在 while 条件本身内实现一个中断 while 的 if 条件

Question

我有这部分代码：

int arrayBinary_search(int myarray[], int key){

    int selector;
    int low_limit = 0;
    int high_limit = SIZE;
    while (1){
        selector = (low_limit+high_limit)/2;
        printf("The selector is: %d\n", selector);
        if (myarray[selector] == key){
            return 1;
        }
        else {
            if (low_limit==selector || high_limit==selector)
                break;
            if (key < myarray[selector])
                high_limit = selector;
            else
                low_limit = selector;
            printf("The high_limit is: %d\n", high_limit);
            printf("The low_limit is: %d\n", low_limit);
        }

    }

}

此代码在数组中进行二进制搜索。它有效，但 while(1) 不好看。我想实施一些条件来暂时替换那个“1”。条件是如果在 while 循环内，那会打破它。我试过： while (!(low_limit==selector) && !(high_limit==selector)) 但它在第一个循环后停止，因为在第一个循环后， "selector" 具有与 "high_limit".

相同的值

完整代码here.

Answer 1

您可以使用 do-while 键匹配时退出的循环：

int arrayBinary_search(int myarray[], int key)
{
    int selector;
    int low_limit = 0;
    int high_limit = SIZE;

    do {
        selector = (low_limit + high_limit) / 2;
        printf("The selector is: %d\n", selector);

        if (low_limit == selector || high_limit == selector)
            return 0;
        if (key < myarray[selector])
            high_limit = selector;
        else
            low_limit = selector;
        printf("The high_limit is: %d\n", high_limit);
        printf("The low_limit is: %d\n", low_limit);

    } while (myarray[selector] != key);

    return 1;
}

另一方面，arrayGenerator 函数没有 return 和 int，您也不需要它，您应该 void.

在 while 条件本身内实现一个中断 while 的 if 条件

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