uint8_t intFlag = 0;
时间: 2023-08-23 12:04:19 浏览: 183
根据引用和引用中的内容,可以看出uint8_t并不是一个新的数据类型,而是使用typedef给类型起的别名。它实际上是一个8位无符号整型。所以,当我们声明一个变量uint8_t intFlag = 0;时,我们实际上是声明了一个8位无符号整型的变量intFlag,并将其初始化为0。<em>1</em><em>2</em><em>3</em>
#### 引用[.reference_title]
- *1* *2* [【转】uint8_t / uint16_t / uint32_t /uint64_t 是什么数据类型 - 大总结,看完全明白了](https://blog.csdn.net/winniezheng/article/details/8663683)[target="_blank" data-report-click={"spm":"1018.2226.3001.9630","extra":{"utm_source":"vip_chatgpt_common_search_pc_result","utm_medium":"distribute.pc_search_result.none-task-cask-2~all~insert_cask~default-1-null.142^v92^chatsearchT3_1"}} ] [.reference_item]
- *3* [关于uint8_t等类型的定义](https://blog.csdn.net/magic_show_time/article/details/121489424)[target="_blank" data-report-click={"spm":"1018.2226.3001.9630","extra":{"utm_source":"vip_chatgpt_common_search_pc_result","utm_medium":"distribute.pc_search_result.none-task-cask-2~all~insert_cask~default-1-null.142^v92^chatsearchT3_1"}} ] [.reference_item]
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