Foundations

Arrays

An array stores a fixed number of same-typed elements next to each other.

array `int scores[3]` creates storage for three integers.
index Array indexes start at zero, so `scores[0]` is the first element.

Arrays

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

int main(void) {
    int base = 80;
    int scores[3] = {base, base + 10, base + 20};
    int first = scores[0];
    int last = scores[2];
    int total = first + last;

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

#include <stdio.h>

int main(void) {
    int base = 60;
    int scores[3] = {base, base + 10, base + 20};
    int first = scores[0];
    int last = scores[2];
    int total = first + last;

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

#include <stdio.h>

int main(void) {
    int base = 90;
    int scores[3] = {base, base + 10, base + 20};
    int first = scores[0];
    int last = scores[2];
    int total = first + last;

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

  1. base ← 80, scores ← ⟨addr A⟩, first ← 80, last ← 100, total ← 180

    3int main(void) {4    int base→ 80 = 80; //@base=60, 905    int scores→ ⟨addr A⟩[3] = {base80, base + 10, base + 20};6    int first→ 80 = scores[0]80;7    int last→ 100 = scores[2]100;8    int total→ 180 = first80 + last100;910    printf("first=%d\n", first80);11    printf("total=%d\n", total180);12    return 0;13}
    outputfirst=80
total=180

  1. base ← 60, scores ← ⟨addr A⟩, first ← 60, last ← 80, total ← 140

    3int main(void) {4    int base→ 60 = 60;5    int scores→ ⟨addr A⟩[3] = {base60, base + 10, base + 20};6    int first→ 60 = scores[0]60;7    int last→ 80 = scores[2]80;8    int total→ 140 = first60 + last80;910    printf("first=%d\n", first60);11    printf("total=%d\n", total140);12    return 0;13}
    outputfirst=60
total=140

  1. base ← 90, scores ← ⟨addr A⟩, first ← 90, last ← 110, total ← 200

    3int main(void) {4    int base→ 90 = 90;5    int scores→ ⟨addr A⟩[3] = {base90, base + 10, base + 20};6    int first→ 90 = scores[0]90;7    int last→ 110 = scores[2]110;8    int total→ 200 = first90 + last110;910    printf("first=%d\n", first90);11    printf("total=%d\n", total200);12    return 0;13}
    outputfirst=90
total=200

Follow the Array

  1. base starts at 80.
  2. The scores are 80, 90, and 100.
  3. first reads the first score, 80.
  4. last reads the last score, 100.
  5. total = first + last becomes 180, so the program prints first=80 and total=180. | base | scores | first | total | | --- | --- | --- | --- | | 60 | 60, 70, 80 | 60 | 140 | | 80 | 80, 90, 100 | 80 | 180 | | 90 | 90, 100, 110 | 90 | 200 |

Exercise: arrays.c

Reproduce first=80 and total=180, then use the pinned base values 60 and 90 to predict each first score and total.