C 64 位左移失败 [英] C left shift on 64 bits fail

问题描述

我在 C 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
", x, total);        
    }       

打印的内容:

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 我有那些奇怪的值?我在 Ubuntu 10.10 64 位上使用 gcc 4.6.1.

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 是一个 int，32 位，所以 (1 <<27)*27 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，则 1 << 仍然是一个 int，但是对于乘法，它会被转换为 uint64_t，所以不会有溢出.然而，如果 x >= 31，1 << 将是未定义的行为(因为结果值不能用有符号的 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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