This question already has answers here:
Why are these constructs using pre and post-increment undefined behavior?
(14 answers)
Closed 6 years ago.
First of all, I would like to apologize, if this question has been asked before in this forum. I searched, but could't find any similar problem.
I am a beginner in C. I was going through a tutorial and came across a code, the solution of which I can't understand.
Here is the code -
#include <stdio.h>
#define PRODUCT(x) (x*x)
int main()
{
int i=3, j, k;
j = PRODUCT(i++);
k = PRODUCT(++i);
return 1;
}
I tried running the code through compiler and got the solution as "j = 12" and "k = 49".
I know how #define works. It replace every occurrence of PRODUCT(x) by (x*x), but what I can't grasp is how j and k got the values 12 and 49, respectively.
Any help would be appreciated.
Thank you for your time.
Your code will invoke undefined behavior. Anything could happen. The macro in statements
j = PRODUCT(i++);
k = PRODUCT(++i);
will be expanded to
j = x++ * x++;
k = ++x * ++x;
In both statements x is being modified more than once between two sequence points.
Related
This question already has answers here:
Why are these constructs using pre and post-increment undefined behavior?
(14 answers)
Closed 2 years ago.
stumbled upon such a puzzle:
What will be shown on the screen?
#include <stdio.h>
void main()
{
int x = 10;
printf("x = %d, y = %d", x--, x++);
}
Curiously enough, but shown at the screen this: x = 11, y = 10;
But how??
Argument Evaluation Order is undefined in both C and C++. It's important to avoid any code that passes expressions dependent on each other that has to be evaluated before the function is called. It's a strict no.
int f1() { printf("F1") ; return 1;}
int f2() { printf("F2" ) ; return 1;}
printf("%d%d", f1(), f2()) ;
You can check out by adding several functions that contain a print statement and pass it to a function to observe this in action. You don't know what's coming, the C standard doesn't specify it, it depends on what compiler you use and how it optimizes your code.
This question already has answers here:
The need for parentheses in macros in C [duplicate]
(8 answers)
Closed 2 years ago.
My code:
#include <stdio.h>
#define PRODUCT(x) (x * x)
int main()
{
int i = 3, j, k, l;
j = PRODUCT(i + 1);
k = PRODUCT(i++);
l = PRODUCT(++i);
printf("%d %d %d %d", i, j, k, l);
return 0;
}
I am not able to comprehend why the output is:
7 7 12 49.
Is there any error in macro or some other problem?
Your code has undefined behavior, operations in i:
k=PRODUCT(i++);
l=PRODUCT(++i);
lack a sequence point.
As for:
j=PRODUCT(i+1);
It expands to i+1*i+1 which is i+i+1 which is 7. I assume it's not the expected result, in the future also include that in your question.
Your macro is incorrect. The following expression:
PRODUCT(i+1)
will expand to
(i+1 * i+1)
which is 2*i+1.
Your macro should be:
#define PRODUCT(x) ((x)*(x))
I strongly suggest you stop using macros for this sort of thing. You could easily write this as a function:
int product(int x)
{
return x * x;
}
Note that this will only work for the example I gave. If you try
PRODUCT(i++)
you will get
( (i++) * (i++) )
which invokes undefined behaviour, as this expression lacks a sequence point between the 2 increments.
This question already has answers here:
Why are these constructs using pre and post-increment undefined behavior?
(14 answers)
Closed 9 years ago.
Code in C language.
#include<stdio.h>
#define PRODUCT(x) ( x * x * x)
int main()
{
int i =5,k;
k = PRODUCT( ++i );
printf("i is :%d ",k);
return 0;
}
My Question is why i is : 392? According to me output should be 336. (because 6 * 7 * 8 = 336)
Am I really messed up here??
Preprocessed code will have
( ++i * ++i * ++i)
which have Lack of sequence point between the two execution on same variable resulting Undefined behaviour.
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Could anyone explain these undefined behaviors (i = i++ + ++i , i = i++, etc…)
There is a code following below and i ma facing a very serious problem in understanding the logic for the code.
#include <stdio.h>
#include <stdlib.h>
int main(void )
{
int i = 1 ;
printf("\n%d %d %d %d\n",++i,i++,i++,++i) ;
return 0 ;
}
I am using gcc compiler under the linux distro named Mandriva. In the above mentioned i have used pre and post increment with a variable in the printf statement.
The output that i am supposed to get is 2 2 3 5, but i am getting a different output.
Please help me in this code.
I am feeling much difficult in this code.
It's undefined behavior. There's no sequence points between the increments of i.
Any result is a correct result (including your hard drive being formatted).
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Could anyone explain these undefined behaviors (i = i++ + ++i , i = i++, etc…)
Undefined Behavior and Sequence Points
#include<stdio.h>
int main(){
int i=5,j=5,y,x;
int m=++i;
int n=++i;
x=m+n;
y=++j + ++j ;
printf("%d %d ",x,y);
return 0;
}
OUTPUT : 13 14
Can any one plz explain why 'y' value is 14 and not 13.
Most compilers will increment j twice before performing the addition and attributing the value to y, that is why the result in your case is 14.
The C standard doesn't specify the order of evaluation of that expression, though, so on another compiler the result could be 13 indeed.
In other words, this is undefined behavior and should be not be used other than in obfuscation contests and puzzles.