This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Could anyone explain these undefined behaviors (i = i++ + ++i , i = i++, etc…)
i tried this simple C program in GCC
#include<stdio.h>
int main(){
int x=5;
printf("%d,%d,%d,%d,%d",++x,x++,x,x++,++x);
return 0;
}
and the output was 9,7,7,6,6
i traced it and assumed that it will print 6,6,7,7,9 but i found my assumption in reverse order, how come!
Because your program has undefined behaviour. There is no sequence point between the evaluations of function arguments, and it is undefined behaviour to mutate the same object more than once without intervening sequence point.
The program is simply ill-formed. It is not a valid C program.
Related
This question already has answers here:
Why are these constructs using pre and post-increment undefined behavior?
(14 answers)
Closed 5 years ago.
here is very simple c program:
int main()
{
int i = 22;
printf("%d %d %d",i,i++,i);
return 0;
}
The result is:
23 22 23
how exactly post increment is working here ?
You cannot test correctly in this context.
The order of parameter evaluation for function parameters is undefined, hence you will have platform-dependent results.
The way i++ works actually is by first returning the old value and later incrementing, but that is a bad test.
Judging by the result you got, i++ is evaluated and returns the pre- incremented value. Then, the value of i, the return value of i++, and the value of i are passed to the print function.
This isn't ever something that you should rely on, as you may get different answers on different compilers or even with different optimization settings. As in the other answer, the order of parameter evaluation is undefined behavior.
This question already has answers here:
Why are these constructs using pre and post-increment undefined behavior?
(14 answers)
Closed 7 years ago.
#include <stdio.h>
int main ()
{
int a=10;
printf("%d %d %d",a,a++,a);
return 0;
}
The output I am getting is "11 10 11".
I thought the output would be "10 10 11".
why a is incrementing like this?
Because there is no guarantee about the order in which a C compiler evaluates the arguments. The only thing guaranteed (by the standard) is that they are all evaluated before doing the call. Therefore, you should never count on the order of evaluation of the arguments. Just consider it as random.
Hence, in general, avoid using auto-increment if the same variable exists more than once in an argument list.
This question already has answers here:
Why are these constructs using pre and post-increment undefined behavior?
(14 answers)
Closed 7 years ago.
I am executing this c program on gcc compiler and getting strange results.
So how is it possible
code:
#include<stdio.h>
int main()
{
int i;
i =10;
printf(" %d %d %d ",i++,i++,i); //output : 11 10 12
return 0;
}
as per me result should be 10 11 12 but I am getting 11 10 12.
How is it possible?
In C++, the order of evaluation of function arguments is undefined so if you use the increment operator multiple times in the argument to a particular function there is no 'correct' answer, they may be evaluated in any arbitrary order.
Please familiarize yourself with the concept of Sequence points. Only at such defined sequence points is guaranteed that all side effects of previous evaluations are performed. There are no sequence points between the list of arguments of a function. So, it leads to undefined behavior.
This question already has answers here:
Why are these constructs using pre and post-increment undefined behavior?
(14 answers)
Closed 8 years ago.
I wrote the following simple program
#include<stdio.h>
int main()
{
int i;
i=1;
printf("%d %d %d",i,i++,++i);
return 0;
}
The above program gave 3 2 3 as output which I am not able to interpret the output. I am using gcc-4.8.1
You have undefined behavior here!!
When there are multiple increments to the same variable in the printf() you can't predict the output.
The order of execution within the printf() is not defined.
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Undefined Behavior and Sequence Points
the output of the programme
#include<stdio.h>
main()
{
int i = 10;
printf("%d %d %d\n", ++i, i++, ++i);
}
is 13 11 13. Can someone please explain this ?
It's the oldest question ever. Why do people find this so fascinating?
This is undefined behavior; you're relying on side-effects without a sequence point between modifications.