我有以下arduino代码,执行递归合并排序。我们必须确定如何计算可以在此8192B SRAM中输入的最大阵列元素数量。数组元素的数量在这个空间设置()最大数组大小
int16_t Test_len = 64;
我会一直爱来解决这个设定自己却被小时后绝望,我错过了这一个演讲,因为我有一个流感。
整个代码的副本。
#include <Arduino.h>
#include <mem_syms.h>
// some formatting routines to indent our messages to make it easier
// to trace the recursion.
uint8_t indent_pos = 0;
const uint8_t indent_amt = 2;
void indent_in() {
if (indent_pos <= 32) {
indent_pos ++;
}
}
void indent_out() {
if (indent_pos >= indent_amt) {
indent_pos --;
}
}
void indent() {
for (uint8_t i=0; i < indent_pos * indent_amt; i++) {
Serial.print(" ");
}
}
// print out memory use info, s is a simple descriptive string
void mem_info(char *s) {
indent();
Serial.print(s);
Serial.print(" Stack: ");
Serial.print(STACK_SIZE);
Serial.print(" Heap: ");
Serial.print(HEAP_SIZE);
Serial.print(" Avail: ");
Serial.print(AVAIL_MEM);
Serial.println();
}
// call this after a malloc to confirm that the malloc worked, and
// if not, display the message s and enter a hard loop
void assert_malloc_ok(void * mem_ptr, char *s) {
if (! mem_ptr) {
Serial.print("Malloc failed. ");
Serial.print(s);
Serial.println();
while (1) { }
}
}
// call this on entry to a procedure to assue that at least required amt of
// memory is available in the free area between stack and heap if not, display
// the message s and enter a hard loop
void assert_free_mem_ok(uint16_t required_amt, char *s) {
if (AVAIL_MEM < required_amt) {
Serial.print("Insufficient Free Memory: ");
Serial.print(s);
Serial.print(" require ");
Serial.print(required_amt);
Serial.print(", have ");
Serial.print(AVAIL_MEM);
Serial.println();
while (1) { }
}
}
void merge(int16_t *Left, int16_t Left_len, int16_t *Right, int16_t Right_len,
int16_t *S) {
// position of next element to be processed
int Left_pos = 0;
int Right_pos = 0;
// position of next element of S to be specified
// note: S_pos = Left_pos+Right_pos
int S_pos = 0;
// false, take from right, true take from left
int pick_from_left = 0;
while (S_pos < Left_len + Right_len) {
// pick the smallest element at the head of the lists
// move smallest of Left[Left_pos] and Right[Right_pos] to S[S_pos]
if (Left_pos >= Left_len) {
pick_from_left = 0;
}
else if (Right_pos >= Right_len) {
pick_from_left = 1;
}
else if (Left[Left_pos] <= Right[Right_pos]) {
pick_from_left = 1;
}
else {
pick_from_left = 0;
}
if (pick_from_left) {
S[S_pos] = Left[Left_pos];
Left_pos++;
S_pos++;
}
else {
S[S_pos] = Right[Right_pos];
Right_pos++;
S_pos++;
}
}
}
// sort in place, i.e. A will be reordered
void merge_sort(int16_t *A, int16_t A_len) {
indent_in();
indent();
Serial.print("Entering merge sort: array addr ");
Serial.print((int) A);
Serial.print(" len ");
Serial.println(A_len);
mem_info("");
assert_free_mem_ok(128, "merge_sort");
if (A_len < 2) {
indent_out();
return;
}
if (A_len == 2) {
if (A[0] > A[1]) {
int temp = A[0];
A[0] = A[1];
A[1] = temp;
}
indent_out();
return;
}
// split A in half, sort left, sort right, then merge
// left half is: A[0], ..., A[split_point-1]
// right half is: A[split_point], ..., A[A_len-1]
int split_point = A_len/2;
indent();
Serial.println("Doing left sort");
merge_sort(A, split_point);
mem_info("After left sort");
indent();
Serial.println("Doing right sort");
merge_sort(A+split_point, A_len-split_point);
mem_info("After right sort");
// don't need the merging array S until this point
int *S = (int *) malloc(A_len * sizeof(int));// source of 10 bytes accumulation in heap
assert_malloc_ok(S, "Cannot get merge buffer");
mem_info("Doing merge");
merge(A, split_point, A+split_point, A_len-split_point, S);
for (int i=0; i < A_len; i++) {
A[i] = S[i];
}
// now we are done with it
free(S);
mem_info("After free");
indent_out();
}
void setup() {
Serial.begin(9600);
// int *bad_news = (int *) malloc(4000);
mem_info("********* THIS IS THE BEGINNING *********");
randomSeed(analogRead(0));
int16_t Test_len = 64;
int16_t Test[Test_len];
Serial.print("In: ");
for (int16_t i=0; i < Test_len; i++) {
Test[i] = random(0, 100);
if (1) {
Serial.print(Test[i]);
Serial.print(" ");
}
}
Serial.println();
merge_sort(Test, Test_len);
if (1) {
Serial.print("Out: ");
for (int16_t i=0; i < Test_len; i++) {
if (i < Test_len-1 && Test[i] > Test[i+1]) {
Serial.print("Out of order!!");
}
Serial.print(Test[i]);
Serial.print(" ");
}
Serial.println();
}
}
void loop() {
}
请具体在您的问题。您提供的代码中是否存在错误?如果是,请提供详细信息。你是否要求在代码中填入空白?你自己在解决这个问题方面有什么想法/想法? – Sheena
可以放在8192个字节的RAM中的'element_t'类型元素的数目是'8192/sizeof(element_t)'。如果这不能回答你的问题,请在你的问题中更具体。 – DevSolar