Algorithms

Count Matches

A counting loop increments a result when an element satisfies a condition.

predicate The condition decides whether one element contributes to the count.
counter The counter changes only when the condition is true.

Count Matches

threshold
count_matches.c
Replay: real traced execution (multi-file project)
#include <stdio.h>

int main(void) {
    int threshold = 5;
    int values[5] = {2, 5, 7, 9, 4};
    int count = 0;

    for (int i = 0; i < 5; i++) {
        if (values[i] >= threshold) {
            count++;
        }
    }

    printf("count=%d\n", count);
    return 0;
}

#include <stdio.h>

int main(void) {
    int threshold = 3;
    int values[5] = {2, 5, 7, 9, 4};
    int count = 0;

    for (int i = 0; i < 5; i++) {
        if (values[i] >= threshold) {
            count++;
        }
    }

    printf("count=%d\n", count);
    return 0;
}

#include <stdio.h>

int main(void) {
    int threshold = 8;
    int values[5] = {2, 5, 7, 9, 4};
    int count = 0;

    for (int i = 0; i < 5; i++) {
        if (values[i] >= threshold) {
            count++;
        }
    }

    printf("count=%d\n", count);
    return 0;
}

  1. threshold ← 5, values ← ⟨addr A⟩, count ← 0

    3int main(void) {4    int threshold→ 5 = 5; //@threshold=3, 85    int values→ ⟨addr A⟩[5] = {2, 5, 7, 9, 4};6    int count→ 0 = 0;

  2. for (int i = 0; i < 5; i++)

    pass 1 of 5
    8for (int i0 = 0; i < 5; i++) {9    if (values[i] >= threshold) {
    All 5 passes — pass 1 is the card above
    passi
    10
    21
    32
    43
    54

  3. count ← 1

    pass 1 of 3
    8for (int i = 0; i < 5; i++) {9    if (values[i]5 >= threshold5) {10        count→ 1++;11    }
    All 3 passes — pass 1 is the card above
    passvalues[i]count
    150 → 1
    271 → 2
    392 → 3

  4. printf("count=%d ", count);

    14    printf("count=%d\n", count3);15    return 0;16}
    outputcount=3

  1. threshold ← 3, values ← ⟨addr A⟩, count ← 0

    3int main(void) {4    int threshold→ 3 = 3;5    int values→ ⟨addr A⟩[5] = {2, 5, 7, 9, 4};6    int count→ 0 = 0;

  2. for (int i = 0; i < 5; i++)

    pass 1 of 5
    8for (int i0 = 0; i < 5; i++) {9    if (values[i] >= threshold) {
    All 5 passes — pass 1 is the card above
    passi
    10
    21
    32
    43
    54

  3. count ← 1

    pass 1 of 4
    8for (int i = 0; i < 5; i++) {9    if (values[i]5 >= threshold3) {10        count→ 1++;11    }
    All 4 passes — pass 1 is the card above
    passvalues[i]count
    150 → 1
    271 → 2
    392 → 3
    443 → 4

  4. printf("count=%d ", count);

    14    printf("count=%d\n", count4);15    return 0;16}
    outputcount=4

  1. threshold ← 8, values ← ⟨addr A⟩, count ← 0

    3int main(void) {4    int threshold→ 8 = 8;5    int values→ ⟨addr A⟩[5] = {2, 5, 7, 9, 4};6    int count→ 0 = 0;

  2. for (int i = 0; i < 5; i++)

    pass 1 of 5
    8for (int i0 = 0; i < 5; i++) {9    if (values[i] >= threshold) {
    All 5 passes — pass 1 is the card above
    passivalues[i]thresholdcount
    10———
    21———
    32———
    43980 → 1
    54———

  3. count ← 1

    8for (int i = 0; i < 5; i++) {9    if (values[i]9 >= threshold8) {10        count→ 1++;11    }

  4. printf("count=%d ", count);

    14    printf("count=%d\n", count1);15    return 0;16}
    outputcount=1