正在发生什么类型转换？

发布于 2024-09-24 23:50:16 字数 328 浏览 6 评论 0原文

#include "stdio.h"

int main()
{
    int x = -13701;
    unsigned int y = 3;
    signed short z = x / y;

    printf("z = %d\n", z);

    return 0;
}

我预计答案是-4567。我得到“z = 17278”。为什么这些数字的提升结果是 17278？

我在 Code Pad 中执行了此操作。

原文

#include "stdio.h"

int main()
{
    int x = -13701;
    unsigned int y = 3;
    signed short z = x / y;

    printf("z = %d\n", z);

    return 0;
}

I would expect the answer to be -4567. I am getting "z = 17278".
Why does a promotion of these numbers result in 17278?

I executed this in Code Pad.

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

甜是你 2024-10-01 23:50:16

隐藏的类型转换是：

signed short z = (signed short) (((unsigned int) x) / y);

当混合有符号和无符号类型时，无符号类型获胜。 x 转换为 unsigned int，除以 3，然后将该结果向下转换为（有符号）short。对于 32 位整数：

(unsigned) -13701         == (unsigned) 0xFFFFCA7B // Bit pattern
(unsigned) 0xFFFFCA7B     == (unsigned) 4294953595 // Re-interpret as unsigned
(unsigned) 4294953595 / 3 == (unsigned) 1431651198 // Divide by 3
(unsigned) 1431651198     == (unsigned) 0x5555437E // Bit pattern of that result
(short) 0x5555437E        == (short) 0x437E        // Strip high 16 bits
(short) 0x437E            == (short) 17278         // Re-interpret as short

顺便说一句，signed 关键字是不必要的。 signed Short 是 short 的更长表达方式。唯一需要显式signed 的类型是char。 char 可以有符号或无符号，具体取决于平台；所有其他类型始终默认签名。

The hidden type conversions are:

signed short z = (signed short) (((unsigned int) x) / y);

When you mix signed and unsigned types the unsigned ones win. x is converted to unsigned int, divided by 3, and then that result is down-converted to (signed) short. With 32-bit integers:

(unsigned) -13701         == (unsigned) 0xFFFFCA7B // Bit pattern
(unsigned) 0xFFFFCA7B     == (unsigned) 4294953595 // Re-interpret as unsigned
(unsigned) 4294953595 / 3 == (unsigned) 1431651198 // Divide by 3
(unsigned) 1431651198     == (unsigned) 0x5555437E // Bit pattern of that result
(short) 0x5555437E        == (short) 0x437E        // Strip high 16 bits
(short) 0x437E            == (short) 17278         // Re-interpret as short

By the way, the signed keyword is unnecessary. signed short is a longer way of saying short. The only type that needs an explicit signed is char. char can be signed or unsigned depending on the platform; all other types are always signed by default.

回复收藏 0 原文

无法回应 2024-10-01 23:50:16

简短的回答：除法首先将x提升为unsigned。只有这样，结果才会被转换回有符号短。

长答案：阅读这个SO线程。

回复收藏 0 原文

子栖 2024-10-01 23:50:16

问题来自于unsigned int y。事实上，x/y 变成无符号的。它适用于：

#include "stdio.h"

int main()
{
    int x = -13701;
    signed int y = 3;
    signed short z = x / y;

    printf("z = %d\n", z);

    return 0;
}

The problems comes from the unsigned int y. Indeed, x/y becomes unsigned. It works with :

#include "stdio.h"

int main()
{
    int x = -13701;
    signed int y = 3;
    signed short z = x / y;

    printf("z = %d\n", z);

    return 0;
}

回复收藏 0 原文

同展鸳鸯锦 2024-10-01 23:50:16

每次在加法和乘法算术运算中混合“大”有符号和无符号值时，无符号类型“获胜”，并且在无符号类型的域中执行计算（“大”意味着 int 和更大））。如果您的原始有符号值是负数，则它首先会根据有符号到无符号转换的规则转换为正无符号值。在您的情况下，-13701 将变为 UINT_MAX + 1 - 13701，结果将用作股息。

请注意，在典型的 32 位 int 平台上，有符号到无符号转换的结果将产生无符号值 4294953595。除以 3 后，您将得到 1431651198。该值太大，无法强制转换为 16 位 short 类型平台上的 short 对象。尝试这样做会导致实现定义的行为。因此，如果您的平台的属性与我的假设相同，那么您的代码会产生实现定义的行为。从形式上来说，您获得的“无意义”17278 值只不过是实现定义的行为的具体表现。如果您在启用溢出检查的情况下编译代码（如果您的编译器支持它们），则它可能会陷入赋值。