C中的2D字符数组初始化 [英] 2D character array initialization in C

问题描述

我正在尝试构建需要传递给期望char **

I am trying to build a list of strings that I need to pass to a function expecting char **

如何构建此数组?我想传递两个选项，每个选项的字符数少于100个.

How do I build this array? I want to pass in two options, each with less than 100 characters.

char **options[2][100];

options[0][0] = 'test1';
options[1][0] = 'test2';

这不能编译.我到底在做什么错?如何在C中创建2D字符数组?

This does not compile. What am I doing wrong exactly? How do I create a 2D character array in C?

推荐答案

C字符串用双引号引起来:

C strings are enclosed in double quotes:

const char *options[2][100];

options[0][0] = "test1";
options[1][0] = "test2";

重新阅读您的问题和评论，尽管我猜您真的 想要做的是这样的事情:

Re-reading your question and comments though I'm guessing that what you really want to do is this:

const char *options[2] = { "test1", "test2" };

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