正确的方法来初始化字符串在C NULL结尾数组 [英] Correct way to initialize a NULL-terminated array of strings in C

问题描述

这是code正确吗？

 的char * argv的[] = {富，酒吧，NULL};
 

解决方案

这是语法正确的，它也创造字符串的NULL结尾数组。

ARGV传递给主为的char * [] （或等价，字符** ），但它的更正确对待字符串作为为const char * 而非的char * 。因此，与这个特殊的例子中，你会想要为const char * argv的[] = {富，酒吧，NULL};

也许你是不是真的有富初始化，但实际上，你将要通过的argv修改修改字符串。在这种情况下，的char * [] 是正确的。这是什么样的事情查尔斯可能是说，code是正确的取决于你用​​它做什么意思。

Is this code correct?

char *argv[] = { "foo", "bar", NULL };

解决方案

It's syntactically correct, and it does create a NULL-terminated array of strings.

argv is passed to main as char*[] (or equivalently, char**), but it's "more correct" to treat string literals as a const char* rather than a char*. So with this particular example you'd want const char *argv[] = {"foo", "bar", NULL };

Maybe you aren't really going to initialise it with "foo", but actually with a modifiable string that you will want to modify via argv. In that case char*[] is right. This is the kind of thing Charles probably means by saying that whether code is "correct" depends on what you do with it.

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