Strange output when using do/while in C programming lang

Question

I am trying to understand the output of this program. If I try to "translate" the code, I believe it should go like this:

• while "j" is smaller than 3 - print "Ha" (this loop goes 3 times, so it gives 3 "Ha")
• do/while -> j is equal to j - 2 hence print "Hi" while ++j - In the end the program prints out "Hi" 4 times.

How does the program prints it 4 times, how does the condition works here?

#include <stdio.h>

int main() {

    int j = 0;
    while(j++ < 3){
        printf( "Ha ");
    }
    do{
        j -= 2;
        printf( "Hi "); 
    }
    while(++j);
    for(j = 1; j <= 3; j++){
        printf( "Ho ");
    }
    printf("\n");
    return 0;
}

The output is:

Ha Ha Ha Hi Hi Hi Hi Ho Ho Ho

Answer 1

The best way to understand what your program does is to watch the value of j at each step of your program.

first j=0 then you enter into the while.

• 1. j=0 ; j<3? ; yes ; j=j+1 ; it print "Ha".
• 2. j=1 ; j<3? ; yes ; j=j+1 ; it print "Ha".
• 3. j=2 ; j<3? ; yes ; j=j+1 ; it print "Ha".
• 4. j=3 ; j<3? ; no ; j=j+1 ; leave the while statement

j=4

then you leave the while to enter into de do, while() statement. at this moment j=4.

in the do , while() :

• 1. j=j-2 ; j=2 ; it print "Hi" ; j+=1 ; j=3; j!=0 ? yes
• 2. j=j-2 ; j=1 ; it print "Hi" ; j+=1 ; j=2; j!=0 ? yes
• 3. j=j-2 ; j=0 ; it print "Hi" ; j+=1 ; j=1; j!=0 ? yes
• 4. j=j-2 ; j=1 ; it print "Hi" ; j+=1 ; j=0; j=0 ? no -> leaving the do while()

at the end you enter in the for statement.

• ++x is called Pre incrementation (the variable is incremented before the expression evaluation)
• x++ is the post incrementation (the variable is incremented after the expression evaluation
• i'm not sure about this, you need to check this informations.

Answer 2

After first loop

j == 4
j -= 2 == 2
Hi
++j == 3
j -= 2 == 1
Hi
++j == 2
j -= 2 == 0
Hi 
++j == 1
j -= 2 == -1
Hi
++j == 0 //end of th loop

Answer 3

After the first while loop

while(j++ < 3){
    printf( "Ha ");
}

j is equal to 4.

So within the first iteration of the do-while loop

do{
    j -= 2;
    printf( "Hi "); 
}
while(++j);

j is equal to 2. After the first iteration j is equal to 3. Within the second iteration j is equal to 1. After the second iteration j is equal to 2. Within the third iteration j is equal to 0. After the third iteration j is equal to 1. Within the forth iteration j is equal to -1. So in this forth iteration in the condition

while(++j);

j is equal to 0 and the control is passed to the next loop. So the do-while loop was executed 4 times.

That is the value of the postfix increment operator is the value of its operand before incrementing. And the value of the pre-increment operator is the value of its operand after incrementing.

Answer 4

The ++j is prefix increment, i.e., the value will be increased and then, the incrased value will be used for the condition check.

I have added a print statement for the ease of understanding:

do{
    j -= 2;
    printf( "Hi "); 
    printf("value of j before the while = %d\n", j);
}
while(++j);

and the output is:

Hi value of j before the while = 2
Hi value of j before the while = 1
Hi value of j before the while = 0
Hi value of j before the while = -1

So, in the

• first iteration, in while (++j), j is 2, and ++j is 3
• second iteration, 3-2 = 1, and ++j is 2.
• third iteration, 2-2 = 0, and ++j is 1.
• Fourth iteration, 1-2 = -1, and ++j is 0 - this make the while chec false and the loop ends.