GPIOB->ODR |= 1<<0;
时间: 2024-04-25 19:24:00 浏览: 139
这行代码的作用是将GPIOB的ODR寄存器的第0位设置为1。根据引用\[2\],GPIOB_ODR |= (1<<0)实际上是将GPIOB_ODR和(1<<0)进行或运算,然后将结果赋值给GPIOB_ODR。其中,1<<0表示将1左移0位,即不移动,结果为00000000 00000000 00000000 00000001。因此,将GPIOB_ODR的第0位和1进行或运算,其它位保持不变,实现了将第0位设置为1的效果。所以,这行代码的作用是将GPIOB的ODR寄存器的第0位设置为1。\[2\]
#### 引用[.reference_title]
- *1* *2* [【嵌入式C语言系列】位运算](https://blog.csdn.net/XYF736162986/article/details/122601986)[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^v91^control_2,239^v3^insert_chatgpt"}} ] [.reference_item]
- *3* [对 GPIOB_ODR|=(1<<10) GPIOB_ODR&=~(1<<10)的理解](https://blog.csdn.net/qq_57064185/article/details/119870019)[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^v91^control_2,239^v3^insert_chatgpt"}} ] [.reference_item]
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