C Unsigned int 提供负值? [英] C Unsigned int providing a negative value?

问题描述

我有一个无符号整数，但是当我使用 %d 打印出来时，有时会出现负值?

I have an unsigned integer but when i print it out using %d there is sometimes a negative value there?

推荐答案

打印 %d 会将整数读取为带符号的十进制数，而不管其定义的类型.

Printing %d will read the integer as a signed decimal number, regardless of its defined type.

要打印无符号数，请使用 %u.

To print unsigned numbers, use %u.

这是因为 C 处理变量参数的方式.编译器只是从堆栈中提取值(类型为 void* 并指向调用堆栈)并且 printf 必须从您提供的格式字符串中找出数据包含的内容它到.

This happens because of C's way to handle variable arguments. The compiler just pulls values from the stack (typed as void* and pointing to the call stack) and printf has to figure out what the data contains from the format string you give it to.

这就是您需要提供格式字符串的原因 - C 没有 RTTI 或基类"(例如，Java 中的 Object)来获取通用或预定义的 toString.

This is why you need to supply the format string - C has no way of RTTI or a 'base class' (Object in Java, for example) to get a generic or predefined toString from.

这篇关于C Unsigned int 提供负值?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！