为什么while循环被执行多次？

Question

I have got a question in the interview for which I had to find out the output of the following code. I tried but it was not correct. Please explain the following code.

#include<stdio.h>
int main()
{
    int x=0,a;
    while(x++ < 5)
    {
        a=x;
        printf("a = %d \n",a);
        static int x=3;
        printf("x = %d \n",x);
        x+=2;
    }
    return 0;
}

Output:

a = 1
x = 3
a = 2
x = 5
a = 3
x = 7
a = 4
x = 9
a = 5
x = 11

Can anyone please explain whats going on here?

Answer 1

The loop conditional expression x++ < 5 uses the x declared outside the loop. The statement x += 2; is not affecting the x declared outside the loop because static int x=3; hides the previous declaration of x .

In other words, all modifications to x after the statement static int x=3; is not affecting the x used in loop controlling expression.

Answer 2

It is because x++ returns the current value of x and then is incremented.

In the first iteration,

while(x++ < 5)

is same as

while(0 < 5)

Then,after the condition has been checked, x will be incremented. Thus the value of a is the incremented value of x . The static x ,shadows(hides) the x declared outside the loop and hence,

x+=2;

affects the static x and not the outer one. The variable x declared in the loop,since it is static exists as long as the program does and will not be lost once it goes out of scope. It will be initialized to 3 and 2 will be added in each iteration of the loop to it.

Answer 3

This is equivalent to:

int x=0,a;
int y=3;
while(x++ < 5)
{
    a=x;
    printf("a = %d \n",a);
    printf("x = %d \n",y);
    y+=2;
}

The second x was hiding the first x .