如何创建一个字符串C中的数组？ [英] How do I create an array of strings in C?

问题描述

我想如果我用这个code创建的字符串在C数组：

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";

海湾合作委员会给了我警告：从分配不兼容的指针类型。什么是做到这一点的正确方法？

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

编辑：我很好奇，为什么这应该给出一个编译器警告，因为如果我这样做的printf（A [1]），则它正确打印哼

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";

当你像这样做，你会分配两个指针数组为const char 。然后，这些指针将被设置为静态字符串的地址嗒嗒和哼。

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".

如果你想能够改变的实际字符串的内容时，你必须做一些像

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");

这将拨出14 字符 s分别连续两个阵列，在此之后，静态字符串的内容将被复制到他们。

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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