Compact Code for Initialization an Array in For-loop

Question

I am reviewing a code with the following snippet for minimal reproducible example. So here the code runs an iteration to initialize an array done as follows:

#include <stdio.h>
int i,j;
N=10;
int done[500];
void main() {

 done[0]=done[N]=1;
 for(i=0;i<N+1;done[i++]=0){
  printf("Done val %d iteration %d\n",done[i],i);

 }
}

The thing I am concerned with that I purpossely initialized the values of done[0]=done[1]=1 However when I run the initialization loop, the values at index 0 and 10 remains unchanged. I would like to understand how is the syntax of done[i++] actually evaluated?

Answer 1

The increment step of a for loop takes place after the loop body is executed. It's equivalent to the following code:

i = 0;
while (i < N+1) {
    printf("Done val %d iteration %d\n",done[i],i);
    done[i++] = 0;
}

As you can see, it's printing the value of done[i] before it changes it. So it prints the original value.

Answer 2

The code snippet you provided will never compile correctly to begin with! Although, I am going to assume you already know the reasons why. Therefore, I will simply answer your question.

Firstly, you did not initialize your indexes the way you expected. You did the following instead:

done[0]=done[N]=1;

Thus, index zero will be set to one, yet index ten is set to one instead of index one. Therefore, you should have done the following:

done[0]=done[1]=1;

As for the for loop itself, you must understand that your increment variable is executed last! A for loop is a three step process that follows a procedural order where,

1. increment variable gets initialized
2. Conditional statement gets checked
3. If conditional statement is true, then execute code body then increment/decrement the increment variable, then go to step two. Otherwise, exit the for loop!

Therefore, done[i++]=0 gets evaluated after your printf() statement in the provided code block! For more information please view C++ for loop - Tutorialspoint