Algorithms

Transform Array

A transform loop writes a derived value for each input element.

input to output Each output element is computed from the matching input element.
factor Changing one scalar factor changes every transformed value.

Transform Array

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

int main(void) {
    int factor = 2;
    int input[3] = {1, 2, 3};
    int output[3] = {0, 0, 0};
    int total = 0;

    for (int i = 0; i < 3; i++) {
        output[i] = input[i] * factor;
        total += output[i];
    }

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

#include <stdio.h>

int main(void) {
    int factor = 3;
    int input[3] = {1, 2, 3};
    int output[3] = {0, 0, 0};
    int total = 0;

    for (int i = 0; i < 3; i++) {
        output[i] = input[i] * factor;
        total += output[i];
    }

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

#include <stdio.h>

int main(void) {
    int factor = 4;
    int input[3] = {1, 2, 3};
    int output[3] = {0, 0, 0};
    int total = 0;

    for (int i = 0; i < 3; i++) {
        output[i] = input[i] * factor;
        total += output[i];
    }

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

  1. factor ← 2, input ← ⟨addr A⟩, output ← ⟨addr B⟩, total ← 0

    3int main(void) {4    int factor→ 2 = 2; //@factor=3, 45    int input→ ⟨addr A⟩[3] = {1, 2, 3};6    int output→ ⟨addr B⟩[3] = {0, 0, 0};7    int total→ 0 = 0;

  2. output[i] ← 2, total ← 2

    pass 1 of 3
    9for (int i0 = 0; i < 3; i++) {10    output[i]→ 2 = input[i]1 * factor2;11    total→ 2 += output[i]2;12}
    All 3 passes — pass 1 is the card above
    passiinput[i]output[i]total
    1010 → 20 → 2
    2120 → 42 → 6
    3230 → 66 → 12

  3. printf("total=%d ", total);

    14    printf("total=%d\n", total12);15    return 0;16}
    outputtotal=12

  1. factor ← 3, input ← ⟨addr A⟩, output ← ⟨addr B⟩, total ← 0

    3int main(void) {4    int factor→ 3 = 3;5    int input→ ⟨addr A⟩[3] = {1, 2, 3};6    int output→ ⟨addr B⟩[3] = {0, 0, 0};7    int total→ 0 = 0;

  2. output[i] ← 3, total ← 3

    pass 1 of 3
    9for (int i0 = 0; i < 3; i++) {10    output[i]→ 3 = input[i]1 * factor3;11    total→ 3 += output[i]3;12}
    All 3 passes — pass 1 is the card above
    passiinput[i]output[i]total
    1010 → 30 → 3
    2120 → 63 → 9
    3230 → 99 → 18

  3. printf("total=%d ", total);

    14    printf("total=%d\n", total18);15    return 0;16}
    outputtotal=18

  1. factor ← 4, input ← ⟨addr A⟩, output ← ⟨addr B⟩, total ← 0

    3int main(void) {4    int factor→ 4 = 4;5    int input→ ⟨addr A⟩[3] = {1, 2, 3};6    int output→ ⟨addr B⟩[3] = {0, 0, 0};7    int total→ 0 = 0;

  2. output[i] ← 4, total ← 4

    pass 1 of 3
    9for (int i0 = 0; i < 3; i++) {10    output[i]→ 4 = input[i]1 * factor4;11    total→ 4 += output[i]4;12}
    All 3 passes — pass 1 is the card above
    passiinput[i]output[i]total
    1010 → 40 → 4
    2120 → 84 → 12
    3230 → 1212 → 24

  3. printf("total=%d ", total);

    14    printf("total=%d\n", total24);15    return 0;16}
    outputtotal=24