无符号短无符号长分配

Question

在从long到short分配时,LSB 2字节为0,其中MSB填充来自堆栈的func1()算法值的值.为什么会发生这种情况,为什么编译器试图将这些垃圾值转换为MSB 2字节？

#include <stdio.h>

unsigned short func1(void); // NB: function prototype !

int main(void)

{

     unsigned long int L = 0;

     unsigned short K = 0;

     L = func1();

      printf("%lu", L); // prints junk values

      K = L; 

      printf("%u", K);  // prints 0

     return 0;
}

unsigned short func1(void)

{

      unsigned short i = 0;

      // Algorithm Logic!!!

      return i; // returns 0
}

Answer 1

说明符unsigned long是lu.那unsigned short就是hu.您通过不使用正确的说明符来调用UB.