C: `const `关键字的行为

Java编译器有少量的流逻辑,允许您初始化 final 声明后的变量。这是合法的Java:

final int something;

if ( today == Friday )
    something = 7;
else
    something = 42;

final int something;

if ( today == Friday )
    something = 7;

if ( today != Friday )
    something = 42;

在ANSI C89中, const extern

const int something = ( today == Friday ) ? 7 : 42;

这

常量

#include<stdio.h>

int main ( void )
{
    printf ( "wibble\n" );

    {
        const int x = 10;

        printf ( "x = %d\n", x );
    }

    return 0;
}

C89 ,您通常可以通过引入一个裸块来增加额外的作用域,从而使定义更接近首次使用的点。之前:

int a, b, c;

a = 12;
// Do some stuff with a

b = 17;
// Do some stuff with a and b

c = 23;
// Do some stuff with a, b, and c

int a = 12;
// Do some stuff with a
{
    int b = 17
    // Do some stuff with a and b
    {
        int c = 23;
        // Do some stuff with a, b and c
    }
}

C99 当然,您可以定义块开头以外的变量:

int a = 12;
// Do some stuff with a

int b = 17
// Do some stuff with a and b

int c = 23;
// Do some stuff with a, b and c

const

error: assignment of read-only variable 'foo'

const int foo;
foo = 4;

const指针也是如此(注意: const int *

int * const foo;
foo = 4;

• 函数参数中的指针(或基于参数的局部变量指针),其中函数遵守不修改所指向数据的约定。const关键字有助于确保函数实现尊重不修改的要求(它需要特殊的努力转换来摆脱const),并允许这一要求通过多个函数调用传播。

void func()
{
    int y;

    // Do assertions
    assert(something);
    {
        int const x = 5;
        // Function body
     }
}

const int x = 2;

const int x;

x = 2;

如果我是你,我会尽力确保我理解你描述的编码规则的意图。我怀疑理智的编码规则会阻止初始化变量(即使是非常量变量)。

const int * p;

不 常量变量的声明。它是一个指向const int的非const指针变量的声明。

extern const int x;

但是在执行代码、断言检查等之后,您仍然无法初始化x。

如果你想抛弃const LHS

const int n = 0;

*((int*)&n) = 23;