size of int ???

is it because sizeof doesn't evaluate its operand....???

 

[Joachim Schmitz]

Hi all,

why printf("....%d",sizeof((int)(double)(char) i)) always gives the
size of int ???

is it because sizeof doesn't evaluate its operand....???

It is because your code is just the longer version of
printf("....%d",sizeof(int))

Bye, Jojo

 

[Ivar]

It is because your code is just the longer version of
printf("....%d",sizeof(int))

Bye, Jojo

By doing printf("....%d",sizeof((int)(double)(char) i)), you're
finally casting i to an int, so it would print the sizeof(int) value.
Correct me if I'm wrong

 

[Peter Nilsson]

why printf("....%d",sizeof((int)(double)(char) i)) always gives
the size of int ???

is it because sizeof doesn't evaluate its operand....???

With the exception of variable length arrays in C99, yes.

The object i doesn't even have to be initialised. Even the
following is valid...

sizeof( (int) (int (*)(void)) -1 )

....even though there is no well defined conversion involved
at any stage.

 

[Joachim Schmitz]

By doing printf("....%d",sizeof((int)(double)(char) i)), you're
finally casting i to an int, so it would print the sizeof(int) value.

Exactly, and that's what I said...

Correct me if I'm wrong

No need

Bye, Jojo