Ç - '的char **'中的间接水平不同于'字符（*）[6]' [英] C - 'char **' differs in levels of indirection from 'char (*)[6]'

问题描述

可有人请向我解释什么地方错了以下内容，更重要的是，为什么？

Can someone please explain to me what's wrong with the following, and more importantly why?

int main( int argc, char *argv[] )
{
    char array[] = "array";
    char **test;
    test = &array;

    *test[0] = 'Z';

    printf( "%s\n", array );

    return 0;
}

修改

我上面的例子是基于这样被撞毁的函数：

My example above was based on a function like this that was crashing:

void apple( char **pp )
{
    *pp = malloc( 123 );
    *pp[0] = 'a'; // technically this is correct but in bad form
    *pp[1] = 'b'; // incorrect but no crash
    *pp[2] = '\0'; // incorrect and crash
}

由于虽然向我指出由沃恩卡托*页[0] ='A'; 不会崩溃它是不好的形式。正确的方式是在括号

As pointed out to me by Vaughn Cato although *pp[0] = 'a'; does not crash it is in bad form. The correct form is the parenthesis

void apple( char **pp )
{
    *pp = malloc( 123 );
    (*pp)[0] = 'a'; // correct
    (*pp)[1] = 'b'; // correct
    (*pp)[2] = '\0'; // correct
}

另外，作为另一个海报MK指出，常见问题包括数组和指针的区别：
http://www.lysator.liu.se/c/c -faq / C-2.HTML

推荐答案

测试=安培;阵列

是错误的，因为测试的类型为的char ** 和＆放大器;阵列是字符（*）[6] 距离字符不同类型的**

is wrong because test is of type char** and &array is a char(*)[6] and is a different type from char**

数组是不是同一类型为的char * 尽管C将隐式数组类型和的char *之间的转换在一些上下文中，但是这不是其中之一。基本上期望的char * 相同数组的类型（例如：的char [6] ）是错了，因此，服用一个数组的地址将导致字符的期​​望** 也是错误的。

An array isn't the same type as char* although C will implicitly convert between an array type and a char* in some contexts, but this isn't one of them. Basically the expectation that char* is the same as the type of an array (e.g: char[6]) is wrong and therefore the expectation that taking the address of an array will result in a char** is also wrong.

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