printf（）的呈现奇怪的结果

Question

我的代码是这个

#include<stdio.h> 
int main(void) 
{ 
    unsigned short height = 0; 
    unsigned short width = 0; 
    const unsigned short MIN_SIZE = 3; 
    printf("Enter the values for the width and the height minimum of %u\n:", 
      MIN_SIZE); 
    scanf(" %hd %hd", &width, &height); 
    if (width < MIN_SIZE) 
    { 
     printf("The value of width %u is too small. I set this to %u \n", 
       width, MIN_SIZE); 
     width = MIN_SIZE; 
    } 
    if (height < MIN_SIZE) 
    { 
     printf 
      ("The value of height %u is too small. I setting this to %u \n"), 
      height, MIN_SIZE; 
     height = MIN_SIZE; 
    } 
    for (unsigned int i = 0; i < width; ++i) 
    { 
     printf("*"); 
    } 
    return 0; 
} 

当我给的7例如一个宽度，和0高度，printf（）的呈现奇异数。你能解释一下为什么会发生这种情况？

Answer 1

这可能会编译一个警告。提供所有参数后，需要保留右括号。

printf 
      ("The value of height %u is too small. I setting this to %u \n"), 
      height, MIN_SIZE; 

也许你的意思是：

printf("The value of height %u is too small. I setting this to %u \n", height, MIN_SIZE); 

的主要问题是，我们应该用 “％虎” 的简称intergers。我想试试这个：

#include<stdio.h> 
int main(void) 
{ 
    unsigned short height = 0; 
    unsigned short width = 0; 
    const unsigned short MIN_SIZE = 3; 
    int i ; 
    printf("Enter the values for the width and the height minimum of %u\n:", MIN_SIZE); 
    scanf(" %hu %hu", &width, &height); 
    if (width < MIN_SIZE) { 
     printf("The value of width %hu is too small. I set this to %hu \n", width, MIN_SIZE); 
     width = MIN_SIZE; 
    } 
    if (height < MIN_SIZE) { 
     printf("The value of height %hu is too small. I setting this to %hu \n", height, MIN_SIZE); 
     height = MIN_SIZE; 
    } 
    for (i = 0; i < width; ++i) 
    { 
     printf("*"); 
    } 
    return 0; 
} 

有对SO这个良好的相关讨论：What is the format specifier for unsigned short int?