C: help me understanding the output - c

Question

can any one help me understanding the output of this code?
#include <stdio.h>
int main()
{
int x = 1, y = 1;
for(; y; printf("%d %d \n", x, y))
{
y = x++ <= 5;
}
return 0;
}
the output is:
2 1
3 1
4 1
5 1
6 1
7 0

Answer 1

A for loop in the form of
for (a; b; c)
{
d;
}
is equivalent to
{
a;
while (b)
{
d;
c;
}
}
Now if we take your loop
for(; y; printf("%d %d \n", x, y))
{
y = x++ <= 5;
}
It is equivalent to
{
// Nothing
// Loop while y is non-zero
while (y)
{
// Check if x is less than or equal to 5, assign that result to y
// Then increase x by one
y = x++ <= 5;
printf("%d %d \n", x, y);
}
}
Now it should hopefully be easier to understand what's going on.
Also: Remember that for boolean results (like what you get as result from a comparison), true is equal to 1, and false is equal to 0.

Answer 2

The code is an obfuscated, ugly version of this:
#include <stdio.h>
#include <stdbool.h>
int main()
{
int x = 1;
bool y = true;
while(y == true)
{
y = (x++ <= 5);
printf("%d %d \n", x, (int)y);
}
return 0;
}
y in the original code serves as a boolean. Back in the ancient days, there were no boolean type in C so it was common to use int instead. The expression y = x++ <= 5; evaluates to 0 or 1, which is equivalent to false or true.
Note:
While the C language allows all manner of crazy stuff, you should never write for loops as in the original code. De facto standard is to write for loops like this:
The first clause of a for loop shall only contain iterator initialization.
The second clause should only contain the loop condition.
The third clause should only contain a change of the loop iterator, such as for example an increment (like i++).
A for loop that doesn't follow the above industry standard rules is badly written, no excuses.

Answer 3

The statement y = x++ <= 5; sets y to 1 if x is less than or equal to 5, or to 0 if x is greater than 5. It then increments x. The loop stops when y is equal to 0, that is, on the iteration after x was incremented above 5. Then it runs the printf() statement on each iteration.
You wouldn’t normally see a loop switch the loop condition and the loop body that way. This seems to be a pathological example of how it’s technically legal. You shouldn’t imitate it.

Answer 4

y is 1 (true) as long as x <= 5 and x increases every iteration.
Rewrite your code:
for (int x = 1, y = 1; y; y = x <= 5, x++) {
printf("%d %d\n", x, y);
}
or using while
int x = 1;
int y = 1;
while (y) {
printf("%d %d\n", x, y);
x++;
y = x <= 5;
}

Answer 5

The first component of the for loop is empty; which means values of variables haven't changed. The condition in the for loop is y. Which means that the loop will run till the condition y is true; which means till the value of y is anything other than 0.
When the value of y becomes equal to 0; the condition will become false and the loop will stop iterating.
Then comes the block execution...
y = x++ <= 5;
Here first the condition x <= 5 is checked. Note that x++ is post increment and so the value of x will be incremented after the execution of the statement. So, if the value of x was 1 before increment; it will check 1 <= 5 and not 2 <= 5 and after the execution of the statement; the value will become 2.
After that comes the third component of for loop which is the increment part. Here,
printf("%d %d \n", x, y)
simply prints x and y.
So, what happens is; when the loop starts, both x and y are 1.
The statement
y = x++ <= 5;
is encountered, thus the condition x++ <= 5 is checked, where x is 1. Since the condition is true; the value of y will be 1 and x will be incremented to 2. The same thing continues.... when x will be equal to 5, x++ <= 5 will check 5 <= 5 and the condition will be true and x will be incremented to 6. Now when it goes in the loop the next time; the condition will be false and y will instead be equal to 0. Thus, now when the loop checks the condition y, the condition will be false and thus the control will flow out of the loop and thus you see the result.

Answer 6

I built and dumped this code in IDA since I understand assembly better :D
This code is a bit obfuscated so here will be the normal version of it:
#include <stdio.h>
int x = 1;
int y = 1;
while(y != 0) {
if(x++ <= 5)
{
y = 1;
} else {
y = 0;
}
printf("%d %d\n",x,y);
}
that for loop in the source:
for(; y; printf("%d %d \n", x, y))
checks if y is a non-zero value and if it is, it will print the string you want. In the body of the loop, the result of the comparison (0 or 1) will be copied to y and the check continues.

Answer 7

Precedence of operator <= is higher than the operator =.
So, the statement:
y = x++ <= 5;
is equivalent to:
y = (x++ <= 5)
Because of post-increment ++ operator, the value of x returned first and then x is incremented.
The initial value of both x and yis 1.
Since y is 1, the for loop condition evaluates to true and the statement in the for loop block executed. But in the for loop of your code, in place of loop iterator update statement you have printf() statement which is printing the value of x and y after executing the loop body statements. The flow goes like this:
1 <= 5 evaluate to 1 which is assigned to y and after this x is 2,
Output : 2 1
2 <= 5 evaluate to 1 which is assigned to y and after this x is 3,
Output : 3 1
3 <= 5 evaluate to 1 which is assigned to y and after this x is 4,
Output : 4 1
4 <= 5 evaluate to 1 which is assigned to y and after this x is 5,
Output : 5 1
5 <= 5 evaluate to 1 which is assigned to y and after this x is 6,
Output : 6 1
6 <= 5 evaluate to 0 which is assigned to y and after this x is 7,
Output : 7 0
Now the value y is 0 and in your code, the for loop condition is y so the loop condition evaluates as false and loop exits.