使用"{}"有什么好处?用C语言进行铸造?

问题描述

我试图理解为什么在ProcessHacker代码中使用这种转换样式.

I am trying to understand why use this casting style in ProcessHacker Code.

 RtlSetHeapInformation(
            PhHeapHandle,
            HeapCompatibilityInformation,
            &(ULONG){ HEAP_COMPATIBILITY_LFH }, // HEAP_COMPATIBILITY_LFH = 2UL
            sizeof(ULONG)
            );

使用"{}"有什么好处?铸造?它可以在C和C ++中使用吗?

What are the advantages of using "{}" for casting? Does it work in C and C++?

&(ULONG){ HEAP_COMPATIBILITY_LFH }, // HEAP_COMPATIBILITY_LFH = 2UL

推荐答案

此

&(ULONG){ HEAP_COMPATIBILITY_LFH },

不是演员.这是一个复合文字.它创建一个具有自动存储持续时间的 ULONG 类型的对象，并使用值 HEAP_COMPATIBILITY_LFH 对其进行初始化.然后获取对象的地址.

is not a casting. It is a compound literal. It creates an object of the type ULONG with automatic storage duration and initializes it with the value HEAP_COMPATIBILITY_LFH. Then the address of the object is taken.

这是一个演示程序.

#include <stdio.h>

int main(void)
{
    unsigned long *p = &( unsigned long ){ 10ul };

    printf( "%lu\n", *p );

    return 0;
}

程序输出为

10

来自C标准(6.5.2.5复合文字)

From the C Standard (6.5.2.5 Compound literals)

您可以通过以下方式想象复合文字的定义.

You may imagine the definition of the compound literal the following way.

例如，您可以使用类似括号的列表来初始化标量变量

For example you may initialize a scalar variable using a braced list like

unsigned long x = { 10ul };

但是复合文字没有名称.这样的构造

But a compound literal has no name. So this construction

( unsigned long ){ 10ul }

实际上看起来像

unsigned long unnamed_literal = { 10ul };
     ^                             ^
     |                             |
( unsigned long )               { 10ul }

请注意，在C ++中，没有像复合文字这样的概念.

Pay attention to that in C++ there is no such a notion as the compound literal.

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