no9t9

Nov 17 2004, 12:55 PM

This smells like a homework assignment... So.. I won't tell you how to do it

Here's a POINTER though... variables can be manipulated using their address in memory...

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switch

Nov 26 2004, 02:38 AM

very smooth.

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Latest Entries

Trap FeedBacker

Apr 30 2008, 01:52 PM

code for this program
C Code, Can U Solve This?

Easy code in 'C' program to prove pallidrome rule

Thanking you,
Vikram









-question by vikram

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tricky77puzzle

Feb 27 2008, 12:41 AM

printf is a standard function in C. You can't modify it.

Try doing the same thing in c++.

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Dark_Prisoner

Feb 26 2008, 07:15 PM

QUOTE(nooc @ Nov 19 2004, 11:02 PM)

CODE

#include <stdio.h>

void fun(void)
{
#define fun() i=10
}

int main()
{
int i = 20;
fun();
printf("i = %d\n", i);
return 0;
}

I think that this won't work :
(i guess) we can't put preprocessor directives on a function
If we can , it won't work also , because the compiler or the processor had completed the line :

CODE

fun();

so he won't replace it .

if the preprocessor works in functions i guess this will be the answer

CODE

void fun(void)
{
#define printf("i = %d\n", i) i=0
}

int main()
{
int i = 20;
fun();
printf("i = %d\n", i);
return 0;
}

What do you think ???

Reply

klusk

Feb 25 2008, 02:26 PM

QUOTE(8bit @ Feb 4 2005, 07:07 AM)

CODE

void fun(void)
{
  int * ptr =&i;
  *ptr=50;
}

Beat that.

This code wont work as the variable i is undeclared for fun()....it would have worked if "i" had been a global variable

There is another way if you wish to use pointers...

CODE

void fun(int *ptr)
{
  *ptr=50;
}


int main()
{
int i = 20;
fun(&i);
printf("i = %d\n", i);
return 0;
}

the output will be 50

Reply

Csshih

Dec 13 2007, 06:12 PM

Oh yea?
void fun(void)
{
randomlineofetext(3249830325)
}
the program won't print anything at all xD

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