CareerCup

The behaviour of this type of self-assignment is undefined and dependent on the compiler implementation. So, the result may be either 0, or 10, or even a not-compilable code. The undefined behaviour happens due to missing sequence point in the evaluation of expression "count = count++;"
You can read it @en.cppreference.com/w/cpp/language/eval_order

Undefined behavior!

As far as I know that code is equivalent to

int count = 0;
for(int i=0; i < 10 ; ++i){ count = count; count = count + 1;}
std::cout << count;

so I expect to see 10 on std::cout;

ouput will be 10
we can write count=count++; as two steps i.e.
count=count;
count=count+1;

I'm a java guy but here goes. Since there are no braces to identify the block of code in the for loop, the count = count++; statement is the only statement part of the for loop. So count goes from 1 (count being initially o, count++ is 1) to 10 (the last value for count is 9, so count++ is 10) The output of the code is 10.

G++ compiler gives output as 10. However I am not sure why people are calling the behavior as undefined

int count = 0;
40117e: c7 44 24 0c 00 00 00 movl $0x0,0xc(%esp)
401185: 00
int res = 0;
401186: c7 44 24 08 00 00 00 movl $0x0,0x8(%esp)
40118d: 00
res = count++;
40118e: 8b 44 24 0c mov 0xc(%esp),%eax
401192: 89 44 24 08 mov %eax,0x8(%esp)
401196: 83 44 24 0c 01 addl $0x1,0xc(%esp)
res = ++count;
40119b: 83 44 24 0c 01 addl $0x1,0xc(%esp)
4011a0: 8b 44 24 0c mov 0xc(%esp),%eax
4011a4: 89 44 24 08 mov %eax,0x8(%esp)
count = count++;
4011a8: 83 44 24 0c 01 addl $0x1,0xc(%esp)
count = ++count;
4011ad: 83 44 24 0c 01 addl $0x1,0xc(%esp)

My friend and I did some research about this question with asm. We found for res = count++, the asm does that:
temp = count;
res = count;
count = count +1;
if we say count = count ++, compiler will optimize this 3 one line into one count = count +1;
I think the tricky reason that confuses ppl comes from how we implement operator ++(int) overload for class. We would copy the object, increase the object, then return the old obj for return and assignment. But for primary type, compiler copy, assign and then increase. That's a good question to understand the primary type's difference from the operator overload for class.

10

This is really a very good question. Thanks for the post.

i think the answer should be 10

gives 10 in VC++2008

Answer should be 10.

I think its coz of precedence. Since ++(post increment) have higher precedence than =(assignment) operator..thus for the first iteration, its picks value 0 increment it to 1 and then assigns to count and it goes on.

case( I )

#include <iostream>
using namespace std;

void main()
{
int count = 0;
for(int i=0; i < 10; ++i){
count = count++;
}

cout<<"Output of this code is: "<<endl;
std::cout << count <<endl;


}


root [0] .L main.C
root [1] main()
Output of this code is:
0

case ( II )

#include <iostream>
using namespace std;

void main()
{
int count = 0;
for(int i=0; i < 10; ++i){
count = ++count;
}

cout<<"Output of this code is: "<<endl;
std::cout << count <<endl;


}


root [0] .L main.C
root [1] main()
Output of this code is:
10

I am using g++ 4.1.2, it gives 10

Answer will be zero,
Because count=count++,
1step: count is zero so..count =0;
2step:count get incremented by 1,
3step:but it does not have 3 step in loop so again count is zero .
so final output is zero.
If count=++count then o/p will be 10.

Answer will be 0. Since value is first used then incremented. So count=0 is tempoarily stored somewhere then incremented so count=1 but in assignment of count=count++ count is again replaced by 0. count would be 10 if ++count is used.

i think it will print from 0 to 9 for count.

c++ standard doesnt allow writing same variable twice in an expression. It is undefined behavior or compiler error