Exercise 03_01

[calbch]

Shouldn't it say "why it's not 10" in the bonus?

Since i is shadowed inside of the scope of the for loop, final value would be the initial random value assigned during it's definition.

What am I missing here?

#include <assert.h>
#include <stdio.h>
#include <stdlib.h>

// DO NOT CHANGE THIS
static int track = 0, count = 0;

// DO NOT CHANGE THIS
void verify_count(int *count, int i) {
    assert(i == track++);
    (*count)++;
}

int main() {
    int i = rand() % 100 + 11; // DO NOT CHANGE THIS

    // YOUR CODE HERE
    for (int i = 0; i < 10; ++i) {
        // END YOUR CODE
        printf("%d\n", i);
        verify_count(&count, i); // DO NOT CHANGE THIS
    }

    // BONUS: print the value of i here and try to understand why it's 10
    printf("final value: %d\n", i);

    // DO NOT CHANGE THIS
    assert(count == 10);
    return 0;
}

[sleaper]

I think you are right, but this exercise was done by @7etsuo. So let's wait for his explanation of what he meant by that.

[danwritecode]

I'll take a look at this tonight.

[7etsuo]

Reference Code:

/** LEARNING
 * Your goal is to understand how the for statement works.
 *
 * The for statement has the following syntax:
 *
 * for (initialization; condition; increment) {
 *      statement;
 *      statement;
 *      ...
 * }
 *
 * The for statement is equivalent to the following while statement:
 *
 * initialization;
 * while (condition) {
 *      statement;
 *      statement;
 *      ...
 *      increment;
 *  }
 */

/** EXERCISE
 * Your job is to write a for statement equivalent to the following 
 * while statement:
 *
 * i = 0; // initialization
 * while (i < 10) { // condition
 *  printf("%d\n", i);
 *  verify_count(&count, i);
 *  i++; // increment
 * }
 */

// ❌ I AM NOT DONE

#include <stdio.h>
#include <stdlib.h>
#include <assert.h>

// DO NOT CHANGE THIS
static int track = 0, count = 0;

// DO NOT CHANGE THIS
void verify_count (int *count, int i) {
    assert(i == track++);
    (*count)++;
}

int main() {
    int i = rand() % 100 + 11; // DO NOT CHANGE THIS

    // YOUR CODE HERE
    for (initialization; condition; increment) {
    // END YOUR CODE
        printf("%d\n", i);
        verify_count(&count, i); // DO NOT CHANGE THIS
    }

    // BONUS: print the value of i here and try to understand why it's 10

    // DO NOT CHANGE THIS
    assert (count == 10);
    return 0;
}

Solution:

/ * i = 0; // initialization
  * while (i < 10) { // condition
  * printf("%d\n", i);
  * verify_count(&count, i);
  * i++; // increment
  * } 
  */
    int i = rand() % 100 + 11; // DO NOT CHANGE THIS

    // YOUR CODE HERE
    for (i=0; i<10; i++) {
            // END YOUR CODE
        printf("%d\n", i);
        verify_count(&count, i); // DO NOT CHANGE THIS
    }

    // BONUS: print the value of i here and try to understand why it's 10
        /* Hint look at the last value i printed in the for loop */
        printf("i=%d\n", i); 
    // DO NOT CHANGE THIS
    assert (count == 10);
    return 0;

[7etsuo]

In the context of C programming, within a for loop, the initialization expression is distinct from variable declaration, although it often includes variable declarations as a convenience.

Here's a breakdown:

Variable Declaration: This involves specifying a type and associating it with a variable name. For example: int i; declares a variable i of type int.

Initialization: This step assigns a value to the variable at the time of its creation or at the start of the loop. In a for loop, the initialization expression is executed only once, before the loop begins. It often includes declaring a variable, but can also be used to set an existing variable to a new value.

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

int i = 0 is both declaring and initializing i. However, if i were declared before the loop:

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

Here, i = 0 in the loop is purely initialization, not declaration.

The details are in the definitions.