英文:
Overflow after multiplying positive int8s and casting to uint32
问题
我正在尝试将8位正整数相乘，并将结果转换为uint32类型。
a := int8(12)
b := uint32(a * a) // 4294967184
在上面的代码中，相乘导致了溢出。但是，如果我在相乘之前将每个int8转换为uint32/uint8，我会得到预期的结果，如下所示：
b := uint32(uint32(a) * uint32(a)) // 12
为什么我需要在相乘之前进行类型转换？
代码是否执行类似于以下方式？
tmp := a * a
b := int32(tmp)
还是还有其他原因？
英文:
I am trying to multiply 8-bit positive integers and cast the result to uint32.
a := int8(12)
b := uint32(a * a) // 4294967184
In the above code multiplying causes an overflow.
But if I cast each int8 to uint32/uint8 before multiplying, I get the expected result like below,
b := uint32(uint32(a) * uint32(a)) // 12
Why do I need to cast before multiplying?
Does the code get executed something like this?
tmp := a*a
b := int32(tmp)
Or is there something else?
答案1
得分: 1
你的问题似乎已经回答了，但为了更清楚起见：
将类型转换的语法视为函数调用的语法。参数内的表达式将被完全求值，然后传递给函数。
如果你这样写：
math.Sin(f*f)
你不会期望它等同于
math.Sin(f) * math.Sin(f)
你需要在乘法之前进行类型转换，因为在有符号的8位值中，12 * 12 不是安全的操作。
英文:
Your question seems to answer itself, but for clarity:
Think of the grammar of a type conversion as being the same as the grammar of a function call. The expression inside the parameters is fully evaluated, and then it is passed to the function.
If you do
math.Sin(f*f)
You wouldn't expect it to be equivalent to
math.Sin(f) * math.Sin(f)
You need to cast before multiplying, because 12 * 12 is not a safe operation in signed 8-bit values.
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