如何使用指针将2D数组传递给Cfunction？ [英] How To Pass a 2D array to a Cfunction using Pointers?
问题描述
我在使用数组，函数和指针时遇到了问题。请告诉我使用Pointers将2 D数组传递给C函数。当我编译这样的程序时，我收到错误说功能需要一个如果我给了原型，那么它也会给出一个错误，比如无法将int转换成int *或类似的东西。如果需要我可以发布我遇到问题的特定程序。请帮助我out！
I have got a problem using Arrays,Functions and Pointers together.Please tell me hoe to pass a 2 D Array to a C function using Pointers.When I compile such a Program,I get an error saying "Function needs a prototype".If I give the protype ,then also it gives an error like,"Cannot convert int into int* "or something like that.If needed I can post the particular program with which i had the problem.Please help me out !
推荐答案
我在使用数组，函数和指针时遇到问题。请告诉我锄头通过使用Pointers将二维数组转换为C函数。当我编译这样的程序时，我收到一条错误，上面写着函数需要一个原型。如果我给出了原型，那么它也会给出一个错误，如无法转换为int进入int *或类似的东西。如果需要我可以发布我遇到问题的特定程序。请帮帮我！
I have got a problem using Arrays,Functions and Pointers together.Please tell me hoe to pass a 2 D Array to a C function using Pointers.When I compile such a Program,I get an error saying "Function needs a prototype".If I give the protype ,then also it gives an error like,"Cannot convert int into int* "or something like that.If needed I can post the particular program with which i had the problem.Please help me out !
您无需丢弃整个代码。
发布您声明的原型以及如何你打电话给它。
Raghuram
Hi,
You need not ost the entire code.
Post the prototypw you have declared and how are u calling it.
Raghuram
我宣称的原型是：
void display（int *， int，int）;
函数调用如下：
display（a，3,4）;
函数定义如下：
void display（int * q，int row，int col）
{
}
The prototype I had declared was :
void display(int *,int,int);
The function call is as follows :
display(a,3,4);
The function definition is as follows:
void display(int *q,int row,int col)
{
}
你不需要输入整个代码。
发布你宣布的原型，你怎么称呼它。
Raghuram
Hi,
You need not ost the entire code.
Post the prototypw you have declared and how are u calling it.
Raghuram
我声明的原型是：
void display（int * ，int，int）;
函数调用如下：
display（a，3,4）;
函数定义如下：
void display（int * q，int row，int col）
$
}
The prototype I had declared was :
void display(int *,int,int);
The function call is as follows :
display(a,3,4);
The function definition is as follows:
void display(int *q,int row,int col)
{
}
i认为显示功能是一个二维数组。
如果是是你的原型/定义你已宣布不会工作的情况。
你必须改变calloing部分只传递一个维度到这个功能
检查这个
i think a in display function is a 2-d array.
If that is the case then the prototype/definition u have declared wont work.
U have to change the calloing part to pass only one dimension to this function
check this
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