what is the output of
void main()
{
printf("%d");
}
11  1635 
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The output will be ZERO
Savage
what is the output of
-
void main()
-
{
-
printf("%d");
-
}
-
Try it and find out!
The output will be ZERO
Savage
Umm, I'm pretty sure that's incorrect. Did you try running it?
Umm, I'm pretty sure that's incorrect. Did you try running it?
Yes,I runed it and it gaved me a 0.
Savage
Banfa 9,065
Expert Mod 8TB
The output could be anything, it invokes undefined behaviour on at least 2 counts.
The output could be anything, it invokes undefined behaviour on at least 2 counts.
Let me guess:
1) void main() instead of int main()
2) printf("%d"); is supposed to print an integer value, but no such value is specified.
Hi,
The value of the statement printf("%d") varies depending on the compiler we use.
In Unix, cc compiler will produce a result as some unknown integer value.
In Turbo C compiler the result will be zero.
This depends on the compiler which we use
Regrads,
Chella
Hi,
The value of the statement printf("%d") varies depending on the compiler we use.
In Unix, cc compiler will produce a result as some unknown integer value.
In Turbo C compiler the result will be zero.
This depends on the compiler which we use
Regrads,
Chella
My compiler makes daemons fly out of your nose. As said before: calling the
printf function like this causes undefined behaviour. Every (educated?) guess
what the output would/could be is just guessing. Nothing can be said about it.
kind regards,
Jos
Banfa 9,065
Expert Mod 8TB
1) void main() instead of int main()
2) printf("%d"); is supposed to print an integer value, but no such value is specified.
That is what I was thinking of :D
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