objective-c - 关于一个递归代码的问题～

Question

最近在学习objective-c，教程中有一段代码没弄清楚，是关于递归的，我用截图的形式发上来，给大家看看

void singTheSong(int numberOfBottles)
{
    if (numberOfBottles == 0) {
        printf("there are simply no more bottles of beer on the wall.\n");
    } else {
        printf("%d bottles of beer on the wall. %d bottles of beer.\n",
               numberOfBottles, numberOfBottles);
        int oneFewer = numberOfBottles - 1;
        printf("Take one dowm, pass it around, %d bottles of beer on the wall.\n",
               oneFewer);

        singTheSong(oneFewer);
        printf("put a bottle in the recycling, %d empty bottles in the bin.\n",
               numberOfBottles);
    }
}

int main(int argc, const char * argv[])
{
    singTheSong(99);
    return 0;
}

它的输出结果是这样的：

99 bottles of beer on the wall. 99 bottles of beer.
Take one dowm, pass it around, 98 bottles of beer on the wall.
98 bottles of beer on the wall. 98 bottles of beer.
Take one dowm, pass it around, 97 bottles of beer on the wall.
97 bottles of beer on the wall. 97 bottles of beer.
Take one dowm, pass it around, 96 bottles of beer on the wall.
96 bottles of beer on the wall. 96 bottles of beer.
Take one dowm, pass it around, 95 bottles of beer on the wall.
......(中间重复的省略)
1 bottles of beer on the wall. 1 bottles of beer.
Take one dowm, pass it around, 0 bottles of beer on the wall.
there are simply no more bottles of beer on the wall.
put a bottle in the recycling, 1 empty bottles in the bin.
put a bottle in the recycling, 2 empty bottles in the bin.
......(中间重复的省略)
put a bottle in the recycling, 98 empty bottles in the bin.
put a bottle in the recycling, 99 empty bottles in the bin.
Program ended with exit code: 0

这段代码，我看不懂的地方是，它如何从numberOfBottles等于0的时候，又继续运行了

printf("put a bottle in the recyling, %d empty bottles in the bin.\n",
        numberOfBottles);

这段代码呢？并且numberOfBottles一直在+1

请大家帮我解惑，谢谢了～

Answer 1

Inti utama ialah: Anda perlu memahami bahawa panggilan fungsi akan mempunyai operasi 压栈 dan panggilan rekursif akan kembali 回到sebelum rekursi dan 继续laksanakan kod selepas rekursif panggil.

Ganti singTheSong(99); anda dengan singTheSong(3); dan beritahu saya.

入参为3, 下面两行打印
3 botol bir di dinding 3 botol bir.
Ambil satu dowm, sebarkan, 2 botol bir di dinding.

第一次递归调用singTheSong(2), 入参为2.
注意, 此时singTheSong(oneFewer); 之后的printf打印不会出现, 因为递归调用还没有返回;
后面你看到的 numberOfBottles一直在+1 现象的直接原因就在这.
2 botol bir di dinding 2 botol bir.
Ambil satu dowm, sebarkan, 1 botol bir di dinding.

第二次递归调用singTheSong(1), 入参为1
同样, singTheSong(oneFewer) 之后的printf打印不会出现;
1 botol bir di dinding.
Ambil satu dowm, sebarkan, 0 botol bir di dinding.

第三次递归调用singTheSong(0), 入参为0, 直接打印下面一行, 并退出;
注意, 这里的退出是退回到第二次递归的栈. 并从singTheSong(oneFewer)之后的printf开始执行.
tiada lagi botol bir di dinding.

因为第二次递归入参为1, 所以这里打印下面一行
第二次递归调用退回到 第一次 递归的栈继续执行
letak sebotol dalam kitar semula, 1 botol kosong dalam tong.

因为第一次递归入参为2, 所以这里打印下面一行
第一次递归调用退回到最初的singTheSong(3)继续执行
letak sebotol dalam kitar semula, 2 botol kosong dalam tong.

最初的singTheSong(3)调用入参为3, 所以打印下面一行.
letak sebotol dalam kitar semula, 3 botol kosong dalam tong.

Answer 2

Masalah perintah pelaksanaan, tiada situasi di mana numberOfBottles sentiasa +1
singTheSong(99); perlu dipanggil semasa melaksanakan singTheSong(98); tetapi pada masa ini singTheSong(99); belum selesai melaksanakan

printf("put a bottle in the recyling, %d empty bottles in the bin.\n",
        numberOfBottles);

Kod ini tidak dipanggil sehingga singTheSong(0); selesai Kod singTheSong(1); bawah dalam badan kaedah printf dilaksanakan, dan kemudian secara rekursif, mula melaksanakan 2, 3, 4... 99

.