c - 整数向下转换 [英] c - integer downcast

问题描述

关于 C 中的整数向下转换，例如

一个 int 值 000F'E000 向下转换为 short 或 unsigned short 将成为 E000.
short -> -8192,
unsigned short -> 57344,

所以它只是简单地削减比特吗?

那么向上转换呢?例如.int -10 就是ffffff81，转换成long long的规则是什么?

<小时>

@Update

关于upcasting，根据答案我做了一些测试，发现2的补码有以下规则:

<前>签:正 -> 正:添加 0 作为前缀位，负 -> 负:添加 1 作为前缀位，未签名:添加 0 作为前缀位，

代码:

//整数，向下转换 &向上，#include void downcastTest() {int i = 127<<13;printf("%X, %hX, %hi\n", i, i, i);}void upcastTest() {国际我= 127;int j = -127;printf("%x, %llx\n", i, (long long)i);printf("%x, %llx\n", j, (long long)j);}int main(int argc, char * argv[]) {downcastTest();upcastTest();返回0；}

解决方案

Downcasts

转换为较小的整数类型会丢弃目标类型中不存在的最高有效位(最左边，因为您将完整的二进制整数写在纸上).

向上广播

向上转换为更大的整数更复杂:

• 对于无符号到无符号类型，它添加了足够的零最高有效字节；这总是保留价值.
• 对于signed到signed类型，它对源类型进行符号扩展(即用与源的符号位相等的位打包新字节整数);这总是保留价值，正面或负面
• 对于无符号到有符号类型，它有效地添加了足够的零最高有效字节；这总是保留值，因为在向上转换的性质中，目标中总是有更多位，因此总是有额外的符号位"的空间
• 对于signed 到unsigned 类型，它进行符号扩展，然后强制转换；这不能始终保留该值，因为无法表示负值.

About integer numbers downcasts in C, e.g.

An int value 000F'E000 downcast to short or unsigned short will become E000.
short -> -8192,
unsigned short -> 57344,

So does it simply cut the bits?

And what about upcasts? E.g. int -10 is ffffff81, what is the rule to cast to long long?

---

@Update

About upcasting, according to the answers I did some tests and found that with 2's complement it has the following rules:

    signed:
        positive -> positive: add 0 as prefix bits,
        negative -> negative: add 1 as prefix bits,
    unsigned:
        add 0 as prefix bits,

code:

// integer numbers, downcast & upcast,
#include <stdio.h>

void downcastTest() {
    int i = 127<<13;
    printf("%X, %hX, %hi\n", i, i, i);
}

void upcastTest() {
    int i = 127;
    int j = -127;
    printf("%x, %llx\n", i, (long long)i);
    printf("%x, %llx\n", j, (long long)j);
}

int main(int argc, char * argv[]) {
    downcastTest();
    upcastTest();
    return 0;
}

解决方案

Downcasts

A cast to a smaller integer type discards the most significant (left-most as you'd write the full binary integer on paper) bits that are not present in the destination type.

Upcasts

An upcast to a larger integer is more complex:

• For unsigned to unsigned types, it adds sufficient zero most-significant bytes; this always preserves the value.
• For signed to signed types, it sign-extends the the source type (i.e. packs the new byte(s) with bits equal to the sign bit of the source integer); this always preserves the value, positive or negative
• For unsigned to signed types, it effectively adds sufficient zero-most significant bytes; this always preserves the value, as in the nature of an upcast, there are always more bits in the destination, so there is always space for an extra sign 'bit'
• For signed to unsigned types, it sign-extends, then casts; this cannot always preserve the value, as a negative value cannot be represented.

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