Pointer loop

Learn to loop over an array with a pointer.

Pointer loop

You can walk an array without using indices. Define a start pointer and an end pointer and print every element between them.

int arr[3] = {7, 8, 9};
int *p = arr;       /* start */
int *end = arr + 3; /* end */
while (p < end) { printf("%d", *p); p++; }

p++ jumps to the next element each step. The p < end condition keeps the loop going until the last element.

The end pointer pattern

The pattern end = arr + n gives a pointer to the end of the array. end points one past the last element; it is out of the array and used only for comparison.

int arr[3] = {7, 8, 9};
int *end = arr + 3;  /* arr[3] must not be read */
for (int *p = arr; p < end; p++) {
    printf("%d ", *p);
}

This pattern resembles C++ iterators and keeps loops clean.

Exercise Print with a pointer loop

Define int arr[3] = {7, 8, 9};. Define start and end pointers with int *p = arr; and int *end = arr + 3;. Inside while (p < end) print the element with printf("%d\n", *p); and advance with p++. Output should be 7, 8, 9.

Show example solution

#include <stdio.h>

int main(void) {
    int arr[3] = {7, 8, 9};
    int *p = arr;
    int *end = arr + 3;
    while (p < end) {
        printf("%d\n", *p);
        p++;
    }
    return 0;
}

Test yourself

p is start, end is end. What is the loop condition?
- p < end
- p > end
- p == end
- p = end
Show answer

p < end — p < end keeps the loop going until the last element.
Match loop parts

p++

one forward
p < end

loop condition
*p

current value
end = arr + n

end pointer

Complete the loop step

for (int *p = arr; p < end; ____) { /* ... */ }

Show answer

for (int *p = arr; p < end; p++) { /* ... */ }

Next lesson: Multidim array pointer

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