C语言 将 int 转换为 uint8_t
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converting int to uint8_t
提问by Johan Elmander
is it a correct way to convert an intvalue to uint8_t:
这是将int值转换为的正确方法uint8_t：
int x = 3;
uint8_t y = (uint8_t) x;
assume that x will never be less than 0. Although gccdoes not give any warning for the above lines, I just wanted to be sure if it is correct to do it or is there a better way to convert int to uint8_t?
假设 x 永远不会小于 0。虽然gcc没有对以上几行发出任何警告，但我只是想确定这样做是否正确，或者是否有更好的方法将 int 转换为 uint8_t？
P.S. I use C on Linux if you are going to suggest a standard function
PS 如果您要建议标准功能，我在 Linux 上使用 C
回答by ouah
It is correct but the cast is not necessary:
这是正确的，但演员不是必需的：
uint8_t y = (uint8_t) x;
is equivalent to
相当于
uint8_t y = x;
xis implicitely converted to uint8_tbefore initialization in the declaration above.
xuint8_t在上面的声明中被隐式转换为 before 初始化。
