C语言 C - 'char **' 与 'char (*)[6]' 的间接级别不同
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原文地址: http://stackoverflow.com/questions/7524070/
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C - 'char **' differs in levels of indirection from 'char (*)[6]'
提问by loop
Can someone please explain to me what's wrong with the following, and more importantly why?
有人可以向我解释以下内容有什么问题，更重要的是为什么？
int main( int argc, char *argv[] )
{
char array[] = "array";
char **test;
test = &array;
*test[0] = 'Z';
printf( "%s\n", array );
return 0;
}
EDIT
编辑
My example above was based on a function like this that was crashing:
我上面的例子基于这样一个崩溃的函数：
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] = 'void apple( char **pp )
{
*pp = malloc( 123 );
(*pp)[0] = 'a'; // correct
(*pp)[1] = 'b'; // correct
(*pp)[2] = 'int main( int argc, char *argv[] )
{
char array[] = "array";
char (*test)[6];
test = &array;
(*test)[0] = 'Z';
printf( "%s\n", array );
return 0;
}
'; // correct
}
'; // incorrect and crash
}
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
正如 Vaughn Cato 向我指出的那样，虽然*pp[0] = 'a';没有崩溃，但它的状态很差。正确的形式是括号
int main( int argc, char *argv[] )
{
char array[] = "array";
char *test;
test = array; // same as test = &array[0];
test[0] = 'Z';
printf( "%s\n", array );
return 0;
}
Also as another poster MK pointed out the FAQ covers the difference between arrays and pointers: http://www.lysator.liu.se/c/c-faq/c-2.html
同样作为另一位海报 MK 指出常见问题解答涵盖了数组和指针之间的区别：http: //www.lysator.liu.se/c/c-faq/c-2.html
采纳答案by sashang
test = &array
test = &array
is wrong because test is of type char**and &arrayis a char(*)[6]and is a different type from char**
是错误的，因为 test 属于类型char**并且&array是 achar(*)[6]并且是与char**
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.
数组与char*尽管 C 会char*在某些上下文中隐式地在数组类型和 a 之间进行转换的类型不同，但这不是其中之一。基本上与char*数组类型相同的期望（例如：）char[6]是错误的，因此期望获取数组的地址将导致 achar**也是错误的。
回答by Vaughn Cato
This would be the way to do what you seem to be trying to do:
这将是做你似乎想要做的事情的方法：
##代码##test is a pointer to an array, and an array is different from a pointer, even though C makes it easy to use one like the other in my cases.
test 是一个指向数组的指针，而数组与指针不同，即使在我的情况下，C 可以很容易地使用一个数组。
If you wanted to avoid having to specify a specific sized array, you could use a different approach:
如果您想避免指定特定大小的数组，您可以使用不同的方法：
##代码##回答by cpx
char **test;is a pointer to a pointer, but if you're going to take the address of an entire array then it needs to be a pointer to entire array i.e char (*test)[6];
char **test;是指向指针的指针，但如果您要获取整个数组的地址，则它需要是指向整个数组的指针，即 char (*test)[6];
