通过多维数组作为C函数参数 [英] Passing multidimensional arrays as function arguments in C
问题描述
在C I可以通过一个多维数组给一个函数,当我不知道一个参数是什么阵列的尺寸将是?
In C can I pass a multidimensional array to a function as a single argument when I don't know what the dimensions of the array are going to be ?
另外我多维数组可能包含不是字符串类型。
In addition my multidimensional array may contain types other than strings.
推荐答案
您可以与任何数据类型做到这一点。简单地使它成为一个指针到指针:
You can do this with any data type. Simply make it a pointer-to-pointer:
typedef struct {
int myint;
char* mystring;
} data;
data** array;
但是不要忘了,你还是要在MALLOC变量,它得到一个有点复杂:
But don't forget you still have to malloc the variable, and it does get a bit complex:
//initialize
int x,y,w,h;
w = 10; //width of array
h = 20; //height of array
//malloc the 'y' dimension
array = malloc(sizeof(data*) * h);
//iterate over 'y' dimension
for(y=0;y<h;y++){
//malloc the 'x' dimension
array[y] = malloc(sizeof(data) * w);
//iterate over the 'x' dimension
for(x=0;x<w;x++){
//malloc the string in the data structure
array[y][x].mystring = malloc(50); //50 chars
//initialize
array[y][x].myint = 6;
strcpy(array[y][x].mystring, "w00t");
}
}
在code解除分配结构类似于 - 不要忘了对你的一切调用malloced免费()! (另外,健壮的应用程序,你应该检查的malloc()的返回 。)
现在让我们假设你想将这个传递给函数。您仍然可以使用双指针,因为你可能想指针做数据结构,而不是操纵数据结构的指针:
Now let's say you want to pass this to a function. You can still use the double pointer, because you probably want to do manipulations on the data structure, not the pointer to pointers of data structures:
int whatsMyInt(data** arrayPtr, int x, int y){
return arrayPtr[y][x].myint;
}
调用该函数:
printf("My int is %d.\n", whatsMyInt(array, 2, 4));
输出:
My int is 6.
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