Please explain the output to this code: [closed] - c

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#include<stdio.h>
void main()
{
if(0xA)
if(052)
if('\xeb')
if('\012') //what do the above 3 statements mean?
printf("Hello World!")
else;
else;
else;
else;
}
Output:
Hello World!
so here, what is the meaning of 052, \xeb and \012?
And what is the significance of the multiple else statements?

Answer 1

From the C Standard (6.8.4.1 The if statement)
2 In both forms, the first substatement is executed if the
expression compares unequal to 0. In the else form, the second
substatement is executed if the expression compares equal to 0. If the
first substatement is reached via a label, the second substatement is
not executed.
So as each expression in the if statements
if(0xA)
if(052)
if('\xeb')
if('\012')
are unequal to 0 then the enclosed call of the function printf (provided that a semicolon is present)
printf("Hello World!");
is executed.
So the if statements do not make much sense.
0xA is a hexadecimal integer constant equal to decimal 10
052 is octal integer constant equal to decimal 42
'\xeb' is a character constant specified as a hexadecimal escape sequence
'\012' is a character constant specified as an octal escape sequence
Take into account that according to the C Standard the function main without parameters shall be declared like
int main( void )
Thus as each expression of the if statements is not equal to 0 then the program is in fact equivalent to the following program
#include<stdio.h>
int main( void )
{
printf("Hello World!");
}

Answer 2

This is a trick question. Presumably someone thought you could learn something from it. Presumably this person also thinks that a good way to teach a child to ride a bicycle is to flatten the bicycle's tires, and put large boulders in the way, and randomly give the child a rough shove to the side, because it "builds character", and if you can learn to ride a bicycle in spite of those obstacles, you will eventually be a really good bike rider. But I digress.
Anyway, here's an explanation of what's going on in that code.
0xA is a hexadecimal constant. It has the value 0A16, or 10 (base 10).
052 is an octal constant. It has the value 528, or 42 (base 10).
'\xeb' is a hexadecimal character constant. It is a character with the value EB16, or 235 (base 10) (which in ISO-8859-1 or Unicode ends up being an e with two dots over it, ë).
'\012' is an ordinary, octal character constant. It is a character with the value 128, or 10 (base 10) (which in ASCII ends up being a newline character, '\n').
When you say
if( expr )
in C, the expression is considered "true" if it is nonzero, or "false" if it is zero. Needless to say, all four of those constants are nonzero.
The full syntax of an if statement in C is either
if( expr )
statement-or-block
or
if( expr )
statement-or-block
else
statement-or-block
where statement-or-block is either a single statement, or a "block" (sequence) of multiple statements enclodes in braces { }.
A little-used piece of syntax in C is the "empty statement". If you have nothing, followed by a semicolon, that's an empty statement, and it does... nothing.
It is considered mandatory in polite circles to indent your code to portray its structure. So the normal way of presenting this code would be:
#include <stdio.h>
int main()
{
if(0xA)
if(052)
if('\xeb')
if('\012')
printf("Hello World!")
else
;
else
;
else
;
else
;
}
And, again, all four expressions evaluate immediately to "true". So, in English, this code says "If true, then if true, then if true, then if true, then print 'Hello world!', else do nothing, else do nothing, else do nothing, else do nothing."
Phrased that way it sounds pretty stupid, I know. And in fact, it's just as stupid in C as it is in English.
P.S. Besides the bizarre and unnecessary conditionals, another problem with the code as posted, which I have quietly corrected, is that it declared main as void, which is something else that's not done in polite circles, and throws further doubt on the validity of the book this code came from.

Answer 3

The code you posted is equivalent to
#include<stdio.h>
int main( void ) // void main() is not a standard signature
{
if(0xA)
{
if(052)
{
if('\xeb')
{
if('\012')
{
printf("Hello World!")
}
else
{
}
}
else
{
}
}
else
{
}
}
else
{
}
}
Curly braces can be omitted if there is only a single statement within them:
if ( a )
{
do_something();
}
can be written as
if ( a )
do_something();
although I and others recommend against doing this except in specific cases.
An if statement itself counts as a single statement, so it's possible to write
if ( a )
{
if ( b )
do_something();
}
as
if ( a )
if ( b )
do_something();
And, of course, the compiler doesn't care about indentation, so
if ( a )
if ( b )
do_something();
means the same thing - however, as you've discovered, code written this way is a little hard to understand.
The test condition of an if, while, or for may be any scalar type (integer, floating-point, or pointer value). C didn't originally have a distinct Boolean type, so the rule is any non-zero value in a Boolean context evaluates to "true", while zero evaluates to "false".
Each of 0xA, 052, '\xeb', and '\012' is a scalar expression with a non-zero value (0xA and 052 are integer literals in hexadecimal and octal format, respectively, while '\xeb' and '\012' are escape sequences representing character values in hex and octal format), so the bodies of all the if statements are executed.
You have four nested if statements, and only the innermost one is doing anything interesting. The else clauses are empty and can be omitted entirely:
#include<stdio.h>
int main( void )
{
if(0xA)
{
if(052)
{
if('\xeb')
{
if('\012')
{
printf("Hello World!")
}
}
}
}
And, as I said above, the curly braces may be omitted as well:
#include<stdio.h>
int main( void )
{
if(0xA)
if(052)
if('\xeb')
if('\012')
printf("Hello World!")
}
All of which reduces to
int main( void )
{
printf("Hello World!")
}
A reasonably smart compiler won't bother generating the machine code for the if statements at all, and should generate machine code as though the program had been written as above.