C++数值类型极值输出异常求助：为何显示为非十进制值？

你的问题核心是输出整数类型时用错了格式符，导致有符号整数被当作无符号八进制数输出了——你看到的37777777600和177是八进制表示，对应十进制的-128和127。

常见错误场景 #

比如你可能写了类似这样的代码：

#include <stdio.h>
#include <climits>

int main() {
    printf("char range: %o to %o\n", CHAR_MIN, CHAR_MAX);
    return 0;
}

这里的%o是八进制无符号输出格式符，会把有符号的CHAR_MIN（-128）按照无符号数的八进制形式打印出来，就出现了你看到的异常值。

修正方案 #

1. 用正确的格式符匹配类型：
   • 有符号整数类型（char、short、int、long）用十进制有符号格式符%d或%i：

     printf("char range: %d to %d\n", CHAR_MIN, CHAR_MAX);
     printf("short range: %d to %d\n", SHRT_MIN, SHRT_MAX);
     printf("int range: %d to %d\n", INT_MIN, INT_MAX);
     printf("long range: %ld to %ld\n", LONG_MIN, LONG_MAX);
     

   • 无符号类型用十进制无符号格式符%u：

     printf("unsigned char range: %u to %u\n", 0, UCHAR_MAX);
     printf("unsigned int range: %u to %u\n", 0, UINT_MAX);
     

2. 更推荐的C++方式：
   直接用cout输出，它会自动处理类型的符号和格式，不需要手动指定格式符，更不容易出错：

   #include <iostream>
   #include <climits>
   #include <cfloat>
   
   int main() {
       std::cout << "char range: " << CHAR_MIN << " to " << CHAR_MAX << "\n";
       std::cout << "short range: " << SHRT_MIN << " to " << SHRT_MAX << "\n";
       std::cout << "int range: " << INT_MIN << " to " << INT_MAX << "\n";
       std::cout << "long range: " << LONG_MIN << " to " << LONG_MAX << "\n";
       std::cout << "float range: " << FLT_MIN << " to " << FLT_MAX << "\n";
       std::cout << "double range: " << DBL_MIN << " to " << DBL_MAX << "\n";
       std::cout << "long double range: " << LDBL_MIN << " to " << LDBL_MAX << "\n";
       std::cout << "unsigned char range: 0 to " << UCHAR_MAX << "\n";
       std::cout << "unsigned int range: 0 to " << UINT_MAX << "\n";
       return 0;
   }
   

内容的提问来源于stack exchange，提问作者Felipe Maion