C语言 C 中的二维字符数组初始化
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2D character array initialization in C
提问by Derek
I am trying to build a list of strings that I need to pass to a function expecting char **
我正在尝试构建一个字符串列表，我需要将其传递给期望的函数 char **
How do I build this array? I want to pass in two options, each with less than 100 characters.
我如何构建这个数组？我想传入两个选项，每个选项都少于 100 个字符。
char **options[2][100];
options[0][0] = 'test1';
options[1][0] = 'test2';
This does not compile. What am I doing wrong exactly? How do I create a 2D character array in C?
这不编译。我到底做错了什么？如何在 C 中创建二维字符数组？
回答by Paul R
C strings are enclosed in double quotes:
C 字符串用双引号括起来：
const char *options[2][100];
options[0][0] = "test1";
options[1][0] = "test2";
Re-reading your question and comments though I'm guessing that what you reallywant to do is this:
重新阅读您的问题和评论，尽管我猜您真正想做的是：
const char *options[2] = { "test1", "test2" };
回答by Eric Leschinski
How to create an array size 5 containing pointers to characters:
如何创建一个大小为 5 的包含指向字符的指针的数组：
char *array_of_pointers[ 5 ]; //array size 5 containing pointers to char
char m = 'm'; //character value holding the value 'm'
array_of_pointers[0] = &m; //assign m ptr into the array position 0.
printf("%c", *array_of_pointers[0]); //get the value of the pointer to m
How to create a pointer to an array of characters:
如何创建指向字符数组的指针：
char (*pointer_to_array)[ 5 ]; //A pointer to an array containing 5 chars
char m = 'm'; //character value holding the value 'm'
*pointer_to_array[0] = m; //dereference array and put m in position 0
printf("%c", (*pointer_to_array)[0]); //dereference array and get position 0
How to create an 2D array containing pointers to characters:
如何创建包含指向字符的指针的二维数组：
char *array_of_pointers[5][2];
//An array size 5 containing arrays size 2 containing pointers to char
char m = 'm';
//character value holding the value 'm'
array_of_pointers[4][1] = &m;
//Get position 4 of array, then get position 1, then put m ptr in there.
printf("%c", *array_of_pointers[4][1]);
//Get position 4 of array, then get position 1 and dereference it.
How to create a pointer to an 2D array of characters:
如何创建指向二维字符数组的指针：
char (*pointer_to_array)[5][2];
//A pointer to an array size 5 each containing arrays size 2 which hold chars
char m = 'm';
//character value holding the value 'm'
(*pointer_to_array)[4][1] = m;
//dereference array, Get position 4, get position 1, put m there.
printf("%c", (*pointer_to_array)[4][1]);
//dereference array, Get position 4, get position 1
To help you out with understanding how humans should read complex C/C++ declarations read this: http://www.programmerinterview.com/index.php/c-cplusplus/c-declarations/
为了帮助您理解人类应该如何阅读复杂的 C/C++ 声明，请阅读以下内容：http: //www.programmerinterview.com/index.php/c-cplusplus/c-declarations/
回答by Fred Foo
char **options[2][100];
declares a size-2 array of size-100 arrays of pointers to pointers to char. You'll want to remove one *. You'll also want to put your string literals in double quotes.
声明一个大小为 100 的指针数组的大小为 2 的数组，其中包含指向 的指针char。你会想要删除一个*。您还需要将字符串文字放在双引号中。
回答by Heath Hunnicutt
I think what you originally meant to do was to make an array only of characters, not of pointers:
我认为您最初的意图是制作一个仅包含字符而不是指针的数组：
char options[2][100];
options[0][0]='t';
options[0][1]='e';
options[0][2]='s';
options[0][3]='t';
options[0][4]='1';
options[0][5]='##代码##'; /* NUL termination of C string */
/* A standard C library function which copies strings. */
strcpy(options[1], "test2");
The code above shows two distinct methods of setting the character values in memory you have set aside to contain characters.
上面的代码显示了在内存中设置字符值的两种不同方法，这些方法是您预留的用于包含字符的。
