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.
Related
This question already has answers here:
Why are these constructs using pre and post-increment undefined behavior?
(14 answers)
Closed 2 years ago.
In the following piece of C code, I would expect the printed output to be 5794. But when I compile it with GCC (7.5.0) the output is 5693. Why?
int main()
{
int x=20,y=35;
x=y++ + x++;
y= ++y + ++x;
printf("%d%d",x,y);
}
this program has undefined behavior.which means you can't predict what will happen.
look for more information here
Sequence_point
This question already has answers here:
Why are these constructs using pre and post-increment undefined behavior?
(14 answers)
Closed 6 years ago.
#include <stdio.h>
int main()
{
int i = 6;
printf("%d %d", ++i, i++);//printing
return 0;
}
What will be output of the following code and why?
printf("%d %d",++i ,i++);//printing
Is undefined behavior. The order of argument processing is not specifically defined in the C standard, it is not possible to predict exactly what the output will be. It could be anything at all according to this.
This question already has answers here:
Why are these constructs using pre and post-increment undefined behavior?
(14 answers)
Closed 9 years ago.
When I compile the following code
#include<stdio.h>
#define CUBE(x) (x*x*x)
int main()
{
int a, b=3;
a = CUBE(b++);
printf("%d, %d\n", a, b);
return 0;
}
It gives 27 , 6
But shouldn't the expression a=b++*b++*b++; be calculated as a=3*4*5 and should give 60?
Your expression causes undefined behaviour, so you could get any answer. Trying to modify the same value more than twice between sequence points is bad news.
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.
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Could anyone explain these undefined behaviors (i = i++ + ++i , i = i++, etc…)
Why does sizeof(x++) not increment x?
Here the value of i is 5 but it should be 6 or 7 I guess. What's the reason?
int main(void){
int i=5,j;
j=sizeof(i++ + ++i);
printf("%d %d",i,j);
}