如何在 C 中创建字符串数组? [英] How do I create an array of strings in C?

问题描述

我正在尝试在 C 中创建一个字符串数组.如果我使用此代码:

I am trying to create an array of strings in C. If I use this code:

char (*a[2])[14];
a[0]="blah";
a[1]="hmm";

gcc 给了我警告:从不兼容的指针类型赋值".这样做的正确方法是什么?

gcc gives me "warning: assignment from incompatible pointer type". What is the correct way to do this?

我很好奇为什么这会给出编译器警告，因为如果我执行 printf(a[1]);，它会正确打印hmm".

edit: I am curious why this should give a compiler warning since if I do printf(a[1]);, it correctly prints "hmm".

推荐答案

如果你不想改变字符串，那么你可以简单地做

If you don't want to change the strings, then you could simply do

const char *a[2];
a[0] = "blah";
a[1] = "hmm";

当您这样做时，您将分配一个包含两个指向 const char 的指针的数组.然后将这些指针设置为静态字符串 "blah" 和 "hmm" 的地址.

When you do it like this you will allocate an array of two pointers to const char. These pointers will then be set to the addresses of the static strings "blah" and "hmm".

如果您确实希望能够更改实际的字符串内容，您必须执行类似的操作

If you do want to be able to change the actual string content, the you have to do something like

char a[2][14];
strcpy(a[0], "blah");
strcpy(a[1], "hmm");

这将分配两个连续的 14 个 char 数组，之后静态字符串的内容将被复制到其中.

This will allocate two consecutive arrays of 14 chars each, after which the content of the static strings will be copied into them.

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