如何将 3D 字符数组传递给函数 [英] How to pass 3D array of chars to a function
问题描述
我有一个 chars table[][][] 的 3D 数组，我想将它传递给一个 void 函数，以便它可以对其进行更改.我怎样才能做到这一点?
I have a 3D array of chars table[][][] and I want to pass it to a void function so it can make changes to it. How can I do this?
void make(char minor[][][]);
.....
char greater[20][30][50];
make(greater);
我想这行不通.
与此相关的另一个问题:假设我想创建一个复制函数来将字符串复制到数组中 - 我应该如何在函数中调用 strcpy?
Another question connected with this: Say I want to make a copy function to copy a string into the array - how should I call the strcpy in the function?
void copy(char (*minor)[][])
{ char m[50] = "asdasdasd";
strcpy(minor[][],m);
}
推荐答案
如果你想传递一个数组，就像那个数组一样，而不是你用 动态分配的数组malloc(这将是几级指针，而不是真正的数组)，以下任何函数原型都可以工作:
If you wanted to pass an array, just like that one, as opposed to one you've dynamically allocated with malloc (which would be a few levels of pointers, not real arrays), any of the below function prototypes will work:
void make(char minor[20][30][50])void make(char minor[][30][50])void make(char (*minor)[30][50])
你不能有类似 void make(char ***minor) 的原因是因为 minor 只会衰减为指向数组数组的指针，这样说不会衰减不止一次.
The reason being you can't have something like void make(char ***minor) is because minor will only decay into a pointer to an array of arrays, it won't decay more than once so to say.
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