传递一个字符指针数组功能 [英] Passing a char pointer array to a function

问题描述

我已经写了下面的示例code来证明我的问题

I have written following sample code to demonstrate my problem

#include <iostream>
#include <string.h>

using namespace std;

void f (char*** a)
{
    *a = new (char*[2]);
    *a[0] = new char[10];
    *a[1] = new char[10];
    strcpy(*a[0], "abcd");
    strcpy(*a[1],"efgh");
}

int main ()
{
    char** b = NULL;
    f(&b);
    cout<<b[0]<<"\n";
    cout<<b[1]<<"\n";
    return 1;
}

在此code，我发现的 =一个新的（字符的[2]）;不分配的内存*一个[1]。

In this code i found that a = new (char[2]); not allocating memory of *a[1].

在gdb中我得到以下赛格故障。

In gdb i am getting following seg fault.

Program received signal SIGSEGV, Segmentation fault.
0x0000000000400926 in f (a=0x7fffffffdfe8) at array1.cpp:10
10          *a[1] = new char[10];
(gdb) p *a[1]
Cannot access memory at address 0x0

这是真的困惑我。有人可以解释我要去的地方错了。

This is really confusing me. Can someone explain where i am going wrong.

我知道我可以通过像无效˚F参数（字符**放大器;一），并通过调用函数f（b）和工作原理。但我想知道要发生的事情，如果我使用CHAR ***一个。它也应该工作。很抱歉，如果这是一个愚蠢的问题。结果
对上述问题的任何教程或引用将AP preciated。结果
提前致谢。

I know i can pass argument like void f (char**& a) and by calling function f(b) and this works . But i want to know want happening if i use char*** a. It should also work. Sorry if this is an stupid question.
Any tutorial or reference on above problem will be appreciated.
Thanks in advance.

推荐答案

这是因为运算符precedence ，数组索引运算符 [] 有更高的precedence比引用操作 * 。

It's because of the operator precedence, where the array-indexing operator [] have higher precedence than the dereference operator *.

所以前pression * A [0] 是真的，从编译器的角度来看，同为 *（一[0]），这是不是你想要的。

So the expression *a[0] is really, from the compilers point of view, the same as *(a[0]), which is not what you want.

您必须明确添加括号改变precedence：

You have to explicitly add parentheses to change the precedence:

(*a)[0] = ...

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