C语言 打印程序，请说明输出

Question

#include <stdio.h>
void myPrint (int n) {
    printf("%d", n/2);
    if(n > 0)
        myPrint (n - 1);
    printf("%d", n);
}
int main (void) {
    int count = 4;
    myPrint (count);
    return 0;
}

这个简单的打印程序打印2110001234，有人请解释为什么它在最后打印01234。我不确定为什么它每次都加 1。

Answer 1

最后，

您会看到对myPrint的调用与发生顺序相反。

以下是查看每个递归调用期间发生的情况的一种方法。

myPrint(4)
  printf("%d", n/2)  // Prints 2 because 4/2 = 2
  myPrint(n - 1) // Calls myPrint(3)
    printf("%d", n/2)  // Prints 1 because 3/2 = 1
    myPrint(n - 1) // Calls myPrint(2)
      printf("%d", n/2)  // Prints 1 because 2/2 = 1
      myPrint(n - 1) // Calls myPrint(1)
        printf("%d", n/2)  // Prints 0 because 1/2 = 0
        myPrint(n - 1) // Calls myPrint(0)
          printf("%d", n/2)  // Prints 0 because 0/2 = 0
          // Does not execute if statement
          printf("%d", n);  // Prints 0 because n = 0 at this call
        printf("%d", n);  // Prints 1 because n = 1 at this call
      printf("%d", n);  // Prints 2 because n = 2 at this call
    printf("%d", n);  // Prints 3 because n = 3 at this call
  printf("%d", n);  // Prints 4 because n = 4 at this call

Answer 2

myPrint是递归函数，它打印以下语法：

myPrint(n) -> printf(n/2) ++ myPrint(n-1) ++ printf(n)

对于每个n>0，而对于n=0语法是

myPrint(0) -> printf("00")

所以对于n = 4：

myPrint(4) -> 
-> "2" ++ myPrint(3) ++ "4" ->
-> "2" ++ ("1" ++ myPrint(2) ++ "3") ++ "4" ->
-> "21" ++ ("1" ++ myPrint(1) ++ "2") ++ "34" ->
-> "211" ++ ("0" ++ myPrint(0) ++ "1") ++ "234" ->
-> "2110" ++ ("00") ++ "1234" ->
-> "2110001234"

这里++表示串联。