多维数组作为平面数组问题

发布于 2024-10-25 07:20:00 字数 315 浏览 5 评论 0原文

我正在查看某人编写的两行代码，第二行中有一个异常，但我不明白为什么。

char** array = (char**) new char [2] [6];

std_strlprintf(array[0],6,"[%d]", num);

std_strlprintf 是一个 Brew 函数，它将格式化输出写入字符串。（num是一个整数值，为0）

为什么这段代码会出现异常，将数组的第一个元素作为buff[0]访问有什么问题？

编辑：抱歉，我最初的帖子中有一个拼写错误。现在已更正。这是有异常的代码。

原文

I'm looking at two lines of code that somebody wrote and there is an exception in the 2nd one, however I don't understand why.

char** array = (char**) new char [2] [6];

std_strlprintf(array[0],6,"[%d]", num);

std_strlprintf is a Brew function that writes formatted output to a string. (num is an integral value which is 0)

Why is there an exception with this code, what's wrong with accessing the first elelment of the array as buff[0]?

EDIT: sorry there was a typo in my initial posting. Its corrected now.
THis is the code that has the exception.

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自控 2024-11-01 07:20:00

二维数组与指针数组不同。

您的第一条语句创建两个包含六个字符的数组，每个数组作为单个内存块。将该语句替换为：

char (*array)[6] = new char [2][6];

然后您就可以完成第二个语句了。不要忘记

delete [] array;

编辑 0：

呵呵，我应该知道:) 对于评论中的问题：

随后我应该如何传递数组
到一个以 char** 作为参数的函数
参数？

你不知道。不是这种形式。如果您正在为某些 C API 构建参数列表，例如 execve(2)，你必须一路进行两阶段初始化：

// prototype of the function to call
void my_fancy_func( int argc, char* const argv[] );

char** my_argv = new char*[my_argc];

for ( i = 0; i < my_argc; i++ ) {
    my_argv[i] = new char[arg_buffer_size];
    snprintf( my_argv[i], arg_buffer_size, "%d", i );
}

my_fancy_func( my_argc, my_argv );

Two-dimensional array is not the same as array of pointers.

Your first statement creates two arrays of six chars each as a single memory block. Replace that statement with:

char (*array)[6] = new char [2][6];

and you'll be all set with your second statement. Don't forget to

delete [] array;

Edit 0:

Huh, I should've known :) To your question in the comment:

How should I subsequently pass array
to a function that takes a char** as a
parameter?

You don't. Not in this form. If you are building a list of parameters to some C API like execve(2), you have to go all the way with two-stage initialization:

// prototype of the function to call
void my_fancy_func( int argc, char* const argv[] );

char** my_argv = new char*[my_argc];

for ( i = 0; i < my_argc; i++ ) {
    my_argv[i] = new char[arg_buffer_size];
    snprintf( my_argv[i], arg_buffer_size, "%d", i );
}

my_fancy_func( my_argc, my_argv );

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