问题描述
我在C语言中有此代码(仅供学习):
I have this code in C (it's for study only):
char x;
uint64_t total = 0;
for(x = 20; x < 30; x++){
total = (((((1 << x) * x) / 64) + 1) * sizeof(uint64_t));
printf("%d - %llu\n", x, total);
}
打印内容:
20 - 2621448
21 - 5505032
22 - 11534344
23 - 24117256
24 - 50331656
25 - 104857608
26 - 218103816
27 - 18446744073625665544
28 - 18446744073575333896
29 - 18446744073508225032
为什么在x> 26时我有那些奇怪的值?我在Ubuntu 10.10 64位上的gcc 4.6.1上。
Why at x > 26 do I have those strange values? I'm at gcc 4.6.1 on Ubuntu 10.10 64 bits.
推荐答案
因为 1
是32位的 int
,因此(1 溢出。使用
1ull
。
Because
1
is an int
, 32 bits, so (1 << 27)*27
overflows. Use 1ull
.
关于您的评论,如果
x
是 uint64_t
,然后 1<< x
仍然是 int
,但是对于乘法,它将被强制转换为 uint64_t
,这样就不会溢出。但是,如果 x> = 31
,则 1<< x
是未定义的行为(因为结果值不能用带符号的32位整数类型表示)。
Regarding your comment, if
x
is a uint64_t
, then 1 << x
is still an int
, but for the multiplication it would be cast to uint64_t
, so there'd be no overflow. However, if x >= 31
, 1 << x
would be undefined behaviour (as the resulting value cannot be represented by a signed 32 bit integer type).
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