C ++使用指针来选择排序函数 [英] C++ Using pointers for selection sort function

问题描述

我有一个任务通过传递指针来编写一些排序函数。不幸的是，指针只是我的大脑似乎不理解的那些概念之一。

I've got a task to code some sorting function by passing pointers. Unfortunately, pointers are just one of those concepts my brain doesn't seem to comprehend.

这是调用：

int size = 10000;
int* data = new int[size]; 
//omitted code that populates array for the sake of space
selectionSort(data, data+size); 

这是我对函数的不正确尝试：

And here's my incorrect attempt at the function:

void selectionSort(int* first, int* last) { 
for (int* i = first; i < last-1; i++) {
    int* min = i;
    for (int* j = i+1; j < last; j++) {
        if (j < min) {
            min = j; 
        }
        int* temp = i; 
        i = min; 
        min = temp; 
    }
}

}

我有麻烦弄清楚当我比较一个指针与另一个，或调整指针时会发生什么。是调整/比较它指向的值还是比较实际指针本身？

Basically, I'm having trouble figuring out what happens when I'm comparing one pointer with another, or adjusting a pointer. Is it adjusting/comparing the value it's pointing too or is it comparing the actual pointers themselves?

任何帮助。干杯。

推荐答案

指针有时是一个难以理解的概念。也许它会帮助把它们看作一个值的引用，而不是值本身（它是邮箱的地址，而不是邮箱的内容）。

Pointers are sometimes a difficult concept to grok. Perhaps it will help to think of them as a reference to a value, rather than the value itself (it's an address of the mailbox, not the contents of the mailbox).

在你的代码'int * min = i;'中设置min到同一地址（引用）为'i'。所以稍后在你的'if'语句if（j< min）'你比较引用，而不是值。你需要取消引用你的指针来获取值，像这样：'if（* j< * min）'。

In your code 'int* min = i;' sets min to the same address (reference) as 'i'. So later in your 'if' statement 'if (j < min)' you are comparing references, not values. You need to "dereference" your pointer to get at the values, like this: 'if (*j < *min)'.

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